chi square test null hypothesis rejected

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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B.A., History and Science, Harvard University

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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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S.4 chi-square tests, chi-square test of independence section  .

Do you remember how to test the independence of two categorical variables? This test is performed by using a Chi-square test of independence.

Recall that we can summarize two categorical variables within a two-way table, also called an r × c contingency table, where r = number of rows, c = number of columns. Our question of interest is “Are the two variables independent?” This question is set up using the following hypothesis statements:

 \[E=\frac{\text{row total}\times\text{column total}}{\text{sample size}}\]

We will compare the value of the test statistic to the critical value of \(\chi_{\alpha}^2\) with the degree of freedom = ( r - 1) ( c - 1), and reject the null hypothesis if \(\chi^2 \gt \chi_{\alpha}^2\).

Example S.4.1 Section  

Is gender independent of education level? A random sample of 395 people was surveyed and each person was asked to report the highest education level they obtained. The data that resulted from the survey are summarized in the following table:

Question : Are gender and education level dependent at a 5% level of significance? In other words, given the data collected above, is there a relationship between the gender of an individual and the level of education that they have obtained?

Here's the table of expected counts:

So, working this out, \(\chi^2= \dfrac{(60−50.886)^2}{50.886} + \cdots + \dfrac{(57 − 48.132)^2}{48.132} = 8.006\)

The critical value of \(\chi^2\) with 3 degrees of freedom is 7.815. Since 8.006 > 7.815, we reject the null hypothesis and conclude that the education level depends on gender at a 5% level of significance.

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11.1: Chi-Square Tests for Independence

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Learning Objectives

• To understand what chi-square distributions are.
• To understand how to use a chi-square test to judge whether two factors are independent.

Chi-Square Distributions

As you know, there is a whole family of \(t\)-distributions, each one specified by a parameter called the degrees of freedom, denoted \(df\). Similarly, all the chi-square distributions form a family, and each of its members is also specified by a parameter \(df\), the number of degrees of freedom. Chi is a Greek letter denoted by the symbol \(\chi\) and chi-square is often denoted by \(\chi^2\).

Figure \(\PageIndex{1}\) shows several \(\chi\)-square distributions for different degrees of freedom. A chi-square random variable is a random variable that assumes only positive values and follows a \(\chi\)-square distribution.

Definition: critical value

The value of the chi-square random variable \(\chi^2\) with \(df=k\) that cuts off a right tail of area \(c\) is denoted \(\chi_c^2\) and is called a critical value (Figure \(\PageIndex{2}\)).

Figure \(\PageIndex{3}\) below gives values of \(\chi_c^2\) for various values of \(c\) and under several chi-square distributions with various degrees of freedom.

Tests for Independence

Hypotheses tests encountered earlier in the book had to do with how the numerical values of two population parameters compared. In this subsection we will investigate hypotheses that have to do with whether or not two random variables take their values independently, or whether the value of one has a relation to the value of the other. Thus the hypotheses will be expressed in words, not mathematical symbols. We build the discussion around the following example.

There is a theory that the gender of a baby in the womb is related to the baby’s heart rate: baby girls tend to have higher heart rates. Suppose we wish to test this theory. We examine the heart rate records of \(40\) babies taken during their mothers’ last prenatal checkups before delivery, and to each of these \(40\) randomly selected records we compute the values of two random measures: 1) gender and 2) heart rate. In this context these two random measures are often called factors. Since the burden of proof is that heart rate and gender are related, not that they are unrelated, the problem of testing the theory on baby gender and heart rate can be formulated as a test of the following hypotheses:

\[H_0: \text{Baby gender and baby heart rate are independent}\\ vs. \\ H_a: \text{Baby gender and baby heart rate are not independent} \nonumber \]

The factor gender has two natural categories or levels: boy and girl. We divide the second factor, heart rate, into two levels, low and high, by choosing some heart rate, say \(145\) beats per minute, as the cutoff between them. A heart rate below \(145\) beats per minute will be considered low and \(145\) and above considered high. The \(40\) records give rise to a \(2\times 2\) contingency table . By adjoining row totals, column totals, and a grand total we obtain the table shown as Table \(\PageIndex{1}\). The four entries in boldface type are counts of observations from the sample of \(n = 40\). There were \(11\) girls with low heart rate, \(17\) boys with low heart rate, and so on. They form the core of the expanded table.

In analogy with the fact that the probability of independent events is the product of the probabilities of each event, if heart rate and gender were independent then we would expect the number in each core cell to be close to the product of the row total \(R\) and column total \(C\) of the row and column containing it, divided by the sample size \(n\). Denoting such an expected number of observations \(E\), these four expected values are:

• 1 st row and 1 st column: \(E=(R\times C)/n = 18\times 28 /40 = 12.6\)
• 1 st row and 2 nd column: \(E=(R\times C)/n = 18\times 12 /40 = 5.4\)
• 2 nd row and 1 st column: \(E=(R\times C)/n = 22\times 28 /40 = 15.4\)
• 2 nd row and 2 nd column: \(E=(R\times C)/n = 22\times 12 /40 = 6.6\)

We update Table \(\PageIndex{1}\) by placing each expected value in its corresponding core cell, right under the observed value in the cell. This gives the updated table Table \(\PageIndex{2}\).

A measure of how much the data deviate from what we would expect to see if the factors really were independent is the sum of the squares of the difference of the numbers in each core cell, or, standardizing by dividing each square by the expected number in the cell, the sum \(\sum (O-E)^2 / E\). We would reject the null hypothesis that the factors are independent only if this number is large, so the test is right-tailed. In this example the random variable \(\sum (O-E)^2 / E\) has the chi-square distribution with one degree of freedom. If we had decided at the outset to test at the \(10\%\) level of significance, the critical value defining the rejection region would be, reading from Figure \(\PageIndex{3}\), \(\chi _{\alpha }^{2}=\chi _{0.10 }^{2}=2.706\), so that the rejection region would be the interval \([2.706,\infty )\). When we compute the value of the standardized test statistic we obtain

\[\sum \frac{(O-E)^2}{E}=\frac{(11-12.6)^2}{12.6}+\frac{(7-5.4)^2}{5.4}+\frac{(17-15.4)^2}{15.4}+\frac{(5-6.6)^2}{6.6}=1.231 \nonumber \]

Since \(1.231 < 2.706\), the decision is not to reject \(H_0\). See Figure \(\PageIndex{4}\). The data do not provide sufficient evidence, at the \(10\%\) level of significance, to conclude that heart rate and gender are related.

Fig ure \(\PageIndex{4}\) : Baby Gender Prediction

H 0 vs .   H a : : Baby   gender   and   baby   heart   rate   are   independent Baby   gender   and   baby   heart   rate   are   n o t   independent H 0 vs .   H a : : Baby   gender   and   baby   heart   rate   are   independent Baby   gender   and   baby   heart   rate   are   n o t   independent

With this specific example in mind, now turn to the general situation. In the general setting of testing the independence of two factors, call them Factor \(1\) and Factor \(2\), the hypotheses to be tested are

\[H_0: \text{The two factors are independent}\\ vs. \\ H_a: \text{The two factors are not independent} \nonumber \]

As in the example each factor is divided into a number of categories or levels. These could arise naturally, as in the boy-girl division of gender, or somewhat arbitrarily, as in the high-low division of heart rate. Suppose Factor \(1\) has \(I\) levels and Factor \(2\) has \(J\) levels. Then the information from a random sample gives rise to a general \(I\times J\) contingency table, which with row totals, column totals, and a grand total would appear as shown in Table \(\PageIndex{3}\). Each cell may be labeled by a pair of indices \((i,j)\). \(O_{ij}\) stands for the observed count of observations in the cell in row \(i\) and column \(j\), \(R_i\) for the \(i^{th}\) row total and \(C_j\) for the \(j^{th}\) column total. To simplify the notation we will drop the indices so Table \(\PageIndex{3}\) becomes Table \(\PageIndex{4}\). Nevertheless it is important to keep in mind that the \(Os\), the \(Rs\) and the \(Cs\), though denoted by the same symbols, are in fact different numbers.

As in the example, for each core cell in the table we compute what would be the expected number \(E\) of observations if the two factors were independent. \(E\) is computed for each core cell (each cell with an \(O\) in it) of Table \(\PageIndex{4}\) by the rule applied in the example:

\[E=R×Cn \nonumber \]

where \(R\) is the row total and \(C\) is the column total corresponding to the cell, and \(n\) is the sample size

Here is the test statistic for the general hypothesis based on Table \(\PageIndex{5}\), together with the conditions that it follow a chi-square distribution.

Test Statistic for Testing the Independence of Two Factors

\[\chi^2=\sum (O−E)^2E \nonumber \]

where the sum is over all core cells of the table.

• the two study factors are independent, and
• the observed count \(O\) of each cell in Table \(\PageIndex{5}\) is at least \(5\),

then \(\chi ^2\) approximately follows a chi-square distribution with \(df=(I-1)\times (J-1)\) degrees of freedom.

The same five-step procedures, either the critical value approach or the \(p\)-value approach, that were introduced in Section 8.1 and Section 8.3 are used to perform the test, which is always right-tailed.

Example \(\PageIndex{1}\)

A researcher wishes to investigate whether students’ scores on a college entrance examination (\(CEE\)) have any indicative power for future college performance as measured by \(GPA\). In other words, he wishes to investigate whether the factors \(CEE\) and \(GPA\) are independent or not. He randomly selects \(n = 100\) students in a college and notes each student’s score on the entrance examination and his grade point average at the end of the sophomore year. He divides entrance exam scores into two levels and grade point averages into three levels. Sorting the data according to these divisions, he forms the contingency table shown as Table \(\PageIndex{6}\), in which the row and column totals have already been computed.

Test, at the \(1\%\) level of significance, whether these data provide sufficient evidence to conclude that \(CEE\) scores indicate future performance levels of incoming college freshmen as measured by \(GPA\).

We perform the test using the critical value approach, following the usual five-step method outlined at the end of Section 8.1.

• Step 1 . The hypotheses are \[H_0:\text{CEE and GPA are independent factors}\\ vs.\\ H_a:\text{CEE and GPA are not independent factors} \nonumber \]
• Step 2 . The distribution is chi-square.
• 1 st row and 1 st column: \(E=(R\times C)/n=41\times 52/100=21.32\)
• 1 st row and 2 nd column: \(E=(R\times C)/n=36\times 52/100=18.72\)
• 1 st row and 3 rd column: \(E=(R\times C)/n=23\times 52/100=11.96\)
• 2 nd row and 1 st column: \(E=(R\times C)/n=41\times 48/100=19.68\)
• 2 nd row and 2 nd column: \(E=(R\times C)/n=36\times 48/100=17.28\)
• 2 nd row and 3 rd column: \(E=(R\times C)/n=23\times 48/100=11.04\)

Table \(\PageIndex{6}\) is updated to Table \(\PageIndex{6}\).

The test statistic is

\[\begin{align*} \chi^2 &= \sum \frac{(O-E)^2}{E}\\ &= \frac{(35-21.32)^2}{21.32}+\frac{(12-18.72)^2}{18.72}+\frac{(5-11.96)^2}{11.96}+\frac{(6-19.68)^2}{19.68}+\frac{(24-17.28)^2}{17.28}+\frac{(18-11.04)^2}{11.04}\\ &= 31.75 \end{align*} \nonumber \]

• Step 4 . Since the \(CEE\) factor has two levels and the \(GPA\) factor has three, \(I = 2\) and \(J = 3\). Thus the test statistic follows the chi-square distribution with \(df=(2-1)\times (3-1)=2\) degrees of freedom.

Since the test is right-tailed, the critical value is \(\chi _{0.01}^{2}\). Reading from Figure 7.1.6 "Critical Values of Chi-Square Distributions", \(\chi _{0.01}^{2}=9.210\), so the rejection region is \([9.210,\infty )\).

• Step 5 . Since \(31.75 > 9.21\) the decision is to reject the null hypothesis. See Figure \(\PageIndex{5}\). The data provide sufficient evidence, at the \(1\%\) level of significance, to conclude that \(CEE\) score and \(GPA\) are not independent: the entrance exam score has predictive power.

Key Takeaway

• Critical values of a chi-square distribution with degrees of freedom df are found in Figure 7.1.6.
• A chi-square test can be used to evaluate the hypothesis that two random variables or factors are independent.

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Statistics and probability

Course: statistics and probability   >   unit 14.

• Chi-square distribution introduction

Pearson's chi square test (goodness of fit)

• Chi-square statistic for hypothesis testing
• Chi-square goodness-of-fit example
• Expected counts in a goodness-of-fit test
• Conditions for a goodness-of-fit test
• Test statistic and P-value in a goodness-of-fit test
• Conclusions in a goodness-of-fit test

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Video transcript

Statistics Made Easy

Chi-Square Test of Independence: Definition, Formula, and Example

A Chi-Square Test of Independence  is used to determine whether or not there is a significant association between two categorical variables.

This tutorial explains the following:

• The motivation for performing a Chi-Square Test of Independence.
• The formula to perform a Chi-Square Test of Independence.
• An example of how to perform a Chi-Square Test of Independence.

Chi-Square Test of Independence: Motivation

A Chi-Square test of independence can be used to determine if there is an association between two categorical variables in a many different settings. Here are a few examples:

• We want to know if gender is associated with political party preference so we survey 500 voters and record their gender and political party preference.
• We want to know if a person’s favorite color is associated with their favorite sport so we survey 100 people and ask them about their preferences for both.
• We want to know if education level and marital status are associated so we collect data about these two variables on a simple random sample of 50 people.

In each of these scenarios we want to know if two categorical variables are associated with each other. In each scenario, we can use a Chi-Square test of independence to determine if there is a statistically significant association between the variables. 

Chi-Square Test of Independence: Formula

A Chi-Square test of independence uses the following null and alternative hypotheses:

• H 0 : (null hypothesis)  The two variables are independent.
• H 1 : (alternative hypothesis)  The two variables are not independent. (i.e. they are associated)

We use the following formula to calculate the Chi-Square test statistic X 2 :

X 2 = Σ(O-E) 2  / E

• Σ:  is a fancy symbol that means “sum”
• O:  observed value
• E:  expected value

If the p-value that corresponds to the test statistic X 2  with (#rows-1)*(#columns-1) degrees of freedom is less than your chosen significance level then you can reject the null hypothesis.

Chi-Square Test of Independence: Example

Suppose we want to know whether or not gender is associated with political party preference. We take a simple random sample of 500 voters and survey them on their political party preference. The following table shows the results of the survey:

Use the following steps to perform a Chi-Square test of independence to determine if gender is associated with political party preference.

Step 1: Define the hypotheses.

We will perform the Chi-Square test of independence using the following hypotheses:

• H 0 :  Gender and political party preference are independent.
• H 1 : Gender and political party preference are  not independent.

Step 2: Calculate the expected values.

Next, we will calculate the expected values for each cell in the contingency table using the following formula:

Expected value = (row sum * column sum) / table sum.

For example, the expected value for Male Republicans is: (230*250) / 500 =  115 .

We can repeat this formula to obtain the expected value for each cell in the table:

Step 3: Calculate (O-E) 2  / E for each cell in the table.

Next we will calculate  (O-E) 2  / E  for each cell in the table  where:

For example, Male Republicans would have a value of: (120-115) 2 /115 =  0.2174 .

We can repeat this formula for each cell in the table:

Step 4: Calculate the test statistic X 2  and the corresponding p-value.

X 2  = Σ(O-E) 2  / E = 0.2174 + 0.2174 + 0.0676 + 0.0676 + 0.1471 + 0.1471 =  0.8642

According to the Chi-Square Score to P Value Calculator , the p-value associated with X 2  = 0.8642 and (2-1)*(3-1) = 2 degrees of freedom is  0.649198 .

Step 5: Draw a conclusion.

Since this p-value is not less than 0.05, we fail to reject the null hypothesis. This means we do not have sufficient evidence to say that there is an association between gender and political party preference.

Note:  You can also perform this entire test by simply using the Chi-Square Test of Independence Calculator .

Additional Resources

The following tutorials explain how to perform a Chi-Square test of independence using different statistical programs:

How to Perform a Chi-Square Test of Independence in Stata How to Perform a Chi-Square Test of Independence in Excel How to Perform a Chi-Square Test of Independence in SPSS How to Perform a Chi-Square Test of Independence in Python How to Perform a Chi-Square Test of Independence in R Chi-Square Test of Independence on a TI-84 Calculator Chi-Square Test of Independence Calculator

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Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

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what test do I use if there are 2 categorical variables and one categorical DV? as in I want to test political attitudes and beliefs in conspiracies and how they affect Covid conspiracy thinking

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Chi-square Tests (Business)

Contents Toggle Main Menu 1 Chi-Square Tests 2 Goodness-of-fit Tests 2.1 The Method 3 Worked Example 1 4 Test for Association/Independence 4.1 The Method 5 Worked Example 2 6 Test Yourself 7 See Also

Chi-Square Tests

There are two important applications of Chi-Square tests ($\chi^2$ tests) :

• Goodness-of-fit tests. We use $\chi^2$ tests to test the goodness of fit of data to a hypothetical pattern , for example, does the number of customers arriving at a fairground ride follow a Poisson distribution ?
• Test for Association/Independence. For example, is there an association between the ratings by students of their course lecturers and the average exam mark on the courses? This involves qualitative data; the categories that lecturers fall into given by their student rating. The null hypothesis is that there is no association and the $\chi^2$ test examines this hypothesis.

There are rules to follow when using a $\chi^2$ test, listed below.

• No category has a mean of less than $5$.
• The categories should be as natural as possible.
• Keep to the raw data.
• Avoid categories that vastly exceed others.

Goodness-of-fit Tests

To carry out the test by hand the steps are as follows:

• Firstly, we need to identify the null and alternative hypotheses $H_0$ and $H_1$. For example, if we think our data might follow a Uniform distribution , then we would have:

\begin{align} H_0&: \text{Our data follows a Uniform distribution versus}\\ H_1&: \text{Our data does }\textbf{not }\text{follow a Uniform distribution.}\\ \end{align} Note: The Uniform distribution can be replaced with any other probability distribution.

• After collecting data, we calculate the Chi-Square statistic/value:

\begin{equation} \chi^2 = \sum {\frac{(O-E)^2}{E}} \end{equation}

\begin{split} \\ O &&= \text{Observed frequencies} \\ E &&= \text{Expected frequencies} \\ \sum{}&& =\text{ Sum Of} \end{split}

• We then compare this statistic to a $\chi^2$ table on the appropriate degrees of freedom...

\begin{equation} \nu = (\text{number of categories after pooling}) - 1. \end{equation}

• Finally, we make our conclusion. If the Chi-Square value does not exceed the critical value (e.g. $p = 0.05$) then the null hypothesis will be accepted i.e. the data does follow the hypothetical pattern. If it does exceed the critical value, the null hypothesis must be rejected.

Worked Example 1

Worked example - chi-squared goodness-of-fit test.

The number of accidents per day at a chocolate factory was recorded over the period of three months; the results are shown in the table below.

Suggest a distribution that might fit these data, and test to see whether it is appropriate or not.

Since we are looking at the number of accidents in a certain time interval (day) and there is no fixed limit to the number of accidents which could happen, an appropriate distribution would be the Poisson distribution .

To see if the Poisson distribution is consistent with our data, we shall test the null hypothesis:

$H_0$: The number of accidents follow a Poisson distribution

versus the alternative:

$H_1$: The number of accidents does not follow a Poisson distribution.

To calculate our test statistic, we need to calculate our expected values/frequencies based on the Poisson distribution. We use the formula: \begin{equation} \mathrm{P}(X = r) = \dfrac{~\lambda^r \times e^{-\lambda}~}{r!} \end{equation}

to find the expected probabilities , and then multiply these by the total sample size $(92)$ to obtain the corresponding expected frequencies. Before we can use this formula, we need an estimate for $\lambda$. For the Poisson distribution, $\lambda$ is equal to the mean. Thus, we have:

\begin{align} \lambda &= \dfrac{(0 \times 44) + (1 \times 33) + (2 \times 10) + (3 \times 4) + (4 \times 1) + (5 \times 0)}{92}\\ &=0.75\\ \end{align}.

We now have $\lambda$ we can calculate the expected probabilities:

\begin{align} \mathrm{P}(X = 0) &= \dfrac{~0.75^0 \times e^{-0.75}~}{0!}\\ &= \dfrac{0.47237}{1}\\ &= 0.47237 \text{(5 d.p.)}.\\ &\\ \mathrm{P}(X = 1) &= \dfrac{~0.75^1 \times e^{-0.75}~}{1!}\\ &= \dfrac{0.75 \times 0.47237}{1}\\ &= 0.35427 \text{(5 d.p.)}.\\ &\\ \mathrm{P}(X = 2) &= \dfrac{~0.75^2 \times e^{-0.75}~}{2!}\\ &= \dfrac{0.5625 \times 0.47237}{2}\\ &= 0.13285 \text{(5 d.p.)}.\\ &\\ \end{align}

\begin{align} \mathrm{P}(X = 3) &= \dfrac{~0.75^3 \times e^{-0.75}~}{3!}\\ &= \dfrac{0.42188 \times 0.47237}{6}\\ &= 0.03321 \text{(5 d.p.)}.\\ &\\ \end{align}

\begin{align} \mathrm{P}(X = 4) &= \dfrac{~0.75^4 \times e^{-0.75}~}{4!}\\ &= \dfrac{0.31641 \times 0.47237}{24}\\ &= 0.00623 \text{(5 d.p.)}.\\ \end{align}

For the $5$ or more category, we can just add up all the other probabilities and subtract from $1$, since the entire probability distribution should sum to $1$. So we have

\begin{align} \mathrm{P}(X \geq 5) &= 1 - (0.47237 + 0.35427 + 0.13285 + 0.03321 + 0.00623)\\ &= 1 - 0.99894\\ &= 0.00106 \text{(5 d.p.)}.\\ \end{align}.

Thus, the expected frequencies can be found by multiplying the expected probabilities by the total sample size $(92)$ and then we can arrange them into a table:

We can now calculate our test statistic.

\begin{align} \chi^2 &= \sum{\dfrac{(O - E)^2}{E}~}\\ &= 0.00676 + 0.00509 + 0.40403 + 0.29209 + 0.31787 + 0.31787 + 0.09752\\ &=1.12336.\\ \end{align}

We now need to compare our test statistic to a value from a $\chi^2$ table. The degrees of freedom are (number of categories) - (number of parameters estimated) - $1 = 6 - 1 - 1 = 4$.

Thus, we use the following critical values:

Since $1.12336 < 7.779$, the critical value at the $10$% significance level, there is no evidence against the null hypothesis $H_0$ so we cannot reject it. The number of accidents recorded per day at the chocolate factory follows a Poisson distribution.

Watch the following video for how to perform the Chi-Squared Goodness-of-fit test in Minitab (ver. 16):

Test for Association/Independence

Chi-Square tests can also be used to test for the association between attributes/independence (using Contingency Tables ), for example, is there an association between the ability to drive and the distance commuted to work?

• You will need to use a $\chi^2$ test of independence when the data is given in the form of a contingency table .
• The method is very similar to the goodness-of-fit test.
• First, we form our hypotheses. Regardless of what the categorical variables are, if you are testing to see whether they are independent or not (associated) the hypotheses are:

\begin{align} H_0&: \text{There is no association between the categorical variables versus}\\ H_1&: \text{There }\textit{is}\text{ an association between the categorical variables.}\\ \end{align}

• Then we calculate the test statistic, which is the same as for the goodness-of-fit test.

\begin{split} \\ O &&= \text{Observed frequencies} \\ E&& = \text{Expected frequencies} \\ \sum{}&& =\text{ Sum Of} \end{split}

For this test, we need not worry about any probability distributions as we are just testing for independence. We can get our expected frequencies by using the following formula:

\begin{equation} E = \; \dfrac{\text{row total} \times \text{column total}}{\text{overall sample size}} \end{equation}

for each cell in the contingency table.

• Once we have the test statistic, we can find a $p$-value/critical value. The appropriate degrees of freedom to use is given by:

\begin{equation} \nu = (\text{number of rows }- 1) \times (\text{number of columns} - 1) \end{equation}

• Finally we form a conclusion. If the test statistics is greater than the critical value, we reject $H_0$ in favour of $H_1$. This is equivalent to saying there is sufficient evidence to suggest that the categorical variables are associated, or are not independent. Otherwise, we accept $H_0$.

Worked Example 2

Worked example - chi-squared test for association/independence.

The following table includes data on the number of days sick leave taken by managerial and non-managerial employees of the department store, James Lewis, in the past year.

Is there an association between type of employee and number of days sick leave?

Our hypotheses are: \begin{align} H_0&: \text{There is no association between type of employee and number of days sick leave.}\\ H_1&: \text{There is an association between type of employee and number of days sick leave.}\\ \end{align}

We need to calculate the expected frequencies before we can calculate the test statistic.

\begin{align} E_1 &= \; \dfrac{~\text{row total for '0 - 10 days'} \times \text{column total for 'Non-Managerial'}~}{~\text{total number of employees}~}\\ &= \dfrac{57\times115}{165}\\ &= 39.7273 \text{ (4 d.p.)}.\\ &\\ E_2 &= \; \dfrac{~\text{row total for '0 - 10 days'} \times \text{column total for 'Managerial'}~}{~\text{total number of employees}~}\\ &= \dfrac{57\times50}{165}\\ &= 17.2727 \text{ (4 d.p.)}.\\ &\\ E_3 &= \; \dfrac{~\text{row total for '11 - 20 days'} \times \text{column total for 'Non-Managerial'}~}{~\text{total number of employees}~}\\ &= \dfrac{48\times115}{165}\\ &= 33.4545 \text{ (4 d.p.)}.\\ &\\ E_4 &= \; \dfrac{~\text{row total for '11 - 20 days'} \times \text{column total for 'Managerial'}~}{~\text{total number of employees}~}\\ &= \dfrac{48\times50}{165}\\ &= 14.5455\text{ (4 d.p.)}.\\ &\\ E_5 &= \; \dfrac{~\text{row total for '21 or more days'} \times \text{column total for 'Non-Managerial'}~}{~\text{total number of employees}~}\\ &= \dfrac{60\times115}{165}\\ &= 41.8182 \text{ (4 d.p.)}.\\ &\\ E_6 &= \; \dfrac{~\text{row total for '21 or more days'} \times \text{column total for 'Managerial'}~}{~\text{total number of employees}~}\\ &= \dfrac{60\times50}{165}\\ &= 18.1818 \text{ (4 d.p.)}.\\ \end{align}

For convenience we shall arrange the data into a new table to calculate the test statistic.

Thus our test statistic is $\chi^2 = 11.0181$.

We need to compare this to critical values on $(3 - 1) \times (2 -1) = 2$ degrees of freedom.

Since $10.765 > 9.210$ (the critical value at the $1\%$ level), we can conclude there is very significant evidence that there is an association between number of days sick leave and type of employee. We accept $H_1$.

Watch the following video to see how to perform the test in Minitab (ver. 16):

Test Yourself

Try our Numbas test on hypothesis testing: Hypothesis testing and confidence intervals and also two-sample tests .

For more information about the topics covered here see hypothesis testing .

• Communities

Step 5 - Interpreting The Results

We are now ready for the final step, interpreting the results of our chi-square calculation. For this we will need to consult a Chi-Square Distribution Table. This is a probability table of selected values of X 2  (Table 3). 

Table 3: Chi-Square Distribution Table.

Statisticians calculate certain possibilities of occurrence (P values) for a X 2  value depending on  degrees of freedom . Degrees of freedom is simply the number of classes that can vary independently minus one, (n-1). In this case the degrees of freedom = 1 because we have 2 phenotype classes: resistant and susceptible.

The calculated value of X 2  from our results can be compared to the values in the table aligned with the specific degrees of freedom we have. This will tell us the probability that the deviations (between what we  expected  to see and what we actually saw) are due to chance alone and our hypothesis or model can be supported.

In our example, the X 2  value of 1.2335 and degrees of freedom of 1 are associated with a P value of less than 0.50, but greater than 0.25 (follow blue dotted line and arrows in Fig. 5). This means that a  chi-square  value this large or larger (or differences between expected and  observed  numbers this great or greater) would occur simply by chance between 25% and 50% of the time.  By convention biologists often use the 5.0% value (p<0.05) to determine if observed deviations are significant. Any deviations greater than this level would cause us to reject our hypothesis and assume something other than chance was at play  (see red circle on Fig. 5). If your chi-square calculated value is greater than the chi-square critical value, then you reject your null hypothesis. If your chi-square calculated value is less than the chi-square critical value, then you "fail to reject" your null hypothesis. 

Figure 5.  Finding the probability value for a chi-square of 1.2335 with 1 degree of freedom . First read down column 1 to find the 1 degree of freedom row and then go to the right to where 1.2335 would occur. This corresponds to a probability of less than 0.5 but greater than 0.25, as indicated by the blue arrows.

Therefore in our tomato breeding example, we failed to reject our hypothesis that resistance to bacterial spot in this set of crosses is due to a single dominantly inherited gene (Rx-4). We can assume that the deviations we saw between what we expected and actually observed in terms of the number of resistant and susceptible plants could be due to mere chance. We can continue working with our current hypothesis. Remember, we have not “proven” our hypothesis at this point. Further testing in other crosses and populations will be used to provide additional evidence that our hypothesis accurately explains the mode of inheritance for Rx-4.

Launch the video tutorial below showing how to use a chi-square distribution chart, using this tomato breeding example.

Next we will go through a genotyping example, or you skip ahead in this lesson to a discussion about computer programs available when you have extensive data, as well as the strengths and weaknesses of the chi-square test.

In a BC 2 S 4  IBC population (inbred backcross) with 197 tomato lines, you observed the following phenotypic data with regards to bacterial spot disease. Calculate a chi-square value for the hypothesis that resistance is dominantly inherited. Would you reject or fail to reject this hypothesis?

169 susceptible to bacterial spot

6 resistant to bacterial spot