1. CHAPTER 11 Inference for Distributions of Categorical Data
11.2
Inference for Two-Way Tables

2. Comparing Distributions of a Categorical Variable
Market researchers suspect that background music may affect the mood and buying behavior of customers.
One study in a Mediterranean restaurant compared three randomly assigned treatments: no music, French accordion music, and Italian string music.
Under each condition, the researchers recorded the numbers of customers who ordered French, Italian, and other entrees.
Try Exercise 39

3. Expected Counts and the Chi-Square Statistic
Finding Expected Counts
When H0 is true, the expected count in any cell of a two-way table is
Conditions for Performing a Chi-Square Test for Homogeneity
Random: The data come a well-designed random sample or from a randomized experiment.
10%: When sampling without replacement, check that
n ≤ (1/10)N.
Large Counts: All expected counts are greater than 5

4. Expected Counts and the Chi-Square Statistic
Just as we did with the chi-square goodness-of-fit test, we compare the observed counts with the expected counts using the statistic
This time, the sum is over all cells (not including the totals!) in the two-way table.

5. Chi-Square Test for Homogeneity
Suppose the conditions are met. You can use the chi-square test for homogeneity to test
H0: There is no difference in the distribution of a categorical variable for several populations or treatments.
Ha: There is a difference in the distribution of a categorical variable for several populations or treatments.
Start by finding the expected count for each category assuming that H0 is true. Then calculate the chi-square statistic
where the sum is over all cells (not including totals) in the two-way table. If H0 is true, the c2 statistic has approximately a chi-square distribution with degrees of freedom = (number of rows − 1)(number of columns − 1). The P-value is the area to the right of c2 under the corresponding chi-square density curve.

7. The Chi-Square Test for Independence
The 10% and Large Counts conditions for the chi-square test for independence are the same as for the homogeneity test.
There is a slight difference in the Random condition for the two tests: a test for independence uses data from one sample but a test for homogeneity uses data from two or more samples/groups.
Conditions for Performing a Chi-Square Test for Independence
Random: The data come a well-designed random sample or from a randomized experiment.
10%: When sampling without replacement, check that
n ≤ (1/10)N.
Large Counts: All expected counts are greater than 5

8. Chi-Square Test for Independence
Suppose the conditions are met. You can use the chi-square test for independence to test
H0: There is no association between two categorical variables in the population of interest.
Ha: There is an association between two categorical variables in the population of interest.
Start by finding the expected count for each category assuming that H0 is true. Then calculate the chi-square statistic
where the sum is over all cells (not including totals) in the two-way table. If H0 is true, the c2 statistic has approximately a chi-square distribution with degrees of freedom = (number of rows − 1)(number of columns − 1). The P-value is the area to the right of c2 under the corresponding chi-square density curve.

9. Example: Choosing the right type of chi-square test
Are men and women equally likely to suffer lingering fear from watching scary movies as children? Researchers asked a random sample of 117 college students to write narrative accounts of their exposure to scary movies before the age of 13. More than one-fourth of the students said that some of the fright symptoms are still present when they are awake. The following table breaks down these results by gender.
Try Exercise 39

10. Example: Choosing the right type of chi-square test
Minitab output for a chi-square test using these data is shown below.
Try Exercise 39

11. Example: Choosing the right type of chi-square test
Problem: Assume that the conditions for performing inference are met.
(a) Explain why a chi-square test for independence and not a chi-square test for homogeneity should be used in this setting.
The data were produced using a single random sample of college students, who were then classified by gender and whether or not they had lingering fright symptoms.
The chi-square test for homogeneity requires independent random samples from each population.
Try Exercise 39

12. Example: Choosing the right type of chi-square test
Problem: Assume that the conditions for performing inference are met.
(b) State an appropriate pair of hypotheses for researchers to test in this setting.
The null hypothesis is
H0: There is no association between gender and ongoing fright symptoms in the population of college students. The alternative hypothesis is
Ha: There is an association between gender and ongoing fright symptoms in the population of college students.
Try Exercise 39

13. Example: Choosing the right type of chi-square test
Problem: Assume that the conditions for performing inference are met.
(c) Which cell contributes most to the chi-square statistic? In what way does this cell differ from what the null hypothesis suggests?
Men who admit to having lingering fright symptoms account for the largest component of the chi-square statistic: of the total
Far fewer men in the sample admitted to fright symptoms (7) than we would expect if H0 were true (11.69).
Try Exercise 39

14. Example: Choosing the right type of chi-square test
Problem: Assume that the conditions for performing inference are met.
(d) Interpret the P-value in context. What conclusion would you draw at α = 0.01?
If gender and ongoing fright symptoms really are independent in the population of interest, there is a chance of obtaining a random sample of 117 students that gives a chi-square statistic of or higher. Because the P-value, 0.045, is greater than 0.01, we would fail to reject H0. We do not have convincing evidence that there is an association between gender and fright symptoms in the population of college students.
Try Exercise 39

15. Inference for Two-Way Tables
COMPARE conditional distributions for data in a two-way table.
STATE appropriate hypotheses and COMPUTE expected counts for a chi-square test based on data in a two-way table.
CALCULATE the chi-square statistic, degrees of freedom, and P-value for a chi-square test based on data in a two-way table.
PERFORM a chi-square test for homogeneity.
PERFORM a chi-square test for independence.
CHOOSE the appropriate chi-square test.