1. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.1 Chapter 16 Chi-Squared Tests

2. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.2 A Common Theme… What to do?Data Type? Number of Categories? Statistical Technique: Describe a population NominalTwo or more X2 goodness of fit test Compare two populations NominalTwo or more X2 test of a contingency table Compare two or more populations Nominal-- X2 test of a contingency table Analyze relationship between two variables Nominal-- X2 test of a contingency table One data type… …Two techniques

3. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.3 Two Techniques… The first is a goodness-of-fit test applied to data produced by a multinomial experiment, a generalization of a binomial experiment and is used to describe one population of data. The second uses data arranged in a contingency table to determine whether two classifications of a population of nominal data are statistically independent; this test can also be interpreted as a comparison of two or more populations. In both cases, we use the chi-squared ( ) distribution.

4. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.4 The Multinomial Experiment… Unlike a binomial experiment which only has two possible outcomes (e.g. heads or tails), a multinomial experiment: Consists of a fixed number, n, of trials. Each trial can have one of k outcomes, called cells. Each probability p i remains constant. Our usual notion of probabilities holds, namely: p 1 + p 2 + … + p k = 1, and Each trial is independent of the other trials.

5. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.5 Chi-squared Goodness-of-Fit Test… We test whether there is sufficient evidence to reject a specified set of values for p i. To illustrate, our null hypothesis is: H 0 : p 1 = a 1, p 2 = a 2, …, p k = a k (where a 1, a 2, …, a k are the values of interest) Our research hypothesis is: H 1 : At least one p i ≠ a i

6. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.6 Chi-squared Goodness-of-Fit Test… The test builds on comparing actual frequency and the expected frequency of occurrences in all the cells. Example 16.1… We compare market share before and after an advertising campaign to see if there is a difference (i.e. if the advertising was effective in improving market share). H 0 : p 1 = a 1, p 2 = a 2, …, p k = a k Where a i is the market share before the campaign. If there was no change, we’d expect H 0 to not be rejected. If there is evidence to reject H 0 in favor of: H 1 : At least one p i ≠ a i, what’s a logical conclusion?

7. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.7 Example 16.1… Market shares before the advertising campaign… Company A – 45% Company B – 40% All Others – 15 % 200 customers surveyed after the campaign. The results: Company A – 102 customers preferred their product. Company B – 82 customers… All Others – 16 customers. Before the campaign, we’d expect 45% of 200 customers (i.e. 90 customers) to prefer company A’s product. After the campaign, we observe its 102 customers. Does this mean the campaign was effective? (i.e. at a 5% significance level). IDENTIFY

8. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.8 Example 16.1… Observed Frequency A B Expected Frequency A B Are these changes statistically significant?

9. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.9 Example 16.1… Our null hypothesis is: H 0 : p CompanyA =.45, p CompanyB =.40, p Others =.15 (i.e. the market shares pre-campaign), and our alternative hypothesis is: H 1 : At least one p i ≠ a i In order to complete our hypothesis testing we need a test statistic and a rejection region… IDENTIFY

10. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.10 Chi-squared Goodness-of-Fit Test… Our Chi-squared goodness of fit test statistic is given by: Note: this statistic is approximately Chi-squared with k–1 degrees of freedom provided the sample size is large. The rejection region is: observed frequency expected frequency

11. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.11 Example 16.1… In order to calculate our test statistic, we lay-out the data in a tabular fashion for easier calculation by hand: Company Observed Frequency Expected Frequency Delta Summation Component fifi eiei (f i – ei)(f i – e i ) 2 /e i A10290121.60 B828020.05 Others1630-146.53 Total200 8.18 Check that these are equal COMPUTE

12. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.12 Example 16.1… Our rejection region is: Since our test statistic is 8.18 which is greater than our critical value for Chi-squared, we reject H 0 in favor of H 1, that is, “There is sufficient evidence to infer that the proportions have changed since the advertising campaigns were implemented” INTERPRET

13. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.13 Example 16.1… Note: Table 5 in Appendix B does not allow for the direct calculation of, so we have to use Excel: COMPUTE

14. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.14 Example 16.1… Note: There are a couple of different ways to calculate the p-value of the test: p-value Computed manually from our table Computed directly from the data

15. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.15 Required Conditions… In order to use this technique, the sample size must be large enough so that the expected value for each cell is 5 or more. (i.e. n x p i ≥ 5) If the expected frequency is less than five, combine it with other cells to satisfy the condition.

16. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.16 Identifying Factors… Factors that Identify the Chi-Squared Goodness-of-Fit Test: e i =(n)(p i )

17. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.17 Chi-squared Test of a Contingency Table The Chi-squared test of a contingency table is used to: determine whether there is enough evidence to infer that two nominal variables are related, and to infer that differences exist among two or more populations of nominal variables. In order to use use these techniques, we need to classify the data according to two different criteria.

18. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.18 Example 16.2… The demand for an MBA program’s optional courses and majors is quite variable year over year. The research hypothesis is that the academic background of the students (i.e. their undergrad degrees) affects their choice of major. A random sample of data on last year’s MBA students was collected and summarized in a contingency table…data IDENTIFY

19. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.19 Example 16.2 The Data MBA Major Undergrad Degree AccountingFinanceMarketingTotal BA31131660 BEng816731 BBA12101739 Other105722 Total614447152

20. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.20 Example 16.2… Again, we are interesting in determining whether or not the academic background of the students affects their choice of MBA major. Thus our research hypothesis is: H 1 : The two variables are dependent Our null hypothesis then, is: H 0 : The two variables are independent.

21. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.21 Example 16.2… In this case, our test statistic is: (where k is the number of cells in the contingency table, i.e. rows x columns) Our rejection region is: where the number of degrees of freedom is (r–1)(c–1)

22. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.22 Example 16.2… In order to calculate our test statistic, we need the calculate the expected frequencies for each cell… The expected frequency of the cell in row i and column j is: COMPUTE

23. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.23 Contingency Table Set-up…

24. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.24 Example 16.2 COMPUTE MBA Major Undergrad Degree AccountingFinanceMarketingTotal BA31131660 BEng816 31 x 47 152 31 BBA12101739 Other105722 Total614447152 e 23 = (31)(47)/152 = 9.59 — compare this to f 23 = 7 Compute expected frequencies…

25. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.25 Example 16.2… We can now compare observed with expected frequencies… and calculate our test statistic: MBA Major Undergrad Degree AccountingFinanceMarketing BA3124.081317.371618.55 BEng812.44168.9779.59 BBA1215.651011.291712.06 Other108.8356.3776.80

26. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.26 Example 16.2… We compare = 14.70 with: Since our test statistic falls into the rejection region, we reject H 0 : The two variables are independent. in favor of H 1 : The two variables are dependent. That is, there is evidence of a relationship between undergrad degree and MBA major. INTERPRET

27. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.27 Example 16.2… We can also leverage the tools in Excel to process our data:data COMPUTE Tools > Data Analysis Plus > Contingency Table Compare p-value

28. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.28 Required Condition – Rule of Five… In a contingency table where one or more cells have expected values of less than 5, we need to combine rows or columns to satisfy the rule of five. Note: by doing this, the degrees of freedom must be changed as well.

29. Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 16.29 Identifying Factors… Factors that identify the Chi-squared test of a contingency table: