Chapter 12

1. 12-1 Chapter 12 Chi-Square

2. Outline 12-2 • 12-1 Introduction • 12-2 Test for Goodness of Fit • 12-3 Tests Using Contingency Tables

3. Objectives 12-3 • Test a distribution for goodness of fit using chi-square. • Test two variables for independence using chi-square.

4. 12-2 Test for Goodness of Fit 12-4 • When one is testing to see whether a frequency distribution fits a specific pattern, the chi-square goodness-of-fit test is used.

5. 12-2 Test for Goodness of Fit -Example 12-5 • Suppose a market analyst wished to see whether consumers have any preference among five flavors of a new fruit soda. A sample of 100 people provided the following data:

6. 12-2 Test for Goodness of Fit -Example 12-6 • If there were no preference, one would expect that each flavor would be selected with equal frequency. • In this case, the equal frequency is 100/5 = 20. • That is, approximately 20 people would select each flavor.

7. 12-2 Test for Goodness of Fit - Example 12-7 • The frequencies obtained from the sample are called observed frequencies. • The frequencies obtained from calculations are called expected frequencies. • Table for the test is shown next.

8. 12-2 Test for Goodness of Fit -Example 12-8

9. 12-2 Test for Goodness of Fit -Example 12-9 • The observed frequencies will almost always differ from the expected frequencies due to sampling error. • Question: Are these differences significant, or are they due to chance? • The chi-square goodness-of-fit test will enable one to answer this question.

10. 12-2 Test for Goodness of Fit - Example 12-10 • The appropriate hypotheses for this example are: • H0: Consumers show no preference for flavors of the fruit soda. • H1: Consumers show a preference. • The d. f. for this test is equal to the number of categories minus 1.

11. 12-2 Test for Goodness of Fit - Formula 12-11    2 O E    2 E   d . f . number of categories 1  O observed frequency  E expected frequency

12. 12-2 Test for Goodness of Fit -Example 12-12 • Is there enough evidence to reject the claim that there is no preference in the selection of fruit soda flavours? Let  = 0.05. • Step 1: State the hypotheses and identify the claim.

13. 12-2 Test for Goodness of Fit - Example 12-13 • H0: Consumers show no preference for flavours (claim). • H1: Consumers show a preference. • Step 2: Find the critical value. The d. f. are 5 – 1 = 4 and  = 0.05. Hence, the critical value = 9.488.

14. 12-2 Test for Goodness of Fit -Example 12-14 • Step 3: Compute the test value. = (32 – 20)2/20 + (28 is – 20)2/20 + … + (10 – 20)2/20 = 18.0. • Step 4:Make the decision. The decision is to reject the null hypothesis, since 18.0 > 9.488.

15. 12-2 Test for Goodness of Fit - Example 12-15 • Step 5: Summarize the results. There is enough evidence to reject the claim that consumers show no preference for the flavours.

16. 12-2 Test for Goodness of Fit -Example 12-16  9.488

17. 12-2 Test for Goodness of Fit - Example 12-17 • The advisor of an ecology club at a large college believes that the group consists of 10% freshmen, 20% sophomores, 40% juniors and 30% seniors. The membership for the club this year consisted of 14 freshmen, 19 sophomores, 51 juniors and 16 seniors. At  = 0.10, test the advisor’s conjecture.

18. 12-2 Test for Goodness of Fit - Example 12-18 • Step 1: State the hypotheses and identify the claim. • H0: The club consists of 10% freshmen, 20% sophomores, 40% juniors, and 30% seniors (claim) • H1: The distribution is not the same as stated in the null hypothesis.

19. 12-2 Test for Goodness of Fit -Example 12-19 • Step 2: Find the critical value. The d. f. are 4 – 1 = 3 and  = 0.10. Hence, the critical value = 6.251. • Step 3: Compute the test value. = (14 – 10)2/10 + (19 – 20)2/20 + … + (16 – 30)2/30 = 11.208.

20. 12-2 Test for Goodness of Fit -Example 12-20 • Step 4:Make the decision. The decision is to reject the null hypothesis, since 11.208 > 6.251. • Step 5: Summarize the results. There is enough evidence to reject the advisor’s claim.

21. 12-3 Tests Using Contingency Tables 12-21 • When data can be tabulated in table form in terms of frequencies, several types of hypotheses can be tested using the chi-square test. • Two such tests are the independence of variables test and the homogeneity of proportions test.

22. 12-3 Tests Using Contingency Tables 12-22 • The test of independence of variables is used to determine whether two variables are independent when a single sample is selected.

23. 12-3 Test for Independence - Example 12-23 • Suppose a new postoperative procedure is administered to a number of patients in a large hospital. • Question: Do the doctors feel differently about this procedure from the nurses, or do they feel basically the same way? • Data is on the next slide.

24. 12-3 Test for Independence -Example 12-24

25. 12-3 Test for Independence - Example 12-25 • The null and the alternative hypotheses are as follows: • H0: The opinion about the procedure is independent of the profession. • H1: The opinion about the procedure is dependent on the profession.

26. 12-3 Test for Independence -Example 12-26 • If the null hypothesis is not rejected, the test means that both professions feel basically the same way about the procedure, and the differences are due to chance. • If the null hypothesis is rejected, the test means that one group feels differently about the procedure from the other.

27. 12-3 Test for Independence - Example 12-27 • Note: The rejection of the null hypothesis does not mean that one group favors the procedure and the other does not. • The test value is the 2 value (same as the goodness-of-fit test value). • The expected values are computed from: (row sum)(column sum)/(grand total).

28. 12-3 Test for Independence - Example 12-28

29. 12-3 Test for Independence -Example 12-29 • From the MINITAB output, the P-value = 0.000. Hence, the null hypothesis will be rejected (P<0.0005). • If the critical value approach is used, the degrees of freedom for the chi-square critical value will be (number of columns –1)(number of rows – 1). • d.f. = (3 –1)(2 – 1) = 2.