1. 1 Chapter 17 Statistical Inference For Frequency Data IThree Applications of Pearson’s  2 Testing goodness of fit Testing independence Testing equality of proportions

2. 2 A. Testing Goodness of Fit 1.Statistical hypotheses H 0 : O Pop 1 = E Pop 1,..., O Pop k = E Pop k H 1 : O Pop j ≠ E Pop j for some j and j 2.Randomization Plan One random sample of n elements Each element is classified in terms of membership in one of k mutually exclusive categories

3. 3 B. Testing Independence 1.Statistical hypotheses H 0 : p(A and B) = p(A)p(B) H 1 : p(A and B) ≠ p(A)p(B) 2.Randomization Plan One random sample of n elements Each element is classified in terms of two variables, denoted by A and B, where each variable has two or more categories.

4. 4 C. Testing Equality of Proportions 1.Statistical hypotheses H 0 : p 1 = p 2 =... = p c H 1 : p j ≠ p j for some j and j 2.Randomization Plan c random samples, where c ≥ 2 For each sample, elements are classified in terms of membership in one of r = 2 mutually exclusive categories

5. 5 IITesting Goodness of Fit A. Chi-Square Distribution

6. 6 B.Pearson’s chi-square statistic 1.O j and E j denote, respectively, observed and expected frequencies. k denotes the number of categories. 2.Critical value of chi square is with = k – 1 degrees of freedom.

7. 7 C.Grade-Distribution Example 1.Is the distribution of grades for summer-school students in a statistics class different from that for the fall and spring semesters? Fall and SpringSummer Grade Proportion Obs. frequency A.1215 B.2321 C.4730 D.136 F.050 1.0024

8. 8 2. The statistical hypotheses are H 0 : O Pop 1 = E Pop 1,..., O Pop 5 = E Pop 5 H 1 : O Pop j ≠ E Pop j for some j and j 3.Pearson’s chi-square statistic is 4.Critical value of chi square for  =.05, k = 5 categories, and = 5 – 1 = 4 degrees of freedom is

9. 9 Table 1. Computation of Pearson’s Chi-Square for n = 72 Summer-School Students (1)(2) (3) (4) (5)(6) GradeO j p j np j = E j O j – E j A15.1272(.12) = 8.66.44.763 B21.2372(.23) =16.64.41.166 C30.4772(.47) = 33.8–3.80.427 D6.1372(.13) = 9.4–-3.41.230 F0.0572(.05) = 3.6–3.63.600 721.0072.0  2 = 11.186* *p <.025

10. 10 5.Degrees of freedom when e parameters of a theoretical distribution must be estimated is k – 1 – e. D.Practical Significance 1.Cohen’s w where and denote, the observed and expected proportions in the jth category.

11. 11 2.Simpler equivalent formula for Cohen’s 3.Cohen’s guidelines for interpreting w 0.1 is a small effect 0.3 is a medium effect 0.5 is a large effect

12. 12 E.Yates’ Correction 1.When = 1, Yates’ correction can be applied to make the sampling distribution of the test statistic for O j – E j, which is discrete, better approximate the chi-square distribution.

13. 13 F.Assumptions of the Goodness-of-Fit Test 1.Every observation is assigned to one and only one category. 2.The observations are independent 3.If = 1, every expected frequency should be at least 10. If > 1, every expected frequency should be at least 5.

14. 14 IIITesting Independence A. Statistical Hypotheses H 0 : p(A and B) = p(A)p(B) H 1 : p(A and B) ≠ p(A)p(B) B.Chi-Square Statistic for an r  c Contingency Table with i = 1,..., r Rows and j = 1,..., c Columns

15. 15 C. Computational Example: Is Success on an Employment-Test Item Independent of Gender? Observed Expected b 1 b 2 FailPassFailPass a 1 Man841810288.913.1 a 2 Women93810188.112.9 17726203

16. 16 D.Computation of expected frequencies 1.A and B are statistically independent if p(a i and b j ) = p(a i )p(b j ) 2.Expected frequency, for the cell in row i and column j

17. 17 Observed Expected b 1 b 2 a 1 841810288.913.1 a 2 93810188.112.9 17726203

18. 18 E.Degrees of Freedom for an r  c Contingency Table d f = k – 1 – e = rc – 1 – [(r – 1) + (c – 1)] = rc – 1 – r + 1 – c + 1 = rc – r – c + 1 = (r – 1)(c – 1) = (2 – 1)(2 – 1) = 1

19. 19 F.Strength of Association and Practical Significance where s is the smaller of the number of rows and columns. 1. Cramér’s

20. 20 3.For a contingency table, an alternative formula for is 2.Practical significance, Cohen’s ŵ

21. 21 G.Three-By-Three Contingency Table 1. Motivation and education of conscientious objectors during WWII HighGrade CollegeSchoolSchoolTotal Coward12253572 Partly Coward 1923 30 72 Not Coward 715624151 Total10210489295

22. 22 2.Strength of Association, Cramér’s 3.Practical significance

23. 23 H.Assumptions of the Independence Test 1.Every observation is assigned to one and only one cell of the contingency table. 2.The observations are independent 3.If = 1, every expected frequency should be at least 10. If > 1, every expected frequency should be at least 5.

24. 24 IVTesting Equality of c ≥ 2 Proportions A. Statistical Hypotheses H 0 : p 1 = p 2 =... = p c H 1 : p j ≠ p j for some j and j 1.Computational example: three samples of n = 100 residents of nursing homes were surveyed. Variable A was age heterogeneity in the home; variable B was resident satisfaction.

25. 25 Table 2. Nursing Home Data Age Heterogeneity Low b 1 Medium b 2 High b 3 Satisfieda 1 O = 56O = 58O = 38 E = 50.67E = 50.67E = 50.67 Not Satisfieda 2 O = 44O = 42O = 52 E = 49.33E = 49.33E = 49.33

26. 26 B.Assumptions of the Equality of Proportions Test 1.Every observation is assigned to one and only one cell of the contingency table.

27. 27 2.The observations are independent 3.If = 1, every expected frequency should be at least 10. If > 1, every expected frequency should be at least 5. C.Test of Homogeneity of Proportions 1.Extension of the test of equality of proportions when variable A has r > 2 rows

28. 28 2.Statistical hypotheses for columns j and j'