hypothesis chi square test examples

Hypothesis Testing - Chi Squared Test

Lisa Sullivan, PhD

Professor of Biostatistics

Boston University School of Public Health

Introduction

This module will continue the discussion of hypothesis testing, where a specific statement or hypothesis is generated about a population parameter, and sample statistics are used to assess the likelihood that the hypothesis is true. The hypothesis is based on available information and the investigator's belief about the population parameters. The specific tests considered here are called chi-square tests and are appropriate when the outcome is discrete (dichotomous, ordinal or categorical). For example, in some clinical trials the outcome is a classification such as hypertensive, pre-hypertensive or normotensive. We could use the same classification in an observational study such as the Framingham Heart Study to compare men and women in terms of their blood pressure status - again using the classification of hypertensive, pre-hypertensive or normotensive status.  

The technique to analyze a discrete outcome uses what is called a chi-square test. Specifically, the test statistic follows a chi-square probability distribution. We will consider chi-square tests here with one, two and more than two independent comparison groups.

Learning Objectives

After completing this module, the student will be able to:

• Perform chi-square tests by hand
• Appropriately interpret results of chi-square tests
• Identify the appropriate hypothesis testing procedure based on type of outcome variable and number of samples

Tests with One Sample, Discrete Outcome

Here we consider hypothesis testing with a discrete outcome variable in a single population. Discrete variables are variables that take on more than two distinct responses or categories and the responses can be ordered or unordered (i.e., the outcome can be ordinal or categorical). The procedure we describe here can be used for dichotomous (exactly 2 response options), ordinal or categorical discrete outcomes and the objective is to compare the distribution of responses, or the proportions of participants in each response category, to a known distribution. The known distribution is derived from another study or report and it is again important in setting up the hypotheses that the comparator distribution specified in the null hypothesis is a fair comparison. The comparator is sometimes called an external or a historical control.   

In one sample tests for a discrete outcome, we set up our hypotheses against an appropriate comparator. We select a sample and compute descriptive statistics on the sample data. Specifically, we compute the sample size (n) and the proportions of participants in each response

Test Statistic for Testing H 0 : p 1 = p 10 , p 2 = p 20 , ..., p k = p k0

We find the critical value in a table of probabilities for the chi-square distribution with degrees of freedom (df) = k-1. In the test statistic, O = observed frequency and E=expected frequency in each of the response categories. The observed frequencies are those observed in the sample and the expected frequencies are computed as described below. χ 2 (chi-square) is another probability distribution and ranges from 0 to ∞. The test above statistic formula above is appropriate for large samples, defined as expected frequencies of at least 5 in each of the response categories.  

When we conduct a χ 2 test, we compare the observed frequencies in each response category to the frequencies we would expect if the null hypothesis were true. These expected frequencies are determined by allocating the sample to the response categories according to the distribution specified in H 0 . This is done by multiplying the observed sample size (n) by the proportions specified in the null hypothesis (p 10 , p 20 , ..., p k0 ). To ensure that the sample size is appropriate for the use of the test statistic above, we need to ensure that the following: min(np 10 , n p 20 , ..., n p k0 ) > 5.  

The test of hypothesis with a discrete outcome measured in a single sample, where the goal is to assess whether the distribution of responses follows a known distribution, is called the χ 2 goodness-of-fit test. As the name indicates, the idea is to assess whether the pattern or distribution of responses in the sample "fits" a specified population (external or historical) distribution. In the next example we illustrate the test. As we work through the example, we provide additional details related to the use of this new test statistic.  

A University conducted a survey of its recent graduates to collect demographic and health information for future planning purposes as well as to assess students' satisfaction with their undergraduate experiences. The survey revealed that a substantial proportion of students were not engaging in regular exercise, many felt their nutrition was poor and a substantial number were smoking. In response to a question on regular exercise, 60% of all graduates reported getting no regular exercise, 25% reported exercising sporadically and 15% reported exercising regularly as undergraduates. The next year the University launched a health promotion campaign on campus in an attempt to increase health behaviors among undergraduates. The program included modules on exercise, nutrition and smoking cessation. To evaluate the impact of the program, the University again surveyed graduates and asked the same questions. The survey was completed by 470 graduates and the following data were collected on the exercise question:

Based on the data, is there evidence of a shift in the distribution of responses to the exercise question following the implementation of the health promotion campaign on campus? Run the test at a 5% level of significance.

In this example, we have one sample and a discrete (ordinal) outcome variable (with three response options). We specifically want to compare the distribution of responses in the sample to the distribution reported the previous year (i.e., 60%, 25%, 15% reporting no, sporadic and regular exercise, respectively). We now run the test using the five-step approach.  

• Step 1. Set up hypotheses and determine level of significance.

The null hypothesis again represents the "no change" or "no difference" situation. If the health promotion campaign has no impact then we expect the distribution of responses to the exercise question to be the same as that measured prior to the implementation of the program.

H 0 : p 1 =0.60, p 2 =0.25, p 3 =0.15,  or equivalently H 0 : Distribution of responses is 0.60, 0.25, 0.15  

H 1 :   H 0 is false.          α =0.05

Notice that the research hypothesis is written in words rather than in symbols. The research hypothesis as stated captures any difference in the distribution of responses from that specified in the null hypothesis. We do not specify a specific alternative distribution, instead we are testing whether the sample data "fit" the distribution in H 0 or not. With the χ 2 goodness-of-fit test there is no upper or lower tailed version of the test.

• Step 2. Select the appropriate test statistic.  

The test statistic is:

We must first assess whether the sample size is adequate. Specifically, we need to check min(np 0 , np 1, ..., n p k ) > 5. The sample size here is n=470 and the proportions specified in the null hypothesis are 0.60, 0.25 and 0.15. Thus, min( 470(0.65), 470(0.25), 470(0.15))=min(282, 117.5, 70.5)=70.5. The sample size is more than adequate so the formula can be used.

• Step 3. Set up decision rule.  

The decision rule for the χ 2 test depends on the level of significance and the degrees of freedom, defined as degrees of freedom (df) = k-1 (where k is the number of response categories). If the null hypothesis is true, the observed and expected frequencies will be close in value and the χ 2 statistic will be close to zero. If the null hypothesis is false, then the χ 2 statistic will be large. Critical values can be found in a table of probabilities for the χ 2 distribution. Here we have df=k-1=3-1=2 and a 5% level of significance. The appropriate critical value is 5.99, and the decision rule is as follows: Reject H 0 if χ 2 > 5.99.

• Step 4. Compute the test statistic.  

We now compute the expected frequencies using the sample size and the proportions specified in the null hypothesis. We then substitute the sample data (observed frequencies) and the expected frequencies into the formula for the test statistic identified in Step 2. The computations can be organized as follows.

Notice that the expected frequencies are taken to one decimal place and that the sum of the observed frequencies is equal to the sum of the expected frequencies. The test statistic is computed as follows:

• Step 5. Conclusion.  

We reject H 0 because 8.46 > 5.99. We have statistically significant evidence at α=0.05 to show that H 0 is false, or that the distribution of responses is not 0.60, 0.25, 0.15.  The p-value is p < 0.005.  

In the χ 2 goodness-of-fit test, we conclude that either the distribution specified in H 0 is false (when we reject H 0 ) or that we do not have sufficient evidence to show that the distribution specified in H 0 is false (when we fail to reject H 0 ). Here, we reject H 0 and concluded that the distribution of responses to the exercise question following the implementation of the health promotion campaign was not the same as the distribution prior. The test itself does not provide details of how the distribution has shifted. A comparison of the observed and expected frequencies will provide some insight into the shift (when the null hypothesis is rejected). Does it appear that the health promotion campaign was effective?  

Consider the following: 

If the null hypothesis were true (i.e., no change from the prior year) we would have expected more students to fall in the "No Regular Exercise" category and fewer in the "Regular Exercise" categories. In the sample, 255/470 = 54% reported no regular exercise and 90/470=19% reported regular exercise. Thus, there is a shift toward more regular exercise following the implementation of the health promotion campaign. There is evidence of a statistical difference, is this a meaningful difference? Is there room for improvement?

The National Center for Health Statistics (NCHS) provided data on the distribution of weight (in categories) among Americans in 2002. The distribution was based on specific values of body mass index (BMI) computed as weight in kilograms over height in meters squared. Underweight was defined as BMI< 18.5, Normal weight as BMI between 18.5 and 24.9, overweight as BMI between 25 and 29.9 and obese as BMI of 30 or greater. Americans in 2002 were distributed as follows: 2% Underweight, 39% Normal Weight, 36% Overweight, and 23% Obese. Suppose we want to assess whether the distribution of BMI is different in the Framingham Offspring sample. Using data from the n=3,326 participants who attended the seventh examination of the Offspring in the Framingham Heart Study we created the BMI categories as defined and observed the following:

• Step 1.  Set up hypotheses and determine level of significance.

H 0 : p 1 =0.02, p 2 =0.39, p 3 =0.36, p 4 =0.23     or equivalently

H 0 : Distribution of responses is 0.02, 0.39, 0.36, 0.23

H 1 :   H 0 is false.        α=0.05

The formula for the test statistic is:

We must assess whether the sample size is adequate. Specifically, we need to check min(np 0 , np 1, ..., n p k ) > 5. The sample size here is n=3,326 and the proportions specified in the null hypothesis are 0.02, 0.39, 0.36 and 0.23. Thus, min( 3326(0.02), 3326(0.39), 3326(0.36), 3326(0.23))=min(66.5, 1297.1, 1197.4, 765.0)=66.5. The sample size is more than adequate, so the formula can be used.

Here we have df=k-1=4-1=3 and a 5% level of significance. The appropriate critical value is 7.81 and the decision rule is as follows: Reject H 0 if χ 2 > 7.81.

We now compute the expected frequencies using the sample size and the proportions specified in the null hypothesis. We then substitute the sample data (observed frequencies) into the formula for the test statistic identified in Step 2. We organize the computations in the following table.

The test statistic is computed as follows:

We reject H 0 because 233.53 > 7.81. We have statistically significant evidence at α=0.05 to show that H 0 is false or that the distribution of BMI in Framingham is different from the national data reported in 2002, p < 0.005.  

Again, the χ 2   goodness-of-fit test allows us to assess whether the distribution of responses "fits" a specified distribution. Here we show that the distribution of BMI in the Framingham Offspring Study is different from the national distribution. To understand the nature of the difference we can compare observed and expected frequencies or observed and expected proportions (or percentages). The frequencies are large because of the large sample size, the observed percentages of patients in the Framingham sample are as follows: 0.6% underweight, 28% normal weight, 41% overweight and 30% obese. In the Framingham Offspring sample there are higher percentages of overweight and obese persons (41% and 30% in Framingham as compared to 36% and 23% in the national data), and lower proportions of underweight and normal weight persons (0.6% and 28% in Framingham as compared to 2% and 39% in the national data). Are these meaningful differences?

In the module on hypothesis testing for means and proportions, we discussed hypothesis testing applications with a dichotomous outcome variable in a single population. We presented a test using a test statistic Z to test whether an observed (sample) proportion differed significantly from a historical or external comparator. The chi-square goodness-of-fit test can also be used with a dichotomous outcome and the results are mathematically equivalent.  

In the prior module, we considered the following example. Here we show the equivalence to the chi-square goodness-of-fit test.

The NCHS report indicated that in 2002, 75% of children aged 2 to 17 saw a dentist in the past year. An investigator wants to assess whether use of dental services is similar in children living in the city of Boston. A sample of 125 children aged 2 to 17 living in Boston are surveyed and 64 reported seeing a dentist over the past 12 months. Is there a significant difference in use of dental services between children living in Boston and the national data?

We presented the following approach to the test using a Z statistic. 

• Step 1. Set up hypotheses and determine level of significance

H 0 : p = 0.75

H 1 : p ≠ 0.75                               α=0.05

We must first check that the sample size is adequate. Specifically, we need to check min(np 0 , n(1-p 0 )) = min( 125(0.75), 125(1-0.75))=min(94, 31)=31. The sample size is more than adequate so the following formula can be used

This is a two-tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.960 or if Z > 1.960.

We now substitute the sample data into the formula for the test statistic identified in Step 2. The sample proportion is:

We reject H 0 because -6.15 < -1.960. We have statistically significant evidence at a =0.05 to show that there is a statistically significant difference in the use of dental service by children living in Boston as compared to the national data. (p < 0.0001).  

We now conduct the same test using the chi-square goodness-of-fit test. First, we summarize our sample data as follows:

H 0 : p 1 =0.75, p 2 =0.25     or equivalently H 0 : Distribution of responses is 0.75, 0.25 

We must assess whether the sample size is adequate. Specifically, we need to check min(np 0 , np 1, ...,np k >) > 5. The sample size here is n=125 and the proportions specified in the null hypothesis are 0.75, 0.25. Thus, min( 125(0.75), 125(0.25))=min(93.75, 31.25)=31.25. The sample size is more than adequate so the formula can be used.

Here we have df=k-1=2-1=1 and a 5% level of significance. The appropriate critical value is 3.84, and the decision rule is as follows: Reject H 0 if χ 2 > 3.84. (Note that 1.96 2 = 3.84, where 1.96 was the critical value used in the Z test for proportions shown above.)

(Note that (-6.15) 2 = 37.8, where -6.15 was the value of the Z statistic in the test for proportions shown above.)

We reject H 0 because 37.8 > 3.84. We have statistically significant evidence at α=0.05 to show that there is a statistically significant difference in the use of dental service by children living in Boston as compared to the national data.  (p < 0.0001). This is the same conclusion we reached when we conducted the test using the Z test above. With a dichotomous outcome, Z 2 = χ 2 !   In statistics, there are often several approaches that can be used to test hypotheses. 

Tests for Two or More Independent Samples, Discrete Outcome

Here we extend that application of the chi-square test to the case with two or more independent comparison groups. Specifically, the outcome of interest is discrete with two or more responses and the responses can be ordered or unordered (i.e., the outcome can be dichotomous, ordinal or categorical). We now consider the situation where there are two or more independent comparison groups and the goal of the analysis is to compare the distribution of responses to the discrete outcome variable among several independent comparison groups.  

The test is called the χ 2 test of independence and the null hypothesis is that there is no difference in the distribution of responses to the outcome across comparison groups. This is often stated as follows: The outcome variable and the grouping variable (e.g., the comparison treatments or comparison groups) are independent (hence the name of the test). Independence here implies homogeneity in the distribution of the outcome among comparison groups.    

The null hypothesis in the χ 2 test of independence is often stated in words as: H 0 : The distribution of the outcome is independent of the groups. The alternative or research hypothesis is that there is a difference in the distribution of responses to the outcome variable among the comparison groups (i.e., that the distribution of responses "depends" on the group). In order to test the hypothesis, we measure the discrete outcome variable in each participant in each comparison group. The data of interest are the observed frequencies (or number of participants in each response category in each group). The formula for the test statistic for the χ 2 test of independence is given below.

Test Statistic for Testing H 0 : Distribution of outcome is independent of groups

and we find the critical value in a table of probabilities for the chi-square distribution with df=(r-1)*(c-1).

Here O = observed frequency, E=expected frequency in each of the response categories in each group, r = the number of rows in the two-way table and c = the number of columns in the two-way table.   r and c correspond to the number of comparison groups and the number of response options in the outcome (see below for more details). The observed frequencies are the sample data and the expected frequencies are computed as described below. The test statistic is appropriate for large samples, defined as expected frequencies of at least 5 in each of the response categories in each group.  

The data for the χ 2 test of independence are organized in a two-way table. The outcome and grouping variable are shown in the rows and columns of the table. The sample table below illustrates the data layout. The table entries (blank below) are the numbers of participants in each group responding to each response category of the outcome variable.

Table - Possible outcomes are are listed in the columns; The groups being compared are listed in rows.

In the table above, the grouping variable is shown in the rows of the table; r denotes the number of independent groups. The outcome variable is shown in the columns of the table; c denotes the number of response options in the outcome variable. Each combination of a row (group) and column (response) is called a cell of the table. The table has r*c cells and is sometimes called an r x c ("r by c") table. For example, if there are 4 groups and 5 categories in the outcome variable, the data are organized in a 4 X 5 table. The row and column totals are shown along the right-hand margin and the bottom of the table, respectively. The total sample size, N, can be computed by summing the row totals or the column totals. Similar to ANOVA, N does not refer to a population size here but rather to the total sample size in the analysis. The sample data can be organized into a table like the above. The numbers of participants within each group who select each response option are shown in the cells of the table and these are the observed frequencies used in the test statistic.

The test statistic for the χ 2 test of independence involves comparing observed (sample data) and expected frequencies in each cell of the table. The expected frequencies are computed assuming that the null hypothesis is true. The null hypothesis states that the two variables (the grouping variable and the outcome) are independent. The definition of independence is as follows:

 Two events, A and B, are independent if P(A|B) = P(A), or equivalently, if P(A and B) = P(A) P(B).

The second statement indicates that if two events, A and B, are independent then the probability of their intersection can be computed by multiplying the probability of each individual event. To conduct the χ 2 test of independence, we need to compute expected frequencies in each cell of the table. Expected frequencies are computed by assuming that the grouping variable and outcome are independent (i.e., under the null hypothesis). Thus, if the null hypothesis is true, using the definition of independence:

P(Group 1 and Response Option 1) = P(Group 1) P(Response Option 1).

 The above states that the probability that an individual is in Group 1 and their outcome is Response Option 1 is computed by multiplying the probability that person is in Group 1 by the probability that a person is in Response Option 1. To conduct the χ 2 test of independence, we need expected frequencies and not expected probabilities . To convert the above probability to a frequency, we multiply by N. Consider the following small example.

The data shown above are measured in a sample of size N=150. The frequencies in the cells of the table are the observed frequencies. If Group and Response are independent, then we can compute the probability that a person in the sample is in Group 1 and Response category 1 using:

P(Group 1 and Response 1) = P(Group 1) P(Response 1),

P(Group 1 and Response 1) = (25/150) (62/150) = 0.069.

Thus if Group and Response are independent we would expect 6.9% of the sample to be in the top left cell of the table (Group 1 and Response 1). The expected frequency is 150(0.069) = 10.4.   We could do the same for Group 2 and Response 1:

P(Group 2 and Response 1) = P(Group 2) P(Response 1),

P(Group 2 and Response 1) = (50/150) (62/150) = 0.138.

The expected frequency in Group 2 and Response 1 is 150(0.138) = 20.7.

Thus, the formula for determining the expected cell frequencies in the χ 2 test of independence is as follows:

Expected Cell Frequency = (Row Total * Column Total)/N.

The above computes the expected frequency in one step rather than computing the expected probability first and then converting to a frequency.  

In a prior example we evaluated data from a survey of university graduates which assessed, among other things, how frequently they exercised. The survey was completed by 470 graduates. In the prior example we used the χ 2 goodness-of-fit test to assess whether there was a shift in the distribution of responses to the exercise question following the implementation of a health promotion campaign on campus. We specifically considered one sample (all students) and compared the observed distribution to the distribution of responses the prior year (a historical control). Suppose we now wish to assess whether there is a relationship between exercise on campus and students' living arrangements. As part of the same survey, graduates were asked where they lived their senior year. The response options were dormitory, on-campus apartment, off-campus apartment, and at home (i.e., commuted to and from the university). The data are shown below.

Based on the data, is there a relationship between exercise and student's living arrangement? Do you think where a person lives affect their exercise status? Here we have four independent comparison groups (living arrangement) and a discrete (ordinal) outcome variable with three response options. We specifically want to test whether living arrangement and exercise are independent. We will run the test using the five-step approach.  

H 0 : Living arrangement and exercise are independent

H 1 : H 0 is false.                α=0.05

The null and research hypotheses are written in words rather than in symbols. The research hypothesis is that the grouping variable (living arrangement) and the outcome variable (exercise) are dependent or related.   

• Step 2.  Select the appropriate test statistic.  

The condition for appropriate use of the above test statistic is that each expected frequency is at least 5. In Step 4 we will compute the expected frequencies and we will ensure that the condition is met.

The decision rule depends on the level of significance and the degrees of freedom, defined as df = (r-1)(c-1), where r and c are the numbers of rows and columns in the two-way data table.   The row variable is the living arrangement and there are 4 arrangements considered, thus r=4. The column variable is exercise and 3 responses are considered, thus c=3. For this test, df=(4-1)(3-1)=3(2)=6. Again, with χ 2 tests there are no upper, lower or two-tailed tests. If the null hypothesis is true, the observed and expected frequencies will be close in value and the χ 2 statistic will be close to zero. If the null hypothesis is false, then the χ 2 statistic will be large. The rejection region for the χ 2 test of independence is always in the upper (right-hand) tail of the distribution. For df=6 and a 5% level of significance, the appropriate critical value is 12.59 and the decision rule is as follows: Reject H 0 if c 2 > 12.59.

We now compute the expected frequencies using the formula,

Expected Frequency = (Row Total * Column Total)/N.

The computations can be organized in a two-way table. The top number in each cell of the table is the observed frequency and the bottom number is the expected frequency.   The expected frequencies are shown in parentheses.

Notice that the expected frequencies are taken to one decimal place and that the sums of the observed frequencies are equal to the sums of the expected frequencies in each row and column of the table.  

Recall in Step 2 a condition for the appropriate use of the test statistic was that each expected frequency is at least 5. This is true for this sample (the smallest expected frequency is 9.6) and therefore it is appropriate to use the test statistic.

We reject H 0 because 60.5 > 12.59. We have statistically significant evidence at a =0.05 to show that H 0 is false or that living arrangement and exercise are not independent (i.e., they are dependent or related), p < 0.005.  

Again, the χ 2 test of independence is used to test whether the distribution of the outcome variable is similar across the comparison groups. Here we rejected H 0 and concluded that the distribution of exercise is not independent of living arrangement, or that there is a relationship between living arrangement and exercise. The test provides an overall assessment of statistical significance. When the null hypothesis is rejected, it is important to review the sample data to understand the nature of the relationship. Consider again the sample data. 

Because there are different numbers of students in each living situation, it makes the comparisons of exercise patterns difficult on the basis of the frequencies alone. The following table displays the percentages of students in each exercise category by living arrangement. The percentages sum to 100% in each row of the table. For comparison purposes, percentages are also shown for the total sample along the bottom row of the table.

From the above, it is clear that higher percentages of students living in dormitories and in on-campus apartments reported regular exercise (31% and 23%) as compared to students living in off-campus apartments and at home (10% each).  

Test Yourself

 Pancreaticoduodenectomy (PD) is a procedure that is associated with considerable morbidity. A study was recently conducted on 553 patients who had a successful PD between January 2000 and December 2010 to determine whether their Surgical Apgar Score (SAS) is related to 30-day perioperative morbidity and mortality. The table below gives the number of patients experiencing no, minor, or major morbidity by SAS category.  

Question: What would be an appropriate statistical test to examine whether there is an association between Surgical Apgar Score and patient outcome? Using 14.13 as the value of the test statistic for these data, carry out the appropriate test at a 5% level of significance. Show all parts of your test.

In the module on hypothesis testing for means and proportions, we discussed hypothesis testing applications with a dichotomous outcome variable and two independent comparison groups. We presented a test using a test statistic Z to test for equality of independent proportions. The chi-square test of independence can also be used with a dichotomous outcome and the results are mathematically equivalent.  

In the prior module, we considered the following example. Here we show the equivalence to the chi-square test of independence.

A randomized trial is designed to evaluate the effectiveness of a newly developed pain reliever designed to reduce pain in patients following joint replacement surgery. The trial compares the new pain reliever to the pain reliever currently in use (called the standard of care). A total of 100 patients undergoing joint replacement surgery agreed to participate in the trial. Patients were randomly assigned to receive either the new pain reliever or the standard pain reliever following surgery and were blind to the treatment assignment. Before receiving the assigned treatment, patients were asked to rate their pain on a scale of 0-10 with higher scores indicative of more pain. Each patient was then given the assigned treatment and after 30 minutes was again asked to rate their pain on the same scale. The primary outcome was a reduction in pain of 3 or more scale points (defined by clinicians as a clinically meaningful reduction). The following data were observed in the trial.

We tested whether there was a significant difference in the proportions of patients reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) using a Z statistic, as follows. 

H 0 : p 1 = p 2    

H 1 : p 1 ≠ p 2                             α=0.05

Here the new or experimental pain reliever is group 1 and the standard pain reliever is group 2.

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group or that:

In this example, we have

Therefore, the sample size is adequate, so the following formula can be used:

Reject H 0 if Z < -1.960 or if Z > 1.960.

We now substitute the sample data into the formula for the test statistic identified in Step 2. We first compute the overall proportion of successes:

We now substitute to compute the test statistic.

• Step 5.  Conclusion.  

We now conduct the same test using the chi-square test of independence.  

H 0 : Treatment and outcome (meaningful reduction in pain) are independent

H 1 :   H 0 is false.         α=0.05

The formula for the test statistic is:  

For this test, df=(2-1)(2-1)=1. At a 5% level of significance, the appropriate critical value is 3.84 and the decision rule is as follows: Reject H0 if χ 2 > 3.84. (Note that 1.96 2 = 3.84, where 1.96 was the critical value used in the Z test for proportions shown above.)

We now compute the expected frequencies using:

The computations can be organized in a two-way table. The top number in each cell of the table is the observed frequency and the bottom number is the expected frequency. The expected frequencies are shown in parentheses.

A condition for the appropriate use of the test statistic was that each expected frequency is at least 5. This is true for this sample (the smallest expected frequency is 22.0) and therefore it is appropriate to use the test statistic.

(Note that (2.53) 2 = 6.4, where 2.53 was the value of the Z statistic in the test for proportions shown above.)

Chi-Squared Tests in R

The video below by Mike Marin demonstrates how to perform chi-squared tests in the R programming language.

Answer to Problem on Pancreaticoduodenectomy and Surgical Apgar Scores

We have 3 independent comparison groups (Surgical Apgar Score) and a categorical outcome variable (morbidity/mortality). We can run a Chi-Squared test of independence.

H 0 : Apgar scores and patient outcome are independent of one another.

H A : Apgar scores and patient outcome are not independent.

Chi-squared = 14.3

Since 14.3 is greater than 9.49, we reject H 0.

There is an association between Apgar scores and patient outcome. The lowest Apgar score group (0 to 4) experienced the highest percentage of major morbidity or mortality (16 out of 57=28%) compared to the other Apgar score groups.

Chi-Square (Χ²) Test & How To Calculate Formula Equation

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On This Page:

Chi-square (χ2) is used to test hypotheses about the distribution of observations into categories with no inherent ranking.

What Is a Chi-Square Statistic?

The Chi-square test (pronounced Kai) looks at the pattern of observations and will tell us if certain combinations of the categories occur more frequently than we would expect by chance, given the total number of times each category occurred.

It looks for an association between the variables. We cannot use a correlation coefficient to look for the patterns in this data because the categories often do not form a continuum.

There are three main types of Chi-square tests, tests of goodness of fit, the test of independence, and the test for homogeneity. All three tests rely on the same formula to compute a test statistic.

These tests function by deciphering relationships between observed sets of data and theoretical or “expected” sets of data that align with the null hypothesis.

What is a Contingency Table?

Contingency tables (also known as two-way tables) are grids in which Chi-square data is organized and displayed. They provide a basic picture of the interrelation between two variables and can help find interactions between them.

In contingency tables, one variable and each of its categories are listed vertically, and the other variable and each of its categories are listed horizontally.

Additionally, including column and row totals, also known as “marginal frequencies,” will help facilitate the Chi-square testing process.

In order for the Chi-square test to be considered trustworthy, each cell of your expected contingency table must have a value of at least five.

Each Chi-square test will have one contingency table representing observed counts (see Fig. 1) and one contingency table representing expected counts (see Fig. 2).

Figure 1. Observed table (which contains the observed counts).

To obtain the expected frequencies for any cell in any cross-tabulation in which the two variables are assumed independent, multiply the row and column totals for that cell and divide the product by the total number of cases in the table.

Figure 2. Expected table (what we expect the two-way table to look like if the two categorical variables are independent).

To decide if our calculated value for χ2 is significant, we also need to work out the degrees of freedom for our contingency table using the following formula: df= (rows – 1) x (columns – 1).

Formula Calculation

Calculate the chi-square statistic (χ2) by completing the following steps:

• Calculate the expected frequencies and the observed frequencies.
• For each observed number in the table, subtract the corresponding expected number (O — E).
• Square the difference (O —E)².
• Divide the squares obtained for each cell in the table by the expected number for that cell (O – E)² / E.
• Sum all the values for (O – E)² / E. This is the chi-square statistic.
• Calculate the degrees of freedom for the contingency table using the following formula; df= (rows – 1) x (columns – 1).

Once we have calculated the degrees of freedom (df) and the chi-squared value (χ2), we can use the χ2 table (often at the back of a statistics book) to check if our value for χ2 is higher than the critical value given in the table. If it is, then our result is significant at the level given.

Interpretation

The chi-square statistic tells you how much difference exists between the observed count in each table cell to the counts you would expect if there were no relationship at all in the population.

Small Chi-Square Statistic: If the chi-square statistic is small and the p-value is large (usually greater than 0.05), this often indicates that the observed frequencies in the sample are close to what would be expected under the null hypothesis.

The null hypothesis usually states no association between the variables being studied or that the observed distribution fits the expected distribution.

In theory, if the observed and expected values were equal (no difference), then the chi-square statistic would be zero — but this is unlikely to happen in real life.

Large Chi-Square Statistic : If the chi-square statistic is large and the p-value is small (usually less than 0.05), then the conclusion is often that the data does not fit the model well, i.e., the observed and expected values are significantly different. This often leads to the rejection of the null hypothesis.

How to Report

To report a chi-square output in an APA-style results section, always rely on the following template:

χ2 ( degrees of freedom , N = sample size ) = chi-square statistic value , p = p value .

In the case of the above example, the results would be written as follows:

A chi-square test of independence showed that there was a significant association between gender and post-graduation education plans, χ2 (4, N = 101) = 54.50, p < .001.

APA Style Rules

• Do not use a zero before a decimal when the statistic cannot be greater than 1 (proportion, correlation, level of statistical significance).
• Report exact p values to two or three decimals (e.g., p = .006, p = .03).
• However, report p values less than .001 as “ p < .001.”
• Put a space before and after a mathematical operator (e.g., minus, plus, greater than, less than, equals sign).
• Do not repeat statistics in both the text and a table or figure.

p -value Interpretation

You test whether a given χ2 is statistically significant by testing it against a table of chi-square distributions , according to the number of degrees of freedom for your sample, which is the number of categories minus 1. The chi-square assumes that you have at least 5 observations per category.

If you are using SPSS then you will have an expected p -value.

For a chi-square test, a p-value that is less than or equal to the .05 significance level indicates that the observed values are different to the expected values.

Thus, low p-values (p< .05) indicate a likely difference between the theoretical population and the collected sample. You can conclude that a relationship exists between the categorical variables.

Remember that p -values do not indicate the odds that the null hypothesis is true but rather provide the probability that one would obtain the sample distribution observed (or a more extreme distribution) if the null hypothesis was true.

A level of confidence necessary to accept the null hypothesis can never be reached. Therefore, conclusions must choose to either fail to reject the null or accept the alternative hypothesis, depending on the calculated p-value.

The four steps below show you how to analyze your data using a chi-square goodness-of-fit test in SPSS (when you have hypothesized that you have equal expected proportions).

Step 1 : Analyze > Nonparametric Tests > Legacy Dialogs > Chi-square… on the top menu as shown below:

Step 2 : Move the variable indicating categories into the “Test Variable List:” box.

Step 3 : If you want to test the hypothesis that all categories are equally likely, click “OK.”

Step 4 : Specify the expected count for each category by first clicking the “Values” button under “Expected Values.”

Step 5 : Then, in the box to the right of “Values,” enter the expected count for category one and click the “Add” button. Now enter the expected count for category two and click “Add.” Continue in this way until all expected counts have been entered.

Step 6 : Then click “OK.”

The four steps below show you how to analyze your data using a chi-square test of independence in SPSS Statistics.

Step 1 : Open the Crosstabs dialog (Analyze > Descriptive Statistics > Crosstabs).

Step 2 : Select the variables you want to compare using the chi-square test. Click one variable in the left window and then click the arrow at the top to move the variable. Select the row variable and the column variable.

Step 3 : Click Statistics (a new pop-up window will appear). Check Chi-square, then click Continue.

Step 4 : (Optional) Check the box for Display clustered bar charts.

Step 5 : Click OK.

Goodness-of-Fit Test

The Chi-square goodness of fit test is used to compare a randomly collected sample containing a single, categorical variable to a larger population.

This test is most commonly used to compare a random sample to the population from which it was potentially collected.

The test begins with the creation of a null and alternative hypothesis. In this case, the hypotheses are as follows:

Null Hypothesis (Ho) : The null hypothesis (Ho) is that the observed frequencies are the same (except for chance variation) as the expected frequencies. The collected data is consistent with the population distribution.

Alternative Hypothesis (Ha) : The collected data is not consistent with the population distribution.

The next step is to create a contingency table that represents how the data would be distributed if the null hypothesis were exactly correct.

The sample’s overall deviation from this theoretical/expected data will allow us to draw a conclusion, with a more severe deviation resulting in smaller p-values.

Test for Independence

The Chi-square test for independence looks for an association between two categorical variables within the same population.

Unlike the goodness of fit test, the test for independence does not compare a single observed variable to a theoretical population but rather two variables within a sample set to one another.

The hypotheses for a Chi-square test of independence are as follows:

Null Hypothesis (Ho) : There is no association between the two categorical variables in the population of interest.

Alternative Hypothesis (Ha) : There is no association between the two categorical variables in the population of interest.

The next step is to create a contingency table of expected values that reflects how a data set that perfectly aligns the null hypothesis would appear.

The simplest way to do this is to calculate the marginal frequencies of each row and column; the expected frequency of each cell is equal to the marginal frequency of the row and column that corresponds to a given cell in the observed contingency table divided by the total sample size.

Test for Homogeneity

The Chi-square test for homogeneity is organized and executed exactly the same as the test for independence.

The main difference to remember between the two is that the test for independence looks for an association between two categorical variables within the same population, while the test for homogeneity determines if the distribution of a variable is the same in each of several populations (thus allocating population itself as the second categorical variable).

Null Hypothesis (Ho) : There is no difference in the distribution of a categorical variable for several populations or treatments.

Alternative Hypothesis (Ha) : There is a difference in the distribution of a categorical variable for several populations or treatments.

The difference between these two tests can be a bit tricky to determine, especially in the practical applications of a Chi-square test. A reliable rule of thumb is to determine how the data was collected.

If the data consists of only one random sample with the observations classified according to two categorical variables, it is a test for independence. If the data consists of more than one independent random sample, it is a test for homogeneity.

What is the chi-square test?

The Chi-square test is a non-parametric statistical test used to determine if there’s a significant association between two or more categorical variables in a sample.

It works by comparing the observed frequencies in each category of a cross-tabulation with the frequencies expected under the null hypothesis, which assumes there is no relationship between the variables.

This test is often used in fields like biology, marketing, sociology, and psychology for hypothesis testing.

What does chi-square tell you?

The Chi-square test informs whether there is a significant association between two categorical variables. Suppose the calculated Chi-square value is above the critical value from the Chi-square distribution.

In that case, it suggests a significant relationship between the variables, rejecting the null hypothesis of no association.

How to calculate chi-square?

To calculate the Chi-square statistic, follow these steps:

1. Create a contingency table of observed frequencies for each category.

2. Calculate expected frequencies for each category under the null hypothesis.

3. Compute the Chi-square statistic using the formula: Χ² = Σ [ (O_i – E_i)² / E_i ], where O_i is the observed frequency and E_i is the expected frequency.

4. Compare the calculated statistic with the critical value from the Chi-square distribution to draw a conclusion.

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Statistics By Jim

Making statistics intuitive

Chi-Square Test of Independence and an Example

By Jim Frost 88 Comments

The Chi-square test of independence determines whether there is a statistically significant relationship between categorical variables . It is a hypothesis test that answers the question—do the values of one categorical variable depend on the value of other categorical variables? This test is also known as the chi-square test of association.

In this post, I’ll show you how the Chi-square test of independence works. Then, I’ll show you how to perform the analysis and interpret the results by working through the example. I’ll use this test to determine whether wearing the dreaded red shirt in Star Trek is the kiss of death!

If you need a primer on the basics, read my hypothesis testing overview .

Overview of the Chi-Square Test of Independence

The Chi-square test of association evaluates relationships between categorical variables. Like any statistical hypothesis test , the Chi-square test has both a null hypothesis and an alternative hypothesis.

• Null hypothesis: There are no relationships between the categorical variables. If you know the value of one variable, it does not help you predict the value of another variable.
• Alternative hypothesis: There are relationships between the categorical variables. Knowing the value of one variable does help you predict the value of another variable.

The Chi-square test of association works by comparing the distribution that you observe to the distribution that you expect if there is no relationship between the categorical variables. In the Chi-square context, the word “expected” is equivalent to what you’d expect if the null hypothesis is true. If your observed distribution is sufficiently different than the expected distribution (no relationship), you can reject the null hypothesis and infer that the variables are related.

For a Chi-square test, a p-value that is less than or equal to your significance level indicates there is sufficient evidence to conclude that the observed distribution is not the same as the expected distribution. You can conclude that a relationship exists between the categorical variables.

When you have smaller sample sizes, you might need to use Fisher’s exact test instead of the chi-square version. To learn more, read my post, Fisher’s Exact Test: Using and Interpreting .

Star Trek Fatalities by Uniform Colors

We’ll perform a Chi-square test of independence to determine whether there is a statistically significant association between shirt color and deaths. We need to use this test because these variables are both categorical variables. Shirt color can be only blue, gold, or red. Fatalities can be only dead or alive.

The color of the uniform represents each crewmember’s work area. We will statistically assess whether there is a connection between uniform color and the fatality rate. Believe it or not, there are “real” data about the crew from authoritative sources and the show portrayed the deaths onscreen. The table below shows how many crewmembers are in each area and how many have died.

Tip: Because the chi-square test of association assesses the relationship between categorical variables, bar charts are a great way to graph the data. Use clustering or stacking to compare subgroups within the categories.

Related post : Bar Charts: Using, Examples, and Interpreting

Performing the Chi-Square Test of Independence for Uniform Color and Fatalities

For our example, we will determine whether the observed counts of deaths by uniform color are different from the distribution that we’d expect if there is no association between the two variables.

The table below shows how I’ve entered the data into the worksheet. You can also download the CSV dataset for StarTrekFatalities .

You can use the dataset to perform the analysis in your preferred statistical software. The Chi-squared test of independence results are below. As an aside, I use this example in my post about degrees of freedom in statistics . Learn why there are two degrees of freedom for the table below.

In our statistical results, both p-values are less than 0.05. We can reject the null hypothesis and conclude there is a relationship between shirt color and deaths. The next step is to define that relationship.

Describing the relationship between categorical variables involves comparing the observed count to the expected count in each cell of the Dead column. I’ve annotated this comparison in the statistical output above.

Statisticians refer to this type of table as a contingency table. To learn more about them and how to use them to calculate probabilities, read my post Using Contingency Tables to Calculate Probabilities .

Related post : Chi-Square Table

Graphical Results for the Chi-Square Test of Association

Additionally, you can use bar charts to graph each cell’s contribution to the Chi-square statistic, which is below.

Surprise! It’s the blue and gold uniforms that contribute the most to the Chi-square statistic and produce the statistical significance! Red shirts add almost nothing. In the statistical output, the comparison of observed counts to expected counts shows that blue shirts die less frequently than expected, gold shirts die more often than expected, and red shirts die at the expected rate.

The graph below reiterates these conclusions by displaying fatality percentages by uniform color along with the overall death rate.

The Chi-square test indicates that red shirts don’t die more frequently than expected. Hold on. There’s more to this story!

Time for a bonus lesson and a bonus analysis in this blog post!

2 Proportions test to compare Security Red-Shirts to Non-Security Red-Shirts

The bonus lesson is that it is vital to include the genuinely pertinent variables in the analysis. Perhaps the color of the shirt is not the critical variable but rather the crewmember’s work area. Crewmembers in Security, Engineering, and Operations all wear red shirts. Maybe only security guards have a higher death rate?

We can test this theory using the 2 Proportions test. We’ll compare the fatality rates of red-shirts in security to red-shirts who are not in security.

The summary data are below. In the table, the events represent the counts of deaths, while the trials are the number of personnel.

The p-value of 0.000 signifies that the difference between the two proportions is statistically significant. Security has a mortality rate of 20% while the other red-shirts are only at 4%.

Security officers have the highest mortality rate on the ship, closely followed by the gold-shirts. Red-shirts that are not in security have a fatality rate similar to the blue-shirts.

As it turns out, it’s not the color of the shirt that affects fatality rates; it’s the duty area. That makes more sense.

Risk by Work Area Summary

The Chi-square test of independence and the 2 Proportions test both indicate that the death rate varies by work area on the U.S.S. Enterprise. Doctors, scientists, engineers, and those in ship operations are the safest with about a 5% fatality rate. Crewmembers that are in command or security have death rates that exceed 15%!

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Reader Interactions

July 18, 2024 at 10:27 am

I read this chi-squared example your excellent book on hypothesis testing but there a couple of things that I can’t quite reconcile:

You decribed the proportion of observed Red shirts fatalities as being the same as the expected. Relative to the Blue and Gold comparison the Red Shirt fatality rate (24) is much closer to the expected rate (22.23) but it isn’t exactely the same.

How different would it need to be to conclude that it is different as opposed to just being the least important in a context where we have concluded that there is an association between shirt colour and fatality rates? Would this need to be answered by a series of chi squared tests (or 2 proportion test) that considered the combination of one shirt colour compared with the sum of other shirt colours. I have tried this with the following p values resultinh from the chi-squared test in excel.

Red shirt v non-red shirt pvalue = 0.554809933

Blue shirt v non-blue shirt pvalue = 0.04363293

Gold shirt v non-gold shirt = 0.053533022

This would suggest that if the question was “do gold shirts die more frequently than other colours” the answer would be that the data does not rule out the null hypothesis. For blue shirts this test would suggest that the data can rule out the null hypothesis yet in the full three colour test Gold contributed the most to the chi-squared statistic.

I have a similar example from my work which looks at the proportions of customers using two different websites and considers the proportions which are new customers, existing customers and customers returned after a long gap (reactivated).

For that test the chi-squared test p value was sufficient to rule out the null hypothesis with reactivated customers contributing the most to the chi-squared statistic. But no individual test which set each customer group against the sum of the others would be considered significant.

Is this comparison gold shirt versus non-shirt or reactivated customers v other customers not a valid use of this test?

February 6, 2024 at 9:55 pm

Hi Jim. I am using R to calclate a chi sqaure of independence. I have an value of 1.486444 with a P value greater than 0.05. My question is how do I interpet the value of 1.48644? Is this a strong association between two variables or a weak association?

February 6, 2024 at 10:19 pm

You really just look at the p-value. If you assess the chi-square value, you need as the chi-square value in conjunction with a chi-square distribution with the correct degrees of freedom and use that to calculate the probability. But the p-value does that for you!

In your case, the p-value is greater than your significance level. So, you fail to reject the null hypothesis. You have insufficient evidence to conclude that an association exists between your variables.

Also, it’s important to note that this test doesn’t indicate the strength of association. It only tells you whether your sample data provide sufficient evidence to conclude that an association exists in the population. Unfortunately, you can’t conclude that an association exists.

September 1, 2022 at 5:01 am

Thank you this was such a helpful article.

I’m not sure if you check these comments anymore, yet if you do I did have a quick question for you. I was trying to follow along in SPSS to reproduce your example and I managed to do most of it. I put your data in, used Weight Cases by Frequency of Deaths, and then was able to do the Chi Square analysis that achieved the exact same results as yours.

Unfortunately, I am totally stuck on the next part where you do the 2 graphs, especially the Percentage of Fatalities by Shirt Color. The math makes sense – it’s just e.g., Gold deaths / (Gold deaths + Gold Alive). However, I cannot seem to figure out how to create a bar chart like that in SPSS!? I’ve tried every combination of variables and settings I can think of in the Chart Builder and no luck. I’ve also tried the Compute Variable option with various formulas to create a new column with the death percentages by shirt color but can’t find a way to sum the frequencies.. The best I can get is using an IF statement so it only calculates on the rows with a Death statistic and then I can get the first part: Frequency / ???, but can’t sum the 2 frequencies of Deaths & Alive per shirt colour to calculate the figure properly. And I’m not sure what other things I can try.

So basically I’m totally stuck at the moment. If by some chance you see this, is there any chance you might please be able to help me figure out how to do that Percentage of Fatalities by Shirt Color bar graph in SPSS? The only way I can see at the moment is to manually type the calculated figures into a new dataset and graph it. That would work but doesn’t seem a very practical way of doing things if this was a large dataset instead of a small example one. Hence I’m assuming this must be a better way of doing this?

Thank you in advance for any help you can give me.

September 1, 2022 at 3:38 pm

Yes, I definitely check these comments!

Unfortunately, I don’t have much experience using SPSS, so I’ll be of limited help with that. There must be some way to do that in SPSS though. Worst case scenario, calculate the percentages by hand or in Excel and then enter them into SPSS and graph them. That shouldn’t be necessary but would work in a pinch.

Perhaps someone with more SPSS experience can provide some tips?

September 18, 2021 at 6:09 pm

Hi. This comment relates to Warren’s post. The null hypothesis is that there is no statistically significant relationship between “Uniform color” and “Status”. During the summing used to calculate the Chi-squared statistic, each of the (6) contributions are included. (3 Uniform colors x 2 status possibilities) The “Alive” column gives the small contributions that bring the total contribution from 5.6129 up to 6.189. Any reasoning specific to the “Dead” column only begins after the 2-dimensional Chi-squared calculation has been completed.

September 19, 2021 at 12:38 am

Hi Bill, thanks for your clarifications. I got confused with whom you were replying!

September 17, 2021 at 5:53 pm

The chi-square formula is: χ2 = ∑(Oi – Ei)2/Ei, where Oi = observed value (actual value) and Ei = expected value.

September 17, 2021 at 5:56 pm

Hi Bill, thanks. I do cover the formula and example calculations in my other post on the topic, How Chi-Squared Works .

September 16, 2021 at 6:24 pm

Why is the Pearson Chi Square statistic not equal to the sum of the contributions to Chi-Square? I get 5.6129. The p-value for that Chi-Squre statistic is .0604 which is NOT significant in this century OR the 24th.

September 14, 2021 at 8:25 am

Thank you JIm, Excellent concept teaching!

July 15, 2021 at 1:05 pm

Thank you so much for the Star Trek example! As a long-time Trek fan and Stats student, I absolutely love the debunking of the red shirt theory!

July 19, 2021 at 10:19 pm

I’m so glad you liked my example. I’m a life-long Trek fan as well! I found the red shirt question to be interesting. One the one hand, part of the answer of the answer is that red shirts comprise just over 50% of the crew, so of course they’ll have more deaths. And then on the other hand, it’s only certain red shirts that actually have an elevated risk, those in security.

May 16, 2021 at 1:42 pm

Got this response from the gentleman who did the calculation using a Chi Square. Would you mind commenting? “The numbers reported are nominate (counting) numbers not ordinate (measurement) numbers. As such chi-square analysis must be used to statistically compare outcomes. Two-sample student t-tests cannot be used for ordinate numbers. Correlations are also not usually used for ordinate numbers and most importantly correlations do NOT show cause and effect.”

May 16, 2021 at 3:13 pm

I agree with the first comment. However, please note that I recommended the 2-sample proportions test and the other person is mentioning the 2-sample t-test. Very different tests! And, I agree that the t-test is not appropriate for the Pfizer data. Basically, he’s saying you have categorical data and the t-test is for continuous data. That’s all correct. And that’s why I recommended the the proportions test.

As for the other part about “correlations do NOT show cause and effect.” That’s not quite correct. More accurately, you’d say that correlations do not NECESSARILY imply causation. Sometimes they do and sometimes they don’t imply causation. It depends on the context in which the data were collected. Correlations DO suggest causation when you use a randomized controlled trial (RCT) for the experiment and data collection, which is exactly what Pfizer did. Consequently, the Pfizer data DO suggest that the vaccine caused a reduction in the proportion of COVID infections in the vaccine group compared to the control group (no vaccine). RCTs are intentionally designed so you can draw causal inferences, which is why the FDA requires them for vaccine and other medical trials.

If you’re interested, I’ve written an article about why randomized controlled trials allow you to make causal inferences .

May 16, 2021 at 12:41 pm

Mr. Jim Frost…You are Da Man!! Thank you!! Yes, this is the same document I have been looking at, just did not know how to interpret Table 9. Sorry, never intended to ask you for medical advice, just wanted to understand the statistics and feel confident that the calculations were performed correctly. You have made my day! Now just a purely statistics question, assuming I have not worn out your patience with my dumb questions…Can you explain the criteria used to determine when a Chi Square should be used versus a 2-samples proportions test? I think I saw a comment from someone on your website stating that the Chi Sqaure is often misused in the medical field. Fascinating, fascinating field you are in. Thank you so much for sharing your knowledge and expertise.

May 16, 2021 at 3:00 pm

You bet! That’s why I’m here . . . to educate and clarify statistics and statistical analyses!

The chi-squared test of independence (or association) and the two-sample proportions test are related. The main difference is that the chi-squared test is more general while the 2-sample proportions test is more specific. And, it happens that the proportions test it more targeted at specifically the type of data you have.

The chi-squared test handles two categorical variables where each one can have two or more values. And, it tests whether there is an association between the categorical variables. However, it does not provide an estimate of the effect size or a CI. If you used the chi-squared test with the Pfizer data, you’d presumably obtain significant results and know that an association exists, but not the nature or strength of that association.

The two proportions test also works with categorical data but you must have two variables that each have two levels. In other words, you’re dealing with binary data and, hence, the binomial distribution. The Pfizer data you had fits this exactly. One of the variables is experimental group: control or vaccine. The other variable is COVID status: infected or not infected. Where it really shines in comparison to the chi-squared test is that it gives you an effect size and a CI for the effect size. Proportions and percentages are basically the same thing, but displayed differently: 0.75 vs. 75%.

What you’re interested in answering is whether the percentage (or proportion) of infections amongst those in the vaccinated group is significantly different than the percentage of infections for those in control group. And, that’s the exact question that the proportions test answers. Basically, it provides a more germane answer to that question.

With the Pfizer data, the answer is yes, those in the vaccinated group have a significantly lower proportion of infections than those in the control group (no vaccine). Additionally, you’ll see the proportion for each group listed, and the effect size is the difference between the proportion, which you can find on a separate line, along with the CI of the difference.

Compare that more specific and helpful answer to the one that chi-squared provides: yes, there’s an association between vaccinations and infections. Both are correct but because the proportions test is more applicable to the specific data at hand, it gives a more useful answer.

I see you have an additional comment with questions, so I’m off to that one!

May 15, 2021 at 1:00 pm

Hi Jim, So sorry if my response came off as anything but appreciative of your input. I tried to duplicate your results in your Flu Vaccine article using the 2 Proportion test as you recommended. I was able to duplicate your Estimate for Difference of -0.01942, but I could not duplicate your value for Z, so clearly I am not doing the calculation correctly – even when using Z calculators. So since I couldn’t duplicate your correct results for your flu example, I did not have confidence to proceed to Moderna. I was able to calculate effectiveness (the hazard ratio that is widely reported), but as I have reviewed the EUA documents presented to the FDA in December 2020, I know that there is no regression analysis, and most importantly, no data to show an antibody response produced by the vaccine. So they are not showing the vaccine was successful in producing an immune response, just giving simplistic proportions of how many got covid and how many didn’t. And as they did not even factor in the number of people who had had covid prior to vaccine, I just cant understand how these numbers have any significance at all. I mention the PCR test because it too is under an EUA, and has severe limitations. I would think that those limitations would be statistically significant, as are the symptoms which can indicate any bacterial or viral infection. And you state “I’m sure you can find a journal article or documentation that shows the thorough results if you’re interested”. Clearly I am VERY interested, as I love my parents more than life itself, and have seen the VAERS data, and I don’t want them to be the next statistic. But I CANT find the thorough results that you say are so easy to find. If I could I would not be trying to learn to perform statistical calculations. So I went out on a limb, as you are a fellow trekky and seem like a super nice guy, sharing your expertise with others, and thought you might be able to help me understand the statistics so I can help my parents make an informed choice. We are at a point that children and pregnant women are getting these vaccines. Unhealthy, elderly people in nursing homes (all the people excluded in the trials) are getting these vaccines. I simply ask the question…..do these vaccines provide more protection than NOT getting the vaccine? The ENTIRE POPULATION is being forced to get these vaccines. And you tell me “I’m sure you can find a journal article or documentation that shows the thorough results if you’re interested.” I can only ask…how are you NOT interested? This is the most important statistical question of our lifetime, and of your children’s and granchildren’s lifetime. And I find that no physician or statistician able or willing to answer these questions. Respectfully, Chris

May 15, 2021 at 11:00 pm

No worries. On my website, I’m just discussing the statistical nature of Moderna’s study. Of course, everyone is free to make their own determination and decide accordingly.

You’re obviously free to question the methods and analysis, but as a statistician, I’m satisfied that Moderna performed an appropriate clinical trial and followed that up with a rigorous and appropriate statistical analysis. In my opinion, they have demonstrated that their vaccine is safe and effective. The only caveat is that we don’t have long-term safety data because not enough time has gone by. However, most side effects for vaccines show up in the first 45 days. That timeframe occurred during the trial and all side effects were recorded.

However, I’m not going to get into a debate about whether anyone should get the vaccine or not. I run a statistics website and that’s the aspect I’m focusing on. There are other places to debate the merits of being vaccinated.

May 14, 2021 at 8:05 pm

Hi Jim, thanks for the reply. I have to admit the detail of all the statistical methods you mention are over my head, but by scanning the document it appears you did not actually calculate the vaccine’s efficacy, just stated how the analysis should be done. I am referring to comments like “To analyze the COVID-19 vaccine data, statisticians will use a stratified Cox proportional hazard regression model to assess the magnitude of the difference between treatment and control groups using a one-sided 0.025 significance level”. And “The full data and analyses are currently unavailable, but we can evaluate their interim analysis report. Moderna (and Pfizer) are still assessing the data and will present their analyses to Federal agencies in December 2020.” I am looking at the December 2020 reports that both Pfizer and Moderna presented to the FDA, and I see no “stratified Cox proportional hazard regression model”, just the simplistic hazard ratio you mention in your paper. I don’t see how that shows the results are statistically significant and not chance. Also the PCR test does not confirm disease, just presence of virus (dead or alive) and virus presence doesnt indicate disease. And the symptoms are symptoms of any viral or bacterial infection, or cancer. Just sort of suprised to see no statistical analysis in the December 2020 reports. Was hoping you had done the heavy lifting…lol

May 14, 2021 at 11:38 pm

Hi Christine,

You had asked if Chi-square would work for your data and my response was no, but here are two methods that would. No, I didn’t analyze the Moderna data myself. I don’t have access to their complete data that would allow me to replicate their results. However, in my post, I did calculate the effectiveness, which you can do using the numbers I had, but not the significance.

Based on the data you indicated you had, I’d recommend the two-sample proportions test that I illustrate in the flu vaccine post. That won’t replicate the more complex analyses but is doable with the data that you have.

The Cox proportional hazard regression model analyzes the hazard ratio. The hazard ratio is the outcome measure in this context. They’re tied together and it’s the regression analysis that indicate significance. I’d imagine you’d have to read a thorough report to get the nitty gritty details. I got the details of their analysis straight from Moderna.

I’m not sure what your point with the PCR test. But, I’m just reporting how they did their analysis.

Moderna, Pfizer, and the others have done the “heavy lifting.” When I wrote the post about the COVID vaccination, it was before it was approved for emergency use. By this point, I’m sure you can find a journal article or documentation that shows the thorough results if you’re interested.

May 14, 2021 at 2:56 pm

Hi Jim, my parents are looking into getting the Pfizer vaccine, and I was wondering if I could use a chi square analysis to see if its statistically effective. From the EUA document, 17411 people got the Pfizer vaccine, and of those people – 8 got covid, and 17403 did not. Of the control group of 17511 that did not get the vaccine, 162 got covid, and 17349 did not. My calculations show this is not statistically significant, but wasn’t sure if I did my calculation correctly, or if I can even use a chi square for this data. Can you help? PS. As a Trekky family, I love your analysis…but we all know its the new guy with a speaking part that gets axed…lol

May 14, 2021 at 3:28 pm

There are several ways you can analyze the effectiveness. I write about how they assessed the Moderna vaccine’s effectiveness , which uses a special type of regression analysis.

The other approach is to use a two-sample proportions test. I don’t write about that in the COVID context but I show how it works for flu vaccinations . The same ideas apply to COVID vaccinations. You’re dealing comparing the proportion of infections in the control group to the treatment group. Hence, a two-sample proportions test.

A chi-square analysis won’t get you where you want to go. It would tell you if there is an association, but it’s not going to tell you the effect size.

I’d read those two posts that I wrote. They’ll give you a good insight for possible ways to analyze the data. I also show how they calculate effectiveness for both the COVID and flu shots!

I hope that helps!

April 9, 2021 at 2:49 am

thank you so much for your response and advice! I will probably go for the logistic regression then 🙂

All the best for you!

April 10, 2021 at 12:39 am

You’re very welcome! Best of luck with your study! 🙂

April 7, 2021 at 4:18 am

thank you so much for your quick response! This actually helps me a lot and I also already thought about doing a binary logistic regression. However, my supervisor wanted me to use a chi-square test, as he thinks it is easier to perform and less work. So now I am struggling to decide, which option would be more feasible.

Coming back to the chi-square test – could I create a new variable which differentiates between the four experimental conditions and use this as a new ID? Or can I use the DV to weight the frequencies in the chi-square test? – I did that once in a analysis using a continuous DV as weight. Yet, I am not sure if or how that works with a binary variable. Do you have an idea what would work best in the case of a chi-square test?

Thank you so much!!

April 8, 2021 at 11:25 pm

You’re very welcome!

I don’t think either binary logistic regression or chi-square are more less work than the other. However, Chi-square won’t give you the answers you want. You can’t do interaction effects with chi-square. You won’t get nice odds ratios which are a much more intuitive way to interpret the results than chi-square, at least in my opinion. With chi-square, you don’t get a p-value/significance for each variable, just the overall analysis. With logistic regression, you get p-values for each variable and the interaction term if you include it.

I think you can do chi-square analyses with more than one independent variable. You’d essentially have a three dimensional table rather than a two-dimensional table. I’ve never done that myself so I don’t have much advice to offer you there. But, I strongly recommend using logistic regression. You’ll get results that are more useful.

April 6, 2021 at 10:59 am

thank you so much for this helpful post!

April 6, 2021 at 5:36 am

thank you for this very helpful post. Currently, I am working on my master’s thesis and I am struggling with identifying the right way to test my hypothesis as in my case I have three dummy variables (2 independent and 1 dependent).

The experiment was on the topic advice taking. It was a 2×2 between sample design manipulating the source of advice to be a human (0) or an algorithm (1) and the task to be easy (0) or difficult (1). Then, I measured whether the participants followed (1) or not followed (0) the advice. Now, I want to test if there is an interaction effect. In the easy task I expect that the participants rather follow the human advice and in the difficult task the participants rather follow the algorithmic advice.

I want to test this using a chi-square independence test, but I am not sure how to do that with three variables. Should I rather use the variable “Follow/Notfollow” as a weight or should I combine two of the variables so that I have a new variable with four categories, e.g. Easy.Human, Easy.Algorithm, Difficult.Human, Difficult.Algorithm or Human.Follow, Human.NotFollow, Algorithm.Follow, Algorithm.NotFollow

I am not sure, if this is scientifically correct. I would highly appreciate your help and your advice.

Thank you so much in advance! Best, Anni

April 7, 2021 at 1:58 am

I think using binary logistic regression would be your best bet. You can use your dummy DV with that type. And have two dummy IVs also works. You can also include an interaction term, which isn’t possible in chi-square tests. This model would tell you whether source of advice, difficulty of task, and their interaction relate to the probability of participants following the advice.

March 29, 2021 at 12:43 pm

Hi Jim, I want to thank you for all the content that you have posted online. It has been very helpful for me to apply simple principles of statistics at work. I wanted your thoughts on how to approach the following problem, which appeared to be slightly different from the examples that you shared above. We have two groups – test group (exposed to an ad for brand A) and control group (not exposed to any ads for brand A). We asked both groups a qn: Have you heard of brand A? The possible answers were a Y/N. We then did a t-test to determine if the answers were significantly different for the test and control groups (they were) We asked both groups a follow-up qn as well: How likely are you to buy any of the following brands in the next 3 months? The options were as follows (any one could be picked. B,C & D are competing brands with A) 1.A 2.B 3.C 4.D We wanted to check if the responses we received from both groups were statistically different. Based on my reading, it seemed like the Chi-Square test was the right one to run here. However, I wasn’t too sure what the categorical variables would be in this case and how we could run the Chi-square test here. Would like to get our inputs on how to approach this. Thanks

March 29, 2021 at 2:53 pm

For the first question, I’d typically recommend a 2-sample proportions test. You have two groups and the outcome variable is binary, which is good for proportions. Using a 2-sample proportions test will tell you whether the proportion of individuals who have heard of Brand A differs by the two groups (ads and no ads). You could use the chi-squared test of independence for this case but I recommend the proportions test because it’s designed specifically for this scenario. The procedure can also estimate the effect size and a CI for the effect size (depending on your software). A t-test is not appropriate for these data.

For the next question, yes, the chi-square test is good choice as long as they can only pick one of the options. Maybe, which brand are you most likely to purchase in the next several months. The categories must be mutually exclusive to use chi-square. One variable could be exposed to ad with yes and no as levels. The other would be the purchase question with A, B, C, D as levels. That gives you a 2 X 4 table for your chi-squared test of independence.

March 29, 2021 at 5:08 am

I don’t see the relationship between the table of shirt color and status and the tabulated statistics. Sam

March 29, 2021 at 3:39 pm

I show the relationship several ways in this post. The key is to understand how the actual counts compare to the expected counts. The analysis calculates the expected counts under the assumption that there is no relationship between the variables. Consequently, when there are differences between the actual and expected accounts, a relationship potentially exists.

In the Tabulated Statistics output, I circle and explain how the actual counts compare to the expected counts. Blue uniforms have fewer deaths than expected while Gold uniforms have more deaths than expected. Red uniforms equal the expect amount, although I explore that in more detail later in the post. You can also see these relationships in the graph titled Percentage of Fatalities.

Overall, the results show the relationship between uniform color and deaths and the p-value indicates that this relationship is statistically significant.

February 20, 2021 at 8:51 am

Suppose you have two variables that checking out books and means to get to the central library. How might you formulate null hypothesis and alternative hypothesis for the independence test? please answer anyone

February 21, 2021 at 3:15 pm

In this case, the null hypothesis states that there is no relationship between means to get to the library and checking out a book. The alternative hypothesis states that there is a relationship between them.

November 18, 2020 at 12:39 pm

Hi there I’m just wondering if it would be appropriate to use a Chi square test in the following scenario; – A data set of 1000 individuals – Calculate Score A for all 1000 individuals; results are continuous numerical data eg. 2.13, 3.16, which then allow individuals to be placed in categories; low risk (3.86) -Calculate Score B for the same 1000 individuals; results are discrete numerical data eg. 1, 6, 26 ,4 which the allow individuals to be placed in categories; low risk (26). – I then want to compared the two scoring systems A & B ; to see if (1) the individuals are scoring similarly on both scores (2) I have reason to believe one of the scores overestimates the risk, I’d like tot test this.

Thank you, I haven’t been able to find any similar examples and its stressing me out 🙁

November 13, 2020 at 1:53 pm

Would you be able to advise?

My organization is sending out 6 different emails to employees, in which they have to click on a link in the email. We want to see if one variation in language might get a higher click rate rate for the link. So we have 6 between subjects conditions, and the response can either be a ‘clicked on the link’ or ‘NOT clicked on the link’.

Is this a Chi-Square of Independence test? Also, how would I know where the difference lies, if the test is significant? (i.e., what is the non-parametric equivalent of running an ANOVA and followup pairwise comparisons?

Thanks Jim!

October 15, 2020 at 11:05 pm

I am working on the press coverage of civil military relations in Pakistani press from 2008 to 2018, I want to check that whether is a difference of coverage between two tenures ie 2008 to 2013 and 2013 to 2018. Secondly I want to check the difference of coverage between two types of newspapers ie english newspapers and urdu newspapers. furthermore I also want to check the category wise difference of coverage from the tenure 2008 to 2018.

I have divided my data into three different distributions, 1 is pro civilian, 2 is pro military and 3 is neutral.

October 4, 2020 at 4:07 am

Hi thank you so much for this. I would like to ask, if the study Is about whether factors such as pricing, marketing, and brand affects the intention of the buyer to purchase the product. Can I use Chi-test for the statistic treatment? and if it is not can I ask what statistical treatment would you suggest? Thank you so much again.

October 3, 2020 at 2:51 pm

Jim, Thank you for the post. You displayed a lot of creativity linking the two lessons to Star Trek. Your website and ebook offerings are very inspiring to me. Bill

October 4, 2020 at 12:53 am

Thanks so much, Bill. I really appreciate the kind words and I’m happy that the website and ebooks have been helpful!

September 29, 2020 at 7:10 am

Thank-you for your explanation. I am trying to help my son with his final school year investigation. He has raw data which he collected from 21 people of varying experience. They all threw a rugby ball at a target and the accuracy, time of ball in the air and experience (rated from 1-5) were all recorded. He has calculated the speed and the displacement, and used correlation to compare speed versus accuracy and experience versus accuracy. He needs to incrementally increase the difficulty of maths he uses in his analysis and he was thinking of the Chi Square test as a next step, however from your explanation above the current form of his data would not be suitable for this test. Is there a way of re-arranging the data so that we can use the Chi Square test? Thanks!

September 30, 2020 at 4:33 pm

Hi Rhonwen,

The chi-squared test of independence looks for correlation between categorical variables. From your description, I’m not seeing a good pair of categorical variables to test for correlation. To me, the next step for this data appears to be regression analysis.

September 12, 2020 at 5:37 pm

Thank you for the detailed teaching! I think this explains chi square much better than other websites I have found today. Do you mind sharing which software you use to get Expected Count and contribution to Chi square? Thank you for your help.

August 22, 2020 at 1:06 pm

Good day jim! I was wondering what kind of data analysis should i use if i am going to have a research on knowledge, attitude and practices? Looking forward to your reply! Thank you!

June 25, 2020 at 8:43 am

Very informative and easy to understand it. Thank you so much sir

June 2, 2020 at 11:03 am

Hi I wanted to know how the significance probability can be calculated if the significance level wasn’t given. Thank you

June 3, 2020 at 7:39 pm

Hi, you don’t need to know the significance level to be able to calculate the p-value. For calculating the p-value, you must know the null hypothesis, which we do for this example.

However, I do use a significance level of 0.05 for this example, making the results statistically significant.

May 26, 2020 at 5:55 am

What summary statistics can I use to describe the graph of a categorical data? Good presentation by the way. Very Insightful

May 26, 2020 at 8:39 pm

Hi Michael,

For categorical data like the type in this example, which is in a two-way contingency table, you’d often use counts or percentages. A bar chart is often a good choice for graphing counts or percentages by multiple categories. I show an example of graphing data for contingency tables in my Introduction to Statistics ebook .

May 25, 2020 at 10:27 am

Thank you for your answer. I saw online that bar graphs can be used to visualise the data (I guess it would be the percentage of death in my case) with 95% Ci intervals for the error bar. Is this also applicable if I only have a 2×2 contingency table? If not, what could be my error bar?

May 26, 2020 at 8:59 pm

Hi John, you can obtain CIs for proportions, which is basically a percentage. And, bar charts are often good for graphing contingency tables.

May 24, 2020 at 9:34 am

Hi! So I am working on this little project where I am trying to find a relationship between sex and mortality brought by this disease so my variables are: sex (male or female) and status (dead or alive). I am new to statistics so I do not know much. Is there any way to check the normality of categorical data? There is a part wherein our data must be based on data normality but I am not sure it this applies to categorical data. Thank you for your answer!

May 24, 2020 at 4:23 pm

The normal distribution is for continuous data. You have discrete data values–two binary variables to be precise. So, the normal distribution is not applicable to your data.

May 21, 2020 at 11:26 pm

Hi Jim, this was really helpful. I am in the midst of my proposal on a research to determine the association between burnout and physical activity among anaesthesia trainees.

They are both categorial variable physical activity – 3 categories: high, moderate, low burnout – 2 categories: high and low

How do I calculate my sample size for my study?

May 22, 2020 at 2:13 pm

Hi Jaishree,

I suggest you download a free sample size and power calculation program called G*Power . Then do the following:

• In G*Power, under Test Family, choose, χ². Under Statistical test, choose Goodness-of-fit tests: Contingency tables.
• In Effect size w, you’ll need to enter a value. 0.1 = weak. 0.3 medium, and 0.5 large. That’s based on subject area knowledge.
• In β/α ratio, that’s the ratio of the Type II error rate/Type I error rate. They have a default value of 1, but that seems too low. 2-3 might be more appropriate but you can try different values to see how this affects the results.
• Then you need to enter your sample size and DF. Read my post about Degrees of Freedom , which includes a section about calculating it for chi-square tests.
• Click Calculate.

Experiment and adjust values to see how that changes the output. You want to find a sample size that produces sufficient power while incorporating your best estimates of the other parameters (effect size, etc.).

May 16, 2020 at 10:55 am

Learned so much from this post!! This was such a clear example that it is the first time for me that some statistic tests really make sense to me. Thank you so much for sharing your knowledge, Jim!!

May 5, 2020 at 11:46 am

the information that you have given here has been so useful to me – really understand it much better now. So, thank you very much! Just a quick question, how did you graph the contribution to chi-square statistics? Only, I’ve been using stata to do some data analysis and I’m not sure how it is that I would be able to create a graph like that for my own data. Any insight into that, that you can give would be extremely useful.

May 6, 2020 at 1:30 am

I used Minitab statistical software for the graphs. I think graphs often bring the data to life more than just a table of numbers.

March 20, 2020 at 2:38 pm

I have the results of two Exit Satisfaction Surveys related to two cohorts (graduates of 2017-18 and graduates of 2018-19). The information I received was just the “number” of ratings on each of the 5 points on the Likert Scale (e.g., 122 respondents Strongly Agreed to a given item). I changed the raw ratings into percentages for comparison, e.g., for Part A of the Survey (Proficiency and Knowledge in my major field), I calculated the minimum and maximum percentages on the Strongly Agree point and did the same for other points on the scale. My questions are (1) can I report the range of percentages on each point on the scale for each item or is it better to report an overall agreement/disagreement? and (2) what’s the best statistics to compare the satisfaction of the two cohorts in the same survey? The 2017-18 cohorts included 126, and the 2018-19 cohort included 296 graduates.

I checked out your Introduction to Statistics book that I purchased, but I couldn’t decide about the appropriate statistics for the analysis of each of the surveys as well as comparison of both cohorts.

My sincere thanks in advance for your time and advice,

All the best, Ellie

March 20, 2020 at 7:30 am

Thank you for an excellent post! I am myself will soon perform a Chi-square test of independence on survey responses with two variables, and now think it might be good to start with a 2 proportion test (is a Z-test with 2 proportions what you use in this example?). Since you don’t discuss whether the Star Trek data meets the assumptions of the two tests you use, I wonder if they share approximately the same assumptions? I have already made certain that my data may be used with the Chi-square (my data is by the way not necessarily normally distributed, and has unkown mean and variance), can I therefore be comfortable with using a 2 proportions Z-test too? I hope you have the time to help me out here!

February 18, 2020 at 8:53 am

Excellent post. Btw, is it similar to what they called Test of Association that uses contingency table? The way they compute for the expected value is (row total × column total)/(sample total) . And to check if there is a relationship between two variable, check if the calculate chi-squared value is greater that the critical value of the chi-squared. Is it just the same?

February 20, 2020 at 11:09 am

Hi Hephzibah,

Yes, they’re the same test–test of independence and test of association. I’ll add something to that effect to the article to make that more clear.

January 6, 2020 at 9:24 am

Jim, thanks for creating and publishing this great content. In the initial chi-square test for independence we determined that shirt color does have a relationship with death rate. The Pearson ch-square measurement is 6.189, is this number meaningful? How do we interpret this in plain english?

January 6, 2020 at 3:09 pm

There’s really no direct interpretation of the chi-square value. That’s the test statistic, similar to the t-value in t-tests and the F-value in F-tests. These values are placed in the chi-square probability distribution that has the specified degrees of freedom (df=2 for this example). By placing the value into the probability distribution, the procedure can calculate probabilities, such as the p-value. I’ve been meaning to write a post that shows how this works for chi-squared tests. I show how this works for t-tests and F-tests for one-way ANOVA . Read those to get an idea of the process. Of course, for this chi-squared test uses chi-squared as the test statistic and probability distribution.

I’ll write a post soon about how this test works, both in terms of calculating the chi-square value itself and then using it in the probability distribution.

January 5, 2020 at 7:28 am

Would Chi-squared test be the statistical test of choice, for comparing the incidence rates of disease X between two states? Many thanks.

January 6, 2020 at 1:20 am

Hi Michaela,

It sounds like you’d need to use a two-sample proportions test. I show an example of this test using real data in my post about the effective of flu vaccinations . The reason you’d need to use a proportions test is because your observed data are presumably binary (diseased/not diseased).

You could use the chi-squared test, but I think for your case the results are easier to understand using a two-sample proportions test.

June 3, 2019 at 6:57 pm

Lets say the expected salary for a position is 20,000 dollars. In our observed salary we have various figures a little above and below 20,000 and we want to do a hypothesis test. These salaries are ratio, so does that mean we cannot use Chi Square? Do we have to convert? How? In fact, when you run a chi square on the salary data Chi Square turns out to be very high, sort of off the Chi Square Critical Value chart.

June 3, 2019 at 10:28 pm

Chi-square analysis requires two or more categorical (nominal) variables. Salary is a continuous (ratio) variable. Consequently, you can’t use chi-square.

If you have the one continuous variable of salary and you want to determine whether the difference between the mean salary and $20,000 is statistically significant or not, you’d need to use a one-sample t-test. My post about the different forms of t-tests should be helpful for you.

April 13, 2019 at 4:23 am

I don’t know how to thank you for your detailed informative reply. And I am happy that a specialist like you found this study interesting yoohoo 🙂

As to your comment on how we (me and my graduate student whose thesis I am directing) tracked the errors from Sample writing 1 to 5 for each participant, We did it manually through a close content analysis. I had no idea of a better alternative since going through 25 pieces of writing samples needed meticulous comparison for each participant. I advised my student to tabulate the number, frequency, and type of errors for each participant separately so we could keep track of their (lack of) improvement depending on the participant’s proficiency level.

Do you have any suggestion to make it more rigorous?

Very many thanks, Ellie

April 10, 2019 at 11:52 am

Hi, Jim. I first decided to choose chi-square to analyze my data but now I am thinking of poisson regression since my dependent variable is ‘count.’. I want to see if there is any significant difference between Grade 10 students’ perceptions of their writing problems and the frequency of their writing errors in the five paragraphs they wrote. Here is the detailed situation:

1. Five sample paragraphs were collected from 5 students at 5 proficiency levels based on their total marks in English final exam in the previous semester (from Outstanding to Poor). 2. The students participated in an interview and expressed their perceptions of their problem areas in writing. 3. The students submitted their paragraphs every 2 weeks during the semester. 4. The paragraphs were marked based on the school’s marking rubrics. 5. Errors were categorized under five components (e.g., grammar, word choice, etc.). 6. Paragraphs were compared for measuring the students’ improvement by counting errors manually in each and every paragraph. 7. The students’ errors were also compared to their perceived problem areas to study the extent of their awareness of their writing problems. This comparison showed that students were not aware of a major part of their errors while their perceived errors were not necessarily observed in their writing samples. 8. Comparison of Paragraphs 1 and 5 for each student showed decrease in the number of errors in some language components while some errors still persisted. 9. I’m also interested to see if proficiency level has any impact on students’ perceptions of their real problem areas and the frequency of their errors in each language category.

My question is which test should be used to answer Qs 7 and 8? As to Q9, one of the dependent variables is count and the other one is nominal. One correlation I’m thinking is eta squared (interval-nominal) but for the proficiency-frequency I’m not sure.

My sincere apologies for this long query and many thanks for any clues to the right stats.

April 11, 2019 at 12:25 am

That sounds like a very interesting study!

I think that you’re correct to use some form of regression rather than chi-square. The chi-squared test of independence doesn’t work with counts within an observation. Chi-squared looks at the multiple characteristics of an observations and essentially places in a basket for that combination. For example, you have a red shirt/dead basket and a red-shirt/alive basket. The procedure looks at each observation and places it into one of the baskets. Then it counts the observations in each basket.

What you have are counts (of errors) within each observation. You want to understand that IVs that relate to those counts. That’s a regression thing. Now, what form of regression. Because it involves counts, Poisson regression is a good possibility. You might also read up on negative binomial regression, which is related. Sometimes you can have count data that doesn’t meet certain requirements of the Poisson distribution, but you can use Negative Binomial regression. For more information, look on page 321-322 of my ebook that you just bought! 🙂 I talk a bit about regression with counts.

And, there’s a chance that you might be able to use OLS regression. That depends on how you’re handling the multiple assessments and the average number of errors. The Poisson distribution begins to approximate the normal distribution at around a mean of 25-ish. If the number of errors tend to fall around here or higher, OLS might be the ticket! If you’re summing multiple observations together, that might help in this regard.

I don’t understand the design of how you’re tracking changing the number of errors over time, and how you’ll model that. You might included lagged values of errors to explain current errors, along with other possible IVs.

I found point number 7 to be really interesting. Is it that the blind spot allows the error to persist in greater numbers and that awareness of errors had reduced numbers of those types? Your interpretation of that should be very interesting!

Oh, and for the nominal dependent variable, use nominal logistic regression (p. 319-320)!

I hope this helps!

March 27, 2019 at 11:53 am

Thanks for your clear posts, Could you please give some insight like in T test and F test, how can we calculate a chi- square test statistic value and how to convert to p value?

March 29, 2019 at 12:26 am

I have that exact topic in mind for a future blog post! I’ll write one up similar to the t-test and F-test posts in the near future. It’s too much to do in the comments section, but soon an entire post for it! I’ll aim for sometime in the next couple of months. Stay tuned!

November 16, 2018 at 1:47 pm

This was great. 🙂

September 21, 2018 at 10:47 am

thanks i have learnt alot

February 5, 2018 at 4:26 pm

Hello, Thanks for the nice tutorial. Can you please explain how the ‘Expected count’ is being calculated in the table “tabulated statistics: Uniform color, Status” ?

February 5, 2018 at 10:25 pm

Hi Shihab, that’s an excellent question!

You calculate the expected value for each cell by first multiplying the column proportion by the row proportion that are associated with each cell. This calculation produces the expected proportion for that cell. Then, you take the expected proportion and multiply it by the total number of observations to obtain the expected count. Let’s work through an example!

I’ll calculate the expected value for wearing a Blue uniform and being Alive. That’s the top-left cell in the statistical output.

At the bottom of the Alive column, we see that 90.7% of all observations are alive. So, 0.907 is the proportion for the Alive column. The output doesn’t display the proportion for the Blue row, but we can calculate that easily. We can see that there are 136 total counts in the Blue row and there are 430 total crew members. Hence, the proportion for the Blue row is 136/430 = 0.31627.

Next, we multiply 0.907 * 0.31627 = 0.28685689. That’s the expected proportion that should fall in that Blue/Alive cell.

Now, we multiply that proportion by the total number of observations to obtain the expected count for that cell: 0.28685689 * 430 = 123.348

You can see in the statistical output that has been rounded up to 123.35.

You simply repeat that procedure for the rest of the cells.

January 18, 2018 at 2:29 pm

very nice, thanks

January 1, 2018 at 8:51 am

Amazing post!! In the tabulated statistics section, you ran a Pearson Chi Square and a Likelihood Ratio Chi Square test. Are both of these necessary and do BOTH have to fall below the significance level for the null to be rejected? I’m assuming so. I don’t know what the difference is between these two tests but I will look it up. That was the only part that lost me:)

January 2, 2018 at 11:16 am

Thanks again, Jessica! I really appreciate your kind words!

When the two p-values are in agreement (e.g., both significant or insignificant), that’s easy. Fortunately, in my experience, these two p-values usually do agree. And, as the sample size increases, the agreement between them also increases.

I’ve looked into what to do when they disagree and have not found any clear answers. This paper suggests that as long as all expected frequencies are at least 5, use the Pearson Chi-Square test. When it is less than 5, the article recommends an adjusted Chi-square test, which is neither of the displayed tests!

These tests are most likely to disagree when you have borderline results to begin with (near your significance level), and particularly when you have a small sample. Either of these conditions alone make the results questionable. If these tests disagree, I’d take it as a big warning sign that more research is required!

December 8, 2017 at 6:58 am

December 8, 2017 at 11:10 am

December 7, 2017 at 8:18 am

A good presentation. My experience with researchers in health sciences and clinical studies is that very often people do not bother about the hypotheses (null and alternate) but run after a p-value, more so with Chi-Square test of independence!! Your narration is excellent.

December 7, 2017 at 4:08 am

Helpful post. I can understand now

December 6, 2017 at 9:47 pm

Excellent Example, Thank you.

December 6, 2017 at 11:24 pm

You’re very welcome. I’m glad it was helpful!

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Mastering the Chi-Square Test: A Comprehensive Guide

The Chi-Square Test is a statistical method used to determine if there’s a significant association between two categorical variables in a sample data set. It checks the independence of these variables, making it a robust and flexible tool for data analysis.

Introduction to Chi-Square Test

The  Chi-Square Test  of Independence is an important tool in the statistician’s arsenal. Its primary function is determining whether a significant association exists between two categorical variables in a sample data set. Essentially, it’s a test of independence, gauging if variations in one variable can impact another.

This comprehensive guide gives you a deeper understanding of the Chi-Square Test, its mechanics, importance, and correct implementation.

• Chi-Square Test assess the association between two categorical variables.
• Chi-Square Test requires the data to be a random sample.
• Chi-Square Test is designed for categorical or nominal variables.
• Each observation in the Chi-Square Test must be mutually exclusive and exhaustive.
• Chi-Square Test can’t establish causality, only an association between variables.

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Case Study: Chi-Square Test in Real-World Scenario

Let’s delve into a real-world scenario to illustrate the application of the  Chi-Square Test . Picture this: you’re the lead data analyst for a burgeoning shoe company. The company has an array of products but wants to enhance its marketing strategy by understanding if there’s an association between gender (Male, Female) and product preference (Sneakers, Loafers).

To start, you collect data from a random sample of customers, using a survey to identify their gender and their preferred shoe type. This data then gets organized into a contingency table , with gender across the top and shoe type down the side.

Next, you apply the Chi-Square Test to this data. The null hypothesis (H0) is that gender and shoe preference are independent. In contrast, the alternative hypothesis (H1) proposes that these variables are associated. After calculating the expected frequencies and the Chi-Square statistic, you compare this statistic with the critical value from the Chi-Square distribution.

Suppose the Chi-Square statistic is higher than the critical value in our scenario, leading to the rejection of the null hypothesis. This result indicates a significant association between gender and shoe preference. With this insight, the shoe company has valuable information for targeted marketing campaigns.

For instance, if the data shows that females prefer sneakers over loafers, the company might emphasize its sneaker line in marketing materials directed toward women. Conversely, if men show a higher preference for loafers, the company can highlight these products in campaigns targeting men.

This case study exemplifies the power of the Chi-Square Test. It’s a simple and effective tool that can drive strategic decisions in various real-world contexts, from marketing to medical research.

The Mathematics Behind Chi-Square Test

At the heart of the  Chi-Square Test  lies the calculation of the discrepancy between observed data and the expected data under the assumption of variable independence. This discrepancy termed the Chi-Square statistic, is calculated as the sum of squared differences between observed (O) and expected (E) frequencies, normalized by the expected frequencies in each category.

In mathematical terms, the Chi-Square statistic (χ²) can be represented as follows: χ² = Σ [ (Oᵢ – Eᵢ)² / Eᵢ ] , where the summation (Σ) is carried over all categories.

This formula quantifies the discrepancy between our observations and what we would expect if the null hypothesis of independence were true. We can decide on the variables’ independence by comparing the calculated Chi-Square statistic to a critical value from the Chi-Square distribution. Suppose the computed χ² is greater than the critical value. In that case, we reject the null hypothesis, indicating a significant association between the variables.

Step-by-Step Guide to Perform Chi-Square Test

To effectively execute a  Chi-Square Test , follow these methodical steps:

State the Hypotheses:  The null hypothesis (H0) posits no association between the variables — i.e., independent — while the alternative hypothesis (H1) posits an association between the variables.

Construct a Contingency Table:  Create a matrix to present your observations, with one variable defining the rows and the other defining the columns. Each table cell shows the frequency of observations corresponding to a particular combination of variable categories.

Calculate the Expected Values:  For each cell in the contingency table, calculate the expected frequency assuming that H0 is true. This can be calculated by multiplying the sum of the row and column for that cell and dividing by the total number of observations.

Compute the Chi-Square Statistic:  Apply the formula χ² = Σ [ (Oᵢ – Eᵢ)² / Eᵢ ] to compute the Chi-Square statistic.

Compare Your Test Statistic:  Evaluate your test statistic against a Chi-Square distribution to find the p-value, which will indicate the statistical significance of your test. If the p-value is less than your chosen significance level (usually 0.05), you reject H0.

Interpretation of the results should always be in the context of your research question and hypothesis. This includes considering practical significance — not just statistical significance — and ensuring your findings align with the broader theoretical understanding of the topic.

Assumptions, Limitations, and Misconceptions

The  Chi-Square Test , a vital tool in statistical analysis, comes with certain assumptions and distinct limitations. Firstly, it presumes that the data used are a  random sample  from a larger population and that the variables under investigation are nominal or categorical. Each observation must fall into one unique category or cell in the analysis, meaning observations are mutually  exclusive  and  exhaustive .

The Chi-Square Test has limitations when deployed with small sample sizes. The  expected frequency  of any cell in the contingency table should ideally be 5 or more. If it falls short, this can cause distortions in the test findings, potentially triggering a Type I or Type II error.

Misuse and misconceptions about this test often center on its application and interpretability. A standard error is using it for continuous or ordinal data without appropriate  categorization , leading to misleading results. Also, a significant result from a Chi-Square Test indicates an association between variables, but it doesn’t infer  causality . This is a frequent misconception — interpreting the association as proof of causality — while the test doesn’t offer information about whether changes in one variable cause changes in another.

Moreover, more than a significant Chi-Square test is required to comprehensively understand the relationship between variables. To get a more nuanced interpretation, it’s crucial to accompany the test with a measure of  effect size , such as Cramer’s V or Phi coefficient for a 2×2 contingency table. These measures provide information about the strength of the association, adding another dimension to the interpretation of results. This is essential as statistically significant results do not necessarily imply a practically significant effect. An effect size measure is critical in large sample sizes where even minor deviations from independence might result in a significant Chi-Square test.

Conclusion and Further Reading

Mastering the  Chi-Square Test  is vital in any data analyst’s or statistician’s journey. Its wide range of applications and robustness make it a tool you’ll turn to repeatedly.

For further learning, statistical textbooks and online courses can provide more in-depth knowledge and practice. Don’t hesitate to delve deeper and keep exploring the fascinating world of data analysis.

• Effect Size for Chi-Square Tests
• Assumptions for the Chi-Square Test
• Assumptions for Chi-Square Test (Story)
• Chi Square Test – an overview (External Link)
• Understanding the Null Hypothesis in Chi-Square
• What is the Difference Between the T-Test vs. Chi-Square Test?
• How to Report Chi-Square Test Results in APA Style: A Step-By-Step Guide

Frequently Asked Questions (FAQ)

It’s a statistical test used to determine if there’s a significant association between two categorical variables.

The test is suitable for categorical or nominal variables.

No, the test can only indicate an association, not a causal relationship.

The test assumes that the data is a random sample and that observations are mutually exclusive and exhaustive.

It measures the discrepancy between observed and expected data, calculated by χ² = Σ [ (Oᵢ – Eᵢ)² / Eᵢ ].

The result is generally considered statistically significant if the p-value is less than 0.05.

Misuse can lead to misleading results, making it crucial to use it with categorical data only.

Small sample sizes can lead to wrong results, especially when expected cell frequencies are less than 5.

Low expected cell frequencies can lead to Type I or Type II errors.

Results should be interpreted in context, considering the statistical significance and the broader understanding of the topic.

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The Chi-Square Test

What is a chi-square test.

A Chi-square test is a hypothesis testing method. Two common Chi-square tests involve checking if observed frequencies in one or more categories match expected frequencies.

Is a Chi-square test the same as a χ² test?

Yes, χ is the Greek symbol Chi.

What are my choices?

If you have a single measurement variable, you use a Chi-square goodness of fit test . If you have two measurement variables, you use a Chi-square test of independence . There are other Chi-square tests, but these two are the most common.

Types of Chi-square tests

You use a Chi-square test for hypothesis tests about whether your data is as expected. The basic idea behind the test is to compare the observed values in your data to the expected values that you would see if the null hypothesis is true.

There are two commonly used Chi-square tests: the Chi-square goodness of fit test and the Chi-square test of independence . Both tests involve variables that divide your data into categories. As a result, people can be confused about which test to use. The table below compares the two tests.

Visit the individual pages for each type of Chi-square test to see examples along with details on assumptions and calculations.

Table 1: Choosing a Chi-square test

How to perform a Chi-square test

For both the Chi-square goodness of fit test and the Chi-square test of independence , you perform the same analysis steps, listed below. Visit the pages for each type of test to see these steps in action.

• Define your null and alternative hypotheses before collecting your data.
• Decide on the alpha value. This involves deciding the risk you are willing to take of drawing the wrong conclusion. For example, suppose you set α=0.05 when testing for independence. Here, you have decided on a 5% risk of concluding the two variables are independent when in reality they are not.
• Check the data for errors.
• Check the assumptions for the test. (Visit the pages for each test type for more detail on assumptions.)
• Perform the test and draw your conclusion.

Both Chi-square tests in the table above involve calculating a test statistic. The basic idea behind the tests is that you compare the actual data values with what would be expected if the null hypothesis is true. The test statistic involves finding the squared difference between actual and expected data values, and dividing that difference by the expected data values. You do this for each data point and add up the values.

Then, you compare the test statistic to a theoretical value from the Chi-square distribution . The theoretical value depends on both the alpha value and the degrees of freedom for your data. Visit the pages for each test type for detailed examples.

8. The Chi squared tests

The χ²tests.

Table of Contents

The world is constantly curious about the Chi-Square test's application in machine learning and how it makes a difference. Feature selection is a critical topic in machine learning , as you will have multiple features in line and must choose the best ones to build the model. Examining the relationship between the elements, the chi-square test aids in solving feature selection problems. This tutorial will teach you about the chi-square test types, how to perform these tests, their properties, their application, and more. Let's start!

What Is a Chi-Square Test?

The Chi-Square test is a statistical procedure for determining the difference between observed and expected data. This test can also be used to decide whether it correlates to our data's categorical variables. It helps to determine whether a difference between two categorical variables is due to chance or a relationship between them.

A chi-square test or comparable nonparametric test is required to test a hypothesis regarding the distribution of a categorical variable. Categorical variables, which indicate categories such as animals or countries, can be nominal or ordinal. They cannot have a normal distribution since they only have a few particular values.

Chi-Square Test Formula

c = Degrees of freedom

O = Observed Value

E = Expected Value

The degrees of freedom in a  statistical calculation represent the number of variables that can vary. The degrees of freedom can be calculated to ensure that chi-square tests are statistically valid. These tests are frequently used to compare observed data with data expected to be obtained if a particular hypothesis were true.

The Observed values are those you gather yourselves.

The expected values are the anticipated frequencies, based on the null hypothesis. 

Fundamentals of Hypothesis Testing

Hypothesis testing is a technique for interpreting and drawing inferences about a population based on sample data. It aids in determining which sample data best support mutually exclusive population claims.

Null Hypothesis (H0) - The Null Hypothesis is the assumption that the event will not occur. A null hypothesis has no bearing on the study's outcome unless it is rejected.

H0 is the symbol for it, and it is pronounced H-naught.

Alternate Hypothesis(H1 or Ha) - The Alternate Hypothesis is the logical opposite of the null hypothesis. The acceptance of the alternative hypothesis follows the rejection of the null hypothesis. H1 is the symbol for it.

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Types of Chi-Square Tests

There are two main types of Chi-Square tests:

Independence 

• Goodness-of-Fit 

The Chi-Square Test of Independence is a derivable ( also known as inferential ) statistical test which examines whether the two sets of variables are likely to be related with each other or not. This test is used when we have counts of values for two nominal or categorical variables and is considered as non-parametric test. A relatively large sample size and independence of obseravations are the required criteria for conducting this test.

In a movie theatre, suppose we made a list of movie genres. Let us consider this as the first variable. The second variable is whether or not the people who came to watch those genres of movies have bought snacks at the theatre. Here the null hypothesis is that th genre of the film and whether people bought snacks or not are unrelatable. If this is true, the movie genres don’t impact snack sales. 

Goodness-Of-Fit

In statistical hypothesis testing, the Chi-Square Goodness-of-Fit test determines whether a variable is likely to come from a given distribution or not. We must have a set of data values and the idea of the distribution of this data. We can use this test when we have value counts for categorical variables. This test demonstrates a way of deciding if the data values have a “ good enough” fit for our idea or if it is a representative sample data of the entire population. 

Suppose we have bags of balls with five different colours in each bag. The given condition is that the bag should contain an equal number of balls of each colour. The idea we would like to test here is that the proportions of the five colours of balls in each bag must be exact. 

Chi-Square Test Examples

1. Chi-Square Test for Independence

Example: A researcher wants to determine if there is an association between gender (male/female) and preference for a new product (like/dislike). The test can assess whether preferences are independent of gender.

2. Chi-Square Test for Goodness of Fit

Example: A dice manufacturer wants to test if a six-sided die is fair. They roll the die 60 times and expect each face to appear 10 times. The test checks if the observed frequencies match the expected frequencies.

3. Chi-Square Test for Homogeneity

Example: A fast-food chain wants to see if the preference for a particular menu item is consistent across different cities. The test can compare the distribution of preferences in multiple cities to see if they are homogeneous.

4. Chi-Square Test for a Contingency Table

Example: A study investigates whether smoking status (smoker/non-smoker) is related to the presence of lung disease (yes/no). The test can evaluate the relationship between smoking and lung disease in the sample.

5. Chi-Square Test for Population Proportions

Example: A political analyst wants to see if voter preference (candidate A vs. candidate B) is the same across different age groups. The test can determine if the proportions of preferences differ significantly between age groups.

How to Perform a Chi-Square Test?

Let's say you want to know if gender has anything to do with political party preference. You poll 440 voters in a simple random sample to find out which political party they prefer. The results of the survey are shown in the table below:

To see if gender is linked to political party preference, perform a Chi-Square test of independence using the steps below.

Step 1: Define the Hypothesis

H0: There is no link between gender and political party preference.

H1: There is a link between gender and political party preference.

Step 2: Calculate the Expected Values

Now you will calculate the expected frequency.

For example, the expected value for Male Republicans is: 

Similarly, you can calculate the expected value for each of the cells.

Step 3: Calculate (O-E)2 / E for Each Cell in the Table

Now you will calculate the (O - E)2 / E for each cell in the table.

Step 4: Calculate the Test Statistic X2

X2  is the sum of all the values in the last table

 =  0.743 + 2.05 + 2.33 + 3.33 + 0.384 + 1

Before you can conclude, you must first determine the critical statistic, which requires determining our degrees of freedom. The degrees of freedom in this case are equal to the table's number of columns minus one multiplied by the table's number of rows minus one, or (r-1) (c-1). We have (3-1)(2-1) = 2.

Finally, you compare our obtained statistic to the critical statistic found in the chi-square table. As you can see, for an alpha level of 0.05 and two degrees of freedom, the critical statistic is 5.991, which is less than our obtained statistic of 9.83. You can reject our null hypothesis because the critical statistic is higher than your obtained statistic.

This means you have sufficient evidence to say that there is an association between gender and political party preference.

What Are Categorical Variables?

Categorical variables belong to a subset of variables that can be divided into discrete categories. Names or labels are the most common categories. These variables are also known as qualitative variables because they depict the variable's quality or characteristics.

Categorical variables can be divided into two categories:

1. Nominal Variable: A nominal variable's categories have no natural ordering. Example: Gender, Blood groups

2. Ordinal Variable: A variable that allows the categories to be sorted is an ordinal variable. An example is customer satisfaction (Excellent, Very Good, Good, Average, Bad, and so on).

Chi-Square Practice Problems

1. voting patterns.

A researcher wants to know if voting preferences (party A, party B, or party C) and gender (male, female) are related. Apply a chi-square test to the following set of data:

• Male: Party A - 30, Party B - 20, Party C - 50
• Female: Party A - 40, Party B - 30, Party C - 30

To determine if gender influences voting preferences, run a chi-square test of independence.

2. State of Health

In a sample population, a medical study examines the association between smoking status (smoker, non-smoker) and the occurrence of lung disease (yes, no). The information is as follows:

• Smoker: Yes - 90, No - 60
• Non-smoker: Yes - 30, No - 120 

To find out if smoking status is related to the incidence of lung disease, do a chi-square test.

3. Consumer Preferences

Customers are surveyed by a company to determine whether their age group (under 20, 20-40, over 40) and their preferred product category (food, apparel, or electronics) are related. The information gathered is:

• Under 20: Electronic - 50, Clothing - 30, Food - 20
• 20-40: Electronic - 60, Clothing - 70, Food - 50
• Over 40: Electronic - 30, Clothing - 40, Food - 80

Use a chi-square test to investigate the connection between product preference and age group

4. Academic Performance

An educational researcher looks at the relationship between students' success on standardized tests (pass, fail) and whether or not they participate in after-school programs. The information is as follows:

• Yes: Pass - 80, Fail - 20
• No: Pass - 50, Fail - 50

Use a chi-square test to determine if involvement in after-school programs and test scores are connected.

5. Genetic Inheritance

A geneticist investigates how a particular trait is inherited in plants and seeks to ascertain whether the expression of a trait (trait present, trait absent) and the existence of a genetic marker (marker present, marker absent) are significantly correlated. The information gathered is:

• Marker Present: Trait Present - 70, Trait Absent - 30
• Marker Absent: Trait Present - 40, Trait Absent - 60

Do a chi-square test to determine if there is a correlation between the trait's expression and the genetic marker.

How to Solve Chi-Square Problems?

1. state the hypotheses.

• Null hypothesis (H0): There is no association between the variables
• Alternative hypothesis (H1): There is an association between the variables.

2. Calculate the Expected Frequencies

• Use the formula: E=(Row Total×Column Total)Grand TotalE = \frac{(Row \ Total \times Column \ Total)}{Grand \ Total}E=Grand Total(Row Total×Column Total)​

3. Compute the Chi-Square Statistic

• Use the formula: χ2=∑(O−E)2E\chi^2 = \sum \frac{(O - E)^2}{E}χ2=∑E(O−E)2​, where O is the observed frequency and E is the expected frequency.

4. Determine the Degrees of Freedom (df)

• Use the formula: df=(number of rows−1)×(number of columns−1)df = (number \ of \ rows - 1) \times (number \ of \ columns - 1)df=(number of rows−1)×(number of columns−1)

5. Find the Critical Value and Compare

• Use the chi-square distribution table to find the critical value for the given df and significance level (usually 0.05).
• Compare the chi-square statistic to the critical value to decide whether to reject the null hypothesis.

These practice problems help you understand how chi-square analysis tests hypotheses and explores relationships between categorical variables in various fields.

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When to Use a Chi-Square Test?

A Chi-Square Test is used to examine whether the observed results are in order with the expected values. When the data to be analysed is from a random sample, and when the variable is the question is a categorical variable, then Chi-Square proves the most appropriate test for the same. A categorical variable consists of selections such as breeds of dogs, types of cars, genres of movies, educational attainment, male v/s female etc. Survey responses and questionnaires are the primary sources of these types of data. The Chi-square test is most commonly used for analysing this kind of data. This type of analysis is helpful for researchers who are studying survey response data. The research can range from customer and marketing research to political sciences and economics. 

Chi-Square Distribution 

Chi-square distributions (X2) are a type of continuous probability distribution. They're commonly utilized in hypothesis testing, such as the chi-square goodness of fit and independence tests. The parameter k, which represents the degrees of freedom, determines the shape of a chi-square distribution.

Very few real-world observations follow a chi-square distribution. Chi-square distributions aim to test hypotheses, not to describe real-world distributions. In contrast, other commonly used distributions, such as normal and Poisson distributions, may explain important things like birth weights or illness cases per year.

Chi-square distributions are excellent for hypothesis testing because of its close resemblance to the conventional normal distribution. Many essential statistical tests rely on the traditional normal distribution.

In statistical analysis , the Chi-Square distribution is used in many hypothesis tests and is determined by the parameter k degree of freedom. It belongs to the family of continuous probability distributions . The Sum of the squares of the k-independent standard random variables is called the Chi-Squared distribution. Pearson’s Chi-Square Test formula is - 

Where X^2 is the Chi-Square test symbol

Σ is the summation of observations

O is the observed results

E is the expected results 

The shape of the distribution graph changes with the increase in the value of k, i.e., the degree of freedom. 

When k is 1 or 2, the Chi-square distribution curve is shaped like a backwards ‘J’. It means there is a high chance that X^2 becomes close to zero. 

Courtesy: Scribbr

When k is greater than 2, the shape of the distribution curve looks like a hump and has a low probability that X^2 is very near to 0 or very far from 0. The distribution occurs much longer on the right-hand side and shorter on the left-hand side. The probable value of X^2 is (X^2 - 2).

When k is greater than ninety, a normal distribution is seen, approximating the Chi-square distribution.

What is the P-Value in a Chi-Square Test?

The P-Value in a Chi-Square test is a statistical measure that helps to assess the importance of your test results.

Here P denotes the probability; hence for the calculation of p-values, the Chi-Square test comes into the picture. The different p-values indicate different types of hypothesis interpretations. 

• P <= 0.05 (Hypothesis interpretations are rejected)
• P>= 0.05 (Hypothesis interpretations are accepted) 

The concepts of probability and statistics are entangled with Chi-Square Test. Probability is the estimation of something that is most likely to happen. Simply put, it is the possibility of an event or outcome of the sample. Probability can understandably represent bulky or complicated data. And statistics involves collecting and organising, analysing, interpreting and presenting the data. 

Finding P-Value

When you run all of the Chi-square tests, you'll get a test statistic called X2. You have two options for determining whether this test statistic is statistically significant at some alpha level:

• Compare the test statistic X2 to a critical value from the Chi-square distribution table.
• Compare the p-value of the test statistic X2 to a chosen alpha level.

Test statistics are calculated by taking into account the sampling distribution of the test statistic under the null hypothesis, the sample data, and the approach which is chosen for performing the test. 

The p-value will be as mentioned in the following cases.

• A lower-tailed test is specified by: P(TS ts | H0 is true) p-value = cdf (ts)
• Lower-tailed tests have the following definition: P(TS ts | H0 is true) p-value = cdf (ts)
• A two-sided test is defined as follows, if we assume that the test static distribution  of H0 is symmetric about 0. 2 * P(TS |ts| | H0 is true) = 2 * (1 - cdf(|ts|))

P: probability Event

TS: Test statistic is computed observed value of the test statistic from your sample cdf(): Cumulative distribution function of the test statistic's distribution (TS)

Tools and Software for Chi-Square Analysis

Here are some commonly used tools and software for performing Chi-Square analysis:

1. SPSS (Statistical Package for the Social Sciences) is a widely used software for statistical analysis, including Chi-Square tests. It provides an easy-to-use interface for performing Chi-Square tests for independence, goodness of fit, and other statistical analyses.

2. R is a powerful open-source programming language and software environment for statistical computing. The chisq.test() function in R allows for easy conducting of Chi-Square tests.

3. The SAS suite is used for advanced analytics, including Chi-Square tests. It is often used in research and business environments for complex data analysis.

4. Microsoft Excel offers a Chi-Square test function (CHISQ.TEST) for users who prefer working within spreadsheets. It’s a good option for basic Chi-Square analysis with smaller datasets.

5. Python (with libraries like SciPy or Pandas) offers robust tools for statistical analysis. The scipy.stats.chisquare() function can be used to perform Chi-Square tests.

Properties of Chi-Square Test 

• Variance is double the times the number of degrees of freedom.
• Mean distribution is equal to the number of degrees of freedom.
• When the degree of freedom increases, the Chi-Square distribution curve becomes normal.

Limitations of Chi-Square Test

There are two limitations to using the chi-square test that you should be aware of. 

• The chi-square test, for starters, is extremely sensitive to sample size. Even insignificant relationships can appear statistically significant when a large enough sample is used. Keep in mind that "statistically significant" does not always imply "meaningful" when using the chi-square test.
• Be mindful that the chi-square can only determine whether two variables are related. It does not necessarily follow that one variable has a causal relationship with the other. It would require a more detailed analysis to establish causality.

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Advanced Chi-Square Test Techniques

1. Chi-Square Test with Yates' Correction (Continuity Correction)

This technique is used in 2x2 contingency tables to reduce the Chi-Square value and correct for the overestimation of statistical significance when sample sizes are small. The correction is achieved by subtracting 0.5 from the absolute difference between each observed and expected frequency.

2. Mantel-Haenszel Chi-Square Test

This technique is used to assess the association between two variables while controlling for one or more confounding variables. It’s particularly useful in stratified analyses where the goal is to examine the relationship between variables across different strata (e.g., age groups, geographic locations).

3. Chi-Square Test for Trend (Cochran-Armitage Test)

This test is used when the categorical variable is ordinal, and you want to assess whether there is a linear trend in the proportions across the ordered groups. It’s commonly used in epidemiology to analyze trends in disease rates over time or across different exposure levels.

4. Monte Carlo Simulation for Chi-Square Test

When the sample size is very small or when expected frequencies are too low, the Chi-Square distribution may not provide accurate p-values. In such cases, Monte Carlo simulation can be used to generate an empirical distribution of the test statistic, providing a more accurate significance level.

5. Bayesian Chi-Square Test

In Bayesian statistics, the Chi-Square test can be adapted to incorporate prior knowledge or beliefs about the data. This approach is useful when existing information should influence the analysis, leading to potentially more accurate conclusions.

In this tutorial titled ‘The Complete Guide to Chi-square test’, you explored the concept of Chi-square distribution and how to find the related values. You also take a look at how the critical value and chi-square value is related to each other.

If you want to gain more insight, get a work-ready understanding of statistical concepts, and learn how to use them to get into a career in Data Analytics, our Post Graduate Program in Data Analytics in partnership with Purdue University should be your next stop. A comprehensive program with training from top practitioners and in collaboration with IBM will be all you need to kickstart your career in the field. Get started today!

1. What is the chi-square test used for? 

The chi-square test is a statistical method used to determine if there is a significant association between two categorical variables. It helps researchers understand whether the observed distribution of data differs from the expected distribution, allowing them to assess whether any relationship exists between the variables being studied.

2. What is the chi-square test and its types? 

The chi-square test is a statistical test used to analyze categorical data and assess the independence or association between variables. There are two main types of chi-square tests:

a) Chi-square test of independence: This test determines whether there is a significant association between two categorical variables. b) Chi-square goodness-of-fit test: This test compares the observed data to the expected data to assess how well the observed data fit the expected distribution.

3. What is the difference between t-test and chi-square? 

The t-test and the chi-square test are two different statistical tests used for various data types. The t-test compares the means of two groups and is suitable for continuous numerical data. On the other hand, the chi-square test examines the association between two categorical variables and is applicable to discrete categorical data.

4. What alternatives exist to the Chi-Square Test?

Alternatives include Fisher's Exact Test for small sample sizes, the G-test for large datasets, and logistic regression for modelling categorical outcomes.

5. What is the null hypothesis for Chi-Square?

The null hypothesis states no association between the categorical variables, meaning their distributions are independent.

6. How do I handle small sample sizes in a Chi-Square Test?

Use Fisher's Exact Test or apply Yates' continuity correction in 2x2 tables for small sample sizes to reduce the risk of inaccurate results.

7. What is the appropriate way to analyze Chi-Square Test results?

Compare the calculated Chi-Square statistic with the critical value from the Chi-Square distribution table; if it's more significant, reject the null hypothesis.

8. What is the advantage of the Chi-Square Test?

The Chi-Square test is simple to calculate and applies to categorical data, making it versatile for analyzing relationships in contingency tables.

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About the Author

Avijeet is a Senior Research Analyst at Simplilearn. Passionate about Data Analytics, Machine Learning, and Deep Learning, Avijeet is also interested in politics, cricket, and football.

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• Chi-Square Goodness of Fit Test | Formula, Guide & Examples

Chi-Square Goodness of Fit Test | Formula, Guide & Examples

Published on May 24, 2022 by Shaun Turney . Revised on June 22, 2023.

A chi-square (Χ 2 ) goodness of fit test is a type of Pearson’s chi-square test . You can use it to test whether the observed distribution of a categorical variable differs from your expectations.

You recruit a random sample of 75 dogs and offer each dog a choice between the three flavors by placing bowls in front of them. You expect that the flavors will be equally popular among the dogs, with about 25 dogs choosing each flavor.

The chi-square goodness of fit test tells you how well a statistical model fits a set of observations. It’s often used to analyze genetic crosses .

Table of contents

What is the chi-square goodness of fit test, chi-square goodness of fit test hypotheses, when to use the chi-square goodness of fit test, how to calculate the test statistic (formula), how to perform the chi-square goodness of fit test, when to use a different test, practice questions and examples, other interesting articles, frequently asked questions about the chi-square goodness of fit test.

A chi-square (Χ 2 ) goodness of fit test is a goodness of fit test for a categorical variable . Goodness of fit is a measure of how well a statistical model fits a set of observations.

• When goodness of fit is high , the values expected based on the model are close to the observed values.
• When goodness of fit is low , the values expected based on the model are far from the observed values.

The statistical models that are analyzed by chi-square goodness of fit tests are distributions . They can be any distribution, from as simple as equal probability for all groups, to as complex as a probability distribution with many parameters.

• Hypothesis testing

The chi-square goodness of fit test is a hypothesis test . It allows you to draw conclusions about the distribution of a population based on a sample. Using the chi-square goodness of fit test, you can test whether the goodness of fit is “good enough” to conclude that the population follows the distribution.

With the chi-square goodness of fit test, you can ask questions such as: Was this sample drawn from a population that has…

• Equal proportions of male and female turtles?
• Equal proportions of red, blue, yellow, green, and purple jelly beans?
• 90% right-handed and 10% left-handed people?
• Offspring with an equal probability of inheriting all possible genotypic combinations (i.e., unlinked genes)?
• A Poisson distribution of floods per year?
• A normal distribution of bread prices?

To help visualize the differences between your observed and expected frequencies, you also create a bar graph:

The president of the dog food company looks at your graph and declares that they should eliminate the Garlic Blast and Minty Munch flavors to focus on Blueberry Delight. “Not so fast!” you tell him.

You explain that your observations were a bit different from what you expected, but the differences aren’t dramatic. They could be the result of a real flavor preference or they could be due to chance.

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Like all hypothesis tests, a chi-square goodness of fit test evaluates two hypotheses: the null and alternative hypotheses. They’re two competing answers to the question “Was the sample drawn from a population that follows the specified distribution?”

• Null hypothesis ( H 0 ): The population follows the specified distribution.
• Alternative hypothesis ( H a ):   The population does not follow the specified distribution.

These are general hypotheses that apply to all chi-square goodness of fit tests. You should make your hypotheses more specific by describing the “specified distribution.” You can name the probability distribution (e.g., Poisson distribution) or give the expected proportions of each group.

• Null hypothesis ( H 0 ): The dog population chooses the three flavors in equal proportions ( p 1 = p 2 = p 3 ).
• Alternative hypothesis ( H a ): The dog population does not choose the three flavors in equal proportions.

The following conditions are necessary if you want to perform a chi-square goodness of fit test:

• You want to test a hypothesis about the distribution of one categorical variable . If your variable is continuous , you can convert it to a categorical variable by separating the observations into intervals. This process is known as data binning.
• The sample was randomly selected from the population .
• There are a minimum of five observations expected in each group.
• You want to test a hypothesis about the distribution of one categorical variable. The categorical variable is the dog food flavors.
• You recruited a random sample of 75 dogs.
• There were a minimum of five observations expected in each group. For all three dog food flavors, you expected 25 observations of dogs choosing the flavor.

The test statistic for the chi-square (Χ 2 ) goodness of fit test is Pearson’s chi-square:

The larger the difference between the observations and the expectations ( O − E in the equation), the bigger the chi-square will be.

To use the formula, follow these five steps:

Step 1: Create a table

Create a table with the observed and expected frequencies in two columns.

Step 2: Calculate O − E

Add a new column called “ O −  E ”. Subtract the expected frequencies from the observed frequency.

Step 3: Calculate ( O − E ) 2

Add a new column called “( O −  E ) 2 ”. Square the values in the previous column.

Step 4: Calculate ( O − E ) 2 / E

Add a final column called “( O − E )² /  E “. Divide the previous column by the expected frequencies.

Step 5: Calculate Χ 2

Add up the values of the previous column. This is the chi-square test statistic (Χ 2 ).

The chi-square statistic is a measure of goodness of fit, but on its own it doesn’t tell you much. For example, is Χ 2 = 1.52 a low or high goodness of fit?

To interpret the chi-square goodness of fit, you need to compare it to something. That’s what a chi-square test is: comparing the chi-square value to the appropriate chi-square distribution to decide whether to reject the null hypothesis .

To perform a chi-square goodness of fit test, follow these five steps (the first two steps have already been completed for the dog food example):

Step 1: Calculate the expected frequencies

Sometimes, calculating the expected frequencies is the most difficult step. Think carefully about which expected values are most appropriate for your null hypothesis .

In general, you’ll need to multiply each group’s expected proportion by the total number of observations to get the expected frequencies.

Step 2: Calculate chi-square

Calculate the chi-square value from your observed and expected frequencies using the chi-square formula.

Step 3: Find the critical chi-square value

Find the critical chi-square value in a chi-square critical value table or using statistical software. The critical value is calculated from a chi-square distribution. To find the critical chi-square value, you’ll need to know two things:

• The degrees of freedom ( df ): For chi-square goodness of fit tests, the df is the number of groups minus one.
• Significance level (α): By convention, the significance level is usually .05.

Step 4: Compare the chi-square value to the critical value

Compare the chi-square value to the critical value to determine which is larger.

Critical value = 5.99

Step 5: Decide whether the reject the null hypothesis

• The data allows you to reject the null hypothesis and provides support for the alternative hypothesis.
• The data doesn’t allow you to reject the null hypothesis and doesn’t provide support for the alternative hypothesis.

Whether you use the chi-square goodness of fit test or a related test depends on what hypothesis you want to test and what type of variable you have.

When to use the chi-square test of independence

There’s another type of chi-square test, called the chi-square test of independence .

• Use the chi-square goodness of fit test when you have one categorical variable and you want to test a hypothesis about its distribution .
• Use the chi-square test of independence when you have two categorical variables and you want to test a hypothesis about their relationship .

When to use a different goodness of fit test

The Anderson–Darling and Kolmogorov–Smirnov goodness of fit tests are two other common goodness of fit tests for distributions.

• Use the Anderson–Darling or the Kolmogorov–Smirnov goodness of fit test when you have a continuous variable (that you don’t want to bin).
• Use the chi-square goodness of fit test when you have a categorical variable (or a continuous variable that you want to bin).

Do you want to test your knowledge about the chi-square goodness of fit test? Download our practice questions and examples with the buttons below.

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If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Chi square test of independence
• Statistical power
• Descriptive statistics
• Degrees of freedom
• Pearson correlation
• Null hypothesis

Methodology

• Double-blind study
• Case-control study
• Research ethics
• Data collection
• Structured interviews

Research bias

• Hawthorne effect
• Unconscious bias
• Recall bias
• Halo effect
• Self-serving bias
• Information bias

You can use the CHISQ.TEST() function to perform a chi-square goodness of fit test in Excel. It takes two arguments, CHISQ.TEST(observed_range, expected_range), and returns the p value .

You can use the chisq.test() function to perform a chi-square goodness of fit test in R. Give the observed values in the “x” argument, give the expected values in the “p” argument, and set “rescale.p” to true. For example:

chisq.test(x = c(22,30,23), p = c(25,25,25), rescale.p = TRUE)

Chi-square goodness of fit tests are often used in genetics. One common application is to check if two genes are linked (i.e., if the assortment is independent). When genes are linked, the allele inherited for one gene affects the allele inherited for another gene.

Suppose that you want to know if the genes for pea texture (R = round, r = wrinkled) and color (Y = yellow, y = green) are linked. You perform a dihybrid cross between two heterozygous ( RY / ry ) pea plants. The hypotheses you’re testing with your experiment are:

• This would suggest that the genes are unlinked.
• This would suggest that the genes are linked.

You observe 100 peas:

• 78 round and yellow peas
• 6 round and green peas
• 4 wrinkled and yellow peas
• 12 wrinkled and green peas

To calculate the expected values, you can make a Punnett square. If the two genes are unlinked, the probability of each genotypic combination is equal.

The expected phenotypic ratios are therefore 9 round and yellow: 3 round and green: 3 wrinkled and yellow: 1 wrinkled and green.

From this, you can calculate the expected phenotypic frequencies for 100 peas:

Χ 2 = 8.41 + 8.67 + 11.6 + 5.4 = 34.08

Since there are four groups (round and yellow, round and green, wrinkled and yellow, wrinkled and green), there are three degrees of freedom .

For a test of significance at α = .05 and df = 3, the Χ 2 critical value is 7.82.

Χ 2 = 34.08

Critical value = 7.82

The Χ 2 value is greater than the critical value .

The Χ 2 value is greater than the critical value, so we reject the null hypothesis that the population of offspring have an equal probability of inheriting all possible genotypic combinations. There is a significant difference between the observed and expected genotypic frequencies ( p < .05).

The data supports the alternative hypothesis that the offspring do not have an equal probability of inheriting all possible genotypic combinations, which suggests that the genes are linked

The two main chi-square tests are the chi-square goodness of fit test and the chi-square test of independence .

A chi-square distribution is a continuous probability distribution . The shape of a chi-square distribution depends on its degrees of freedom , k . The mean of a chi-square distribution is equal to its degrees of freedom ( k ) and the variance is 2 k . The range is 0 to ∞.

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Chi-Square test

The Chi-square test is a hypothesis test used to determine whether there is a relationship between two categorical variables .

What are categorical variables? Categorical variables are, for example, a person's gender, preferred newspaper, frequency of television viewing, or their highest level of education. So whenever you want to test whether there is a relationship between two categorical variables, you use a Chi 2 test.

Definition:

The chi-square test is a hypothesis test used for categorical variables with nominal or ordinal measurement scale . The chi-square test checks whether the frequencies occurring in the sample differ significantly from the frequencies one would expect. Thus, the observed frequencies are compared with the expected frequencies and their deviations are examined.

Let's say we want to investigate whether there is a connection between gender and the highest level of education. To do this, we create a questionnaire in which the participants tick their gender and what their highest educational level is. The result of the survey is then displayed in a contingency table.

The Chi-square test is used to investigate whether there is a relationship between gender and the highest level of education.

Null hypothesis and alternative hypothesis

The null hypothesis and the alternative hypothesis then result in:

Null hypothesis: there is no relationship between gender and highest educational attainment.

Alternative hypothesis: There is a relation between gender and the highest educational attainment.

Tip: On DATAtab you can calculate the Chi-square test online. Simply visit the Chi-Square Test Calculator .

Applications of the Chi-Square Test

There are various applications of the Chi-square test, it can be used to answer the following questions:

1) Independence test

Are two categorical variables independent of each other? For example, does gender have an impact on whether a person has a Netflix subscription or not?

2) Distribution test

Are the observed values of two categorical variables equal to the expected values? One question could be, is one of the three video streaming services Netflix, Amazon, and Disney subscribed to above average?

3) Homogeneity test

Are two or more samples from the same population? One question could be whether the subscription frequencies of the three video streaming services Netflix, Amazon and Disney differ in different age groups.

Calculate chi-squared

The chi-squared value is calculated via:

To clarify the calculation of the chi-squared value, we refer to the following case: for variables one and two with category A and B , an observation was made or a sample exists. Now we want to check whether the frequencies from the sample correspond to the expected frequencies from the population.

Observed frequency:

Expected frequency:

With the upper equation you can now calculate chi-squared :

After calculating chi-squared the number of degrees of freedom df is needed. This is given by

• p : number of lines
• q : number of columns

From the table of the chi-squared distribution one can now read the critical chi-squared value. For a significance level of 5 % and a df of 1, this results in 3.841. Since the calculated chi-squared value is smaller, there is no significant difference.

As a prerequisite for this test, please note that all expected frequencies must be greater than 5.

Chi-Square Test of Independence

The Chi-Square Test of Independence is used when two categorical variables are to be tested for independence. The aim is to analyze whether the characteristic values of the first variable are influenced by the characteristic values of the second variable and vice versa.

For example, does gender have an influence on whether a person has a Netflix subscription or not? For the two variables gender (male, female) and has Netflix subscription (yes, no), it is tested whether they are independent. If this is not the case, there is a relationship between the characteristics.

The research question that can be answered with the Chi-square test is: Are the characteristics of gender and ownership of a Netflix subscription independent of each other?

In order to calculate the chi-square, an observed and an expected frequency must be given. In the independence test, the expected frequency is the one that results when both variables are independent. If two variables are independent, the expected frequencies of the individual cells are obtained with

where i and j are the rows and columns of the table respectively.

For the fictitious Netflix example, the following tables could be used. On the left is the table with the frequencies observed in the sample, and on the right is the table that would result if perfect independence existed.

Expected frequency if independent:

The Chi-square is then calculated as

From the Chi-square table you can now read the critical value again and compare it with the result.

The assumptions for the Chi-square independence test are that the observations are from a random sample and that the expected frequencies per cell are greater than 5.

Chi-square distribution test

If a variable is present with two or more values, the differences in the frequency of the individual values can be examined.

The Chi-square distribution test , or Goodness-of-fit test , checks whether the frequencies of the individual characteristic values in the sample correspond to the frequencies of a defined distribution. In most cases, this defined distribution corresponds to that of the population. In this case, it is tested whether the sample comes from the respective population.

For market researchers it could be of interest whether there is a difference in the market penetration of the three video streaming services Netflix, Amazon and Disney between Berlin and the whole of Germany. The expected frequency is then the distribution of streaming services throughout Germany and the observed frequency results from a survey in Berlin. In the following tables the fictitious results are shown:

Observed frequency in Berlin:

Expected frequency (all Germany):

The Chi-square then results in

Chi-square homogeneity test

The Chi-square homogeneity test can be used to check whether two or more samples come from the same population? One question could be whether the subscription frequency of three video streaming services Netflix, Amazon and Disney differ in different age groups. As a fictitious example, a survey is made in three age groups with the following result:

As with the Chi-square independence test, this result is compared with the table that would result if the distributions of Streaming providers were independent of age.

Effect size in the Chi-square test

So far we only know whether we can reject the null hypothesis or not, but it is very often of great interest to know how strong the relationship between the two variables is. This can be answered with the help of the effect size.

In the Chi-square test, Cramér's V can be used to calculate the effect size. Here a value of 0.1 is small, a value of 0.3 is medium and a value of 0.5 is large. DATAtab will of course calculate the effect size for you very easily.

Effect size vs. p-value

Please note that the p-value does not tell you anything about the strength of the correlation or the effect and depends on the sample size! The following points should therefore be considered:

• If there is a correlation in the population, the larger the sample, the more clearly this will be shown in the p-value.
• If the sample is very large, very small correlations can also be detected in the population.
• These small correlations may no longer be relevant under certain circumstances.

Therefore, if there is a small sample and a large sample and there is an equally large effect in both samples, the p-values would still differ. The larger the sample, the smaller the p-value and therefore even very small correlations can be confirmed with a very large sample.

This is where the effect size plays an important role. With the effect size in the Chi-square test, differences can be made comparable across several studies.

Example chi-squared test

Independence test.

As an example of a chi-squared test where independence is tested, we consider the use of umbrellas. On a rainy day we counted how many women and how many men come to university with an umbrella.

Is the difference in the use of an umbrella for women and men statistically significant or random?

This is how it works in the online statistics calculator: After you have copied the above table into the hypothesis test calculator , you can calculate the chi-squared test. To do this, simply click on the two variables Gender and Umbrella . As a result, you will get the (1) contingency table, the (2) expected frequency for perfectly independent variables and the (3) chi-squared test

Expected frequencies for perfectly independent variables:

With a significance level of 5% and a degree of freedom of 1, the table of chi-squared values gives a critical value of 3.841. Since the calculated chi-squared value is smaller than the critical value, there is no significant difference in this example and the null hypothesis is not rejected. In terms of content, this means that men and women do not differ in the frequency of their umbrella use.

Distribution test

In one district of Vienna, the party membership of 22 persons was recorded. Now it is to be examined whether the residents of the district (random sample) have the same voting behaviour as the residents of the entire city of Vienna (population).

To calculate the chi-squared test for the example, simply copy the upper table into the Hypothesis Test Calculator .

Party A has a 40% share in Vienna and party C has 35%. You will therefore now receive the following results:

If the significance level is set at 0.05, the p-value calculated at 0.876 is greater than the significance level. Thus, the null hypothesis is not rejected and it can be assumed that the residents of the district have the same voting behavior as the residents of the entire city of Vienna.

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Step-by-step guide to hypothesis testing in statistics

Hypothesis testing in statistics helps us use data to make informed decisions. It starts with an assumption or guess about a group or population—something we believe might be true. We then collect sample data to check if there is enough evidence to support or reject that guess. This method is useful in many fields, like science, business, and healthcare, where decisions need to be based on facts.

Learning how to do hypothesis testing in statistics step-by-step can help you better understand data and make smarter choices, even when things are uncertain. This guide will take you through each step, from creating your hypothesis to making sense of the results, so you can see how it works in practical situations.

What is Hypothesis Testing?

Table of Contents

Hypothesis testing is a method for determining whether data supports a certain idea or assumption about a larger group. It starts by making a guess, like an average or a proportion, and then uses a small sample of data to see if that guess seems true or not.

For example, if a company wants to know if its new product is more popular than its old one, it can use hypothesis testing. They start with a statement like “The new product is not more popular than the old one” (this is the null hypothesis) and compare it with “The new product is more popular” (this is the alternative hypothesis). Then, they look at customer feedback to see if there’s enough evidence to reject the first statement and support the second one.

Simply put, hypothesis testing is a way to use data to help make decisions and understand what the data is really telling us, even when we don’t have all the answers.

Importance Of Hypothesis Testing In Decision-Making And Data Analysis

Hypothesis testing is important because it helps us make smart choices and understand data better. Here’s why it’s useful:

• Reduces Guesswork : It helps us see if our guesses or ideas are likely correct, even when we don’t have all the details.
• Uses Real Data : Instead of just guessing, it checks if our ideas match up with real data, which makes our decisions more reliable.
• Avoids Errors : It helps us avoid mistakes by carefully checking if our ideas are right so we don’t make costly errors.
• Shows What to Do Next : It tells us if our ideas work or not, helping us decide whether to keep, change, or drop something. For example, a company might test a new ad and decide what to do based on the results.
• Confirms Research Findings : It makes sure that research results are accurate and not just random chance so that we can trust the findings.

Here’s a simple guide to understanding hypothesis testing, with an example:

1. Set Up Your Hypotheses

Explanation: Start by defining two statements:

• Null Hypothesis (H0): This is the idea that there is no change or effect. It’s what you assume is true.
• Alternative Hypothesis (H1): This is what you want to test. It suggests there is a change or effect.

Example: Suppose a company says their new batteries last an average of 500 hours. To check this:

• Null Hypothesis (H0): The average battery life is 500 hours.
• Alternative Hypothesis (H1): The average battery life is not 500 hours.

2. Choose the Test

Explanation: Pick a statistical test that fits your data and your hypotheses. Different tests are used for various kinds of data.

Example: Since you’re comparing the average battery life, you use a one-sample t-test .

3. Set the Significance Level

Explanation: Decide how much risk you’re willing to take if you make a wrong decision. This is called the significance level, often set at 0.05 or 5%.

Example: You choose a significance level of 0.05, meaning you’re okay with a 5% chance of being wrong.

4. Gather and Analyze Data

Explanation: Collect your data and perform the test. Calculate the test statistic to see how far your sample result is from what you assumed.

Example: You test 30 batteries and find they last an average of 485 hours. You then calculate how this average compares to the claimed 500 hours using the t-test.

5. Find the p-Value

Explanation: The p-value tells you the probability of getting a result as extreme as yours if the null hypothesis is true.

Example: You find a p-value of 0.0001. This means there’s a very small chance (0.01%) of getting an average battery life of 485 hours or less if the true average is 500 hours.

6. Make Your Decision

Explanation: Compare the p-value to your significance level. If the p-value is smaller, you reject the null hypothesis. If it’s larger, you do not reject it.

Example: Since 0.0001 is much less than 0.05, you reject the null hypothesis. This means the data suggests the average battery life is different from 500 hours.

7. Report Your Findings

Explanation: Summarize what the results mean. State whether you rejected the null hypothesis and what that implies.

Example: You conclude that the average battery life is likely different from 500 hours. This suggests the company’s claim might not be accurate.

Hypothesis testing is a way to use data to check if your guesses or assumptions are likely true. By following these steps—setting up your hypotheses, choosing the right test, deciding on a significance level, analyzing your data, finding the p-value, making a decision, and reporting results—you can determine if your data supports or challenges your initial idea.

Understanding Hypothesis Testing: A Simple Explanation

Hypothesis testing is a way to use data to make decisions. Here’s a straightforward guide:

1. What is the Null and Alternative Hypotheses?

• Null Hypothesis (H0): This is your starting assumption. It says that nothing has changed or that there is no effect. It’s what you assume to be true until your data shows otherwise. Example: If a company says their batteries last 500 hours, the null hypothesis is: “The average battery life is 500 hours.” This means you think the claim is correct unless you find evidence to prove otherwise.
• Alternative Hypothesis (H1): This is what you want to find out. It suggests that there is an effect or a difference. It’s what you are testing to see if it might be true. Example: To test the company’s claim, you might say: “The average battery life is not 500 hours.” This means you think the average battery life might be different from what the company says.

2. One-Tailed vs. Two-Tailed Tests

• One-Tailed Test: This test checks for an effect in only one direction. You use it when you’re only interested in finding out if something is either more or less than a specific value. Example: If you think the battery lasts longer than 500 hours, you would use a one-tailed test to see if the battery life is significantly more than 500 hours.
• Two-Tailed Test: This test checks for an effect in both directions. Use this when you want to see if something is different from a specific value, whether it’s more or less. Example: If you want to see if the battery life is different from 500 hours, whether it’s more or less, you would use a two-tailed test. This checks for any significant difference, regardless of the direction.

3. Common Misunderstandings

• Clarification: Hypothesis testing doesn’t prove that the null hypothesis is true. It just helps you decide if you should reject it. If there isn’t enough evidence against it, you don’t reject it, but that doesn’t mean it’s definitely true.
• Clarification: A small p-value shows that your data is unlikely if the null hypothesis is true. It suggests that the alternative hypothesis might be right, but it doesn’t prove the null hypothesis is false.
• Clarification: The significance level (alpha) is a set threshold, like 0.05, that helps you decide how much risk you’re willing to take for making a wrong decision. It should be chosen carefully, not randomly.
• Clarification: Hypothesis testing helps you make decisions based on data, but it doesn’t guarantee your results are correct. The quality of your data and the right choice of test affect how reliable your results are.

Benefits and Limitations of Hypothesis Testing

• Clear Decisions: Hypothesis testing helps you make clear decisions based on data. It shows whether the evidence supports or goes against your initial idea.
• Objective Analysis: It relies on data rather than personal opinions, so your decisions are based on facts rather than feelings.
• Concrete Numbers: You get specific numbers, like p-values, to understand how strong the evidence is against your idea.
• Control Risk: You can set a risk level (alpha level) to manage the chance of making an error, which helps avoid incorrect conclusions.
• Widely Used: It can be used in many areas, from science and business to social studies and engineering, making it a versatile tool.

Limitations

• Sample Size Matters: The results can be affected by the size of the sample. Small samples might give unreliable results, while large samples might find differences that aren’t meaningful in real life.
• Risk of Misinterpretation: A small p-value means the results are unlikely if the null hypothesis is true, but it doesn’t show how important the effect is.
• Needs Assumptions: Hypothesis testing requires certain conditions, like data being normally distributed . If these aren’t met, the results might not be accurate.
• Simple Decisions: It often results in a basic yes or no decision without giving detailed information about the size or impact of the effect.
• Can Be Misused: Sometimes, people misuse hypothesis testing, tweaking data to get a desired result or focusing only on whether the result is statistically significant.
• No Absolute Proof: Hypothesis testing doesn’t prove that your hypothesis is true. It only helps you decide if there’s enough evidence to reject the null hypothesis, so the conclusions are based on likelihood, not certainty.

Final Thoughts 

Hypothesis testing helps you make decisions based on data. It involves setting up your initial idea, picking a significance level, doing the test, and looking at the results. By following these steps, you can make sure your conclusions are based on solid information, not just guesses.

This approach lets you see if the evidence supports or contradicts your initial idea, helping you make better decisions. But remember that hypothesis testing isn’t perfect. Things like sample size and assumptions can affect the results, so it’s important to be aware of these limitations.

In simple terms, using a step-by-step guide for hypothesis testing is a great way to better understand your data. Follow the steps carefully and keep in mind the method’s limits.

What is the difference between one-tailed and two-tailed tests?

 A one-tailed test assesses the probability of the observed data in one direction (either greater than or less than a certain value). In contrast, a two-tailed test looks at both directions (greater than and less than) to detect any significant deviation from the null hypothesis.

How do you choose the appropriate test for hypothesis testing?

The choice of test depends on the type of data you have and the hypotheses you are testing. Common tests include t-tests, chi-square tests, and ANOVA. You get more details about ANOVA, you may read Complete Details on What is ANOVA in Statistics ?  It’s important to match the test to the data characteristics and the research question.

What is the role of sample size in hypothesis testing?  

Sample size affects the reliability of hypothesis testing. Larger samples provide more reliable estimates and can detect smaller effects, while smaller samples may lead to less accurate results and reduced power.

Can hypothesis testing prove that a hypothesis is true?  

Hypothesis testing cannot prove that a hypothesis is true. It can only provide evidence to support or reject the null hypothesis. A result can indicate whether the data is consistent with the null hypothesis or not, but it does not prove the alternative hypothesis with certainty.

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• Math Article
• Chi Square Test

Chi-Square Test

A chi-squared test  (symbolically represented as  χ 2 ) is basically a data analysis on the basis of observations of a random set of variables. Usually, it is a comparison of two statistical data sets. This test was introduced by Karl Pearson in 1900 for categorical data analysis and distribution . So it was mentioned as Pearson’s chi-squared test .

The chi-square test is used to estimate how likely the observations that are made would be, by considering the assumption of the null hypothesis as true.

A hypothesis is a consideration that a given condition or statement might be true, which we can test afterwards. Chi-squared tests are usually created from a sum of squared falsities or errors over  the sample variance.

Chi-Square Distribution

When we consider, the null speculation is true, the sampling distribution of the test statistic is called as chi-squared distribution . The chi-squared test helps to determine whether there is a notable difference between the normal frequencies and the observed frequencies in one or more classes or categories. It gives the probability of independent variables.

Note: Chi-squared test is applicable only for categorical data, such as men and women falling under the categories of Gender, Age, Height, etc.

Finding P-Value

P stands for probability here. To calculate the p-value, the chi-square test is used in statistics. The different values of p indicates the different hypothesis interpretation, are given below:

• P≤ 0.05; Hypothesis rejected
• P>.05; Hypothesis Accepted

Probability is all about chance or risk or uncertainty. It is the possibility of the outcome of the sample or the occurrence of an event. But when we talk about statistics, it is more about how we handle various data using different techniques. It helps to represent complicated data or bulk data in a very easy and understandable way. It describes the collection, analysis, interpretation, presentation, and organization of data. The concept of both probability and statistics is related to the chi-squared test.

Also, read:

The following are the important properties of the chi-square test:

• Two times the number of degrees of freedom is equal to the variance.
• The number of degree of freedom is equal to the mean distribution
• The chi-square distribution curve approaches the normal distribution when the degree of freedom increases.

The chi-squared test is done to check if there is any difference between the observed value and expected value. The formula for chi-square can be written as;

χ 2  = ∑(O i – E i ) 2 /E i

where O i is the observed value and E i is the expected value.

Chi-Square Test of Independence

The chi-square test of independence also known as the chi-square test of association which is used to determine the association between the categorical variables. It is considered as a non-parametric test . It is mostly used to test statistical independence.

The chi-square test of independence is not appropriate when the categorical variables represent the pre-test and post-test observations. For this test, the data must meet the following requirements:

• Two categorical variables
• Relatively large sample size
• Categories of variables (two or more)
• Independence of observations

Example of Categorical Data

Let us take an example of a categorical data where there is a society of 1000 residents with four neighbourhoods, P, Q, R and S. A random sample of 650 residents of the society is taken whose occupations are doctors, engineers and teachers. The null hypothesis is that each person’s neighbourhood of residency is independent of the person’s professional division. The data are categorised as:

Assume the sample living in neighbourhood P, 150, to estimate what proportion of the whole 1,000 people live in neighbourhood P. In the same way, we take 349/650 to calculate what ratio of the 1,000 are doctors. By the supposition of independence under the hypothesis, we should “expect” the number of doctors in neighbourhood P is;

150 x 349/650  ≈ 80.54

So by  the chi-square test formula for that particular cell in the table, we get;

(Observed – Expected) 2 /Expected Value = (90-80.54) 2 /80.54  ≈ 1.11

Some of the exciting facts about the Chi-square test are given below:

The Chi-square statistic can only be used on numbers. We cannot use them for data in terms of percentages, proportions, means or similar statistical contents. Suppose, if we have 20% of 400 people, we need to convert it to a number, i.e. 80, before running a test statistic.

A chi-square test will give us a p-value. The p-value will tell us whether our test results are significant or not. 

However, to perform a chi-square test and get the p-value, we require two pieces of information:

(1) Degrees of freedom. That’s just the number of categories minus 1.

(2) The alpha level(α). You or the researcher chooses this. The usual alpha level is 0.05 (5%), but you could also have other levels like 0.01 or 0.10.

In elementary statistics, we usually get questions along with the degrees of freedom(DF) and the alpha level. Thus, we don’t usually have to figure out what they are. To get the degrees of freedom, count the categories and subtract 1.

The chi-square distribution table with three probability levels is provided here. The statistic here is used to examine whether distributions of certain variables vary from one another. The categorical variable will produce data in the categories and numerical variables will produce data in numerical form.

The distribution of χ 2 with (r-1)(c-1) degrees of freedom(DF) , is represented in the table given below. Here, r represents the number of rows in the two-way table and c represents the number of columns.

Solved Problem

 A survey on cars had conducted in 2011 and determined that 60% of car owners have  only one car, 28% have two cars, and 12% have three or more. Supposing that you have decided to conduct your own survey and have collected the data below, determine whether your data supports the results of the study.

Use a significance level of 0.05. Also, given that, out of 129 car owners, 73 had one car and 38 had two cars.

Let us state the null and alternative hypotheses.

H 0 : The proportion of car owners with one, two or three cars is 0.60, 0.28 and 0.12 respectively.

H 1 : The proportion of car owners with one, two or three cars does not match the proposed model.

A Chi-Square goodness of fit test is appropriate because we are examining the distribution of a single categorical variable. 

Let’s tabulate the given information and calculate the required values.

Therefore, χ 2  = ∑(O i  – E i ) 2 /E i  = 0.7533

Let’s compare it to the chi-square value for the significance level 0.05. 

The degrees for freedom = 3 – 1 = 2

Using the table, the critical value for a 0.05 significance level with df = 2 is 5.99. 

That means that 95 times out of 100, a survey that agrees with a sample will have a χ 2  value of 5.99 or less. 

The Chi-square statistic is only 0.7533, so we will accept the null hypothesis.

Frequently Asked Questions – FAQs

What is the chi-square test write its formula, how do you calculate chi squared, what is a chi-square test used for, how do you interpret a chi-square test, what is a good chi-square value.

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