1. ©2006 Thomson/South-Western 1 Chapter 12 – Analysis of Categorical Data Slides prepared by Jeff Heyl Lincoln University ©2006 Thomson/South-Western Concise Managerial Statistics Concise Managerial Statistics KVANLI PAVUR KEELING KVANLI PAVUR KEELING

2. ©2006 Thomson/South-Western 2 wheren =sample size and x =the observed number of successes in the sample p=estimate of p =proportion of sample having a specified attribute =^xn Point Estimate for a Population Proportion

3. ©2006 Thomson/South-Western 3 Confidence Interval for a Population Proportion (1-  ) 100% confidence interval for large samples p - Z  /2 to p + Z  /2 ^ ^ p(1 - p) n - 1 ^^ p(1 - p) n - 1 ^^ sp =sp =sp =sp = ^ Estimated standard error p(1 - p) n - 1 ^^

4. ©2006 Thomson/South-Western 4 Choosing the Sample Size With 95% confidence E = 1.96 p(1 - p) n - 1 ^^ With E =.02 E =.02 = 1.96 (.0867)(.9133) n - 1 n = 761.5 (round to 762) n = + 1 Z  /2 p(1 - p) E 2 ^ ^ 2

5. ©2006 Thomson/South-Western 5 Curve of Values Figure 12.1 |.5 1 p ^ p(1 − p) ^^.25 –

6. ©2006 Thomson/South-Western 6 Hypothesis Testing Using a Large Sample H o : p = p o H a : p ≠ p o reject H o if |Z| > Z  /2 Two-Tailed Test where Z =Z =Z =Z = p - p o p o (1 - p o ) n ^

7. ©2006 Thomson/South-Western 7 Hypothesis Testing Using a Large Sample H o : p ≤ p o H a : p > p o reject H o if Z > Z  One-Tailed Test H o : p ≥ p o H a : p < p o reject H o if Z < -Z  Z =Z =Z =Z = (point estimate) - (hypothesized value) (stanard deviation of point estimator)

8. ©2006 Thomson/South-Western 8 Z Curve Showing p-Value Z * = 2.41 Z p-value= area =.5 −.4920 =.008 Figure 12.2

9. ©2006 Thomson/South-Western 9 Z Curve Showing p-Value Z * = 2.92 Z p-value= 2(area) = 2(.5 −.4982) = 2(.0018) =.0036 Figure 12.3

10. ©2006 Thomson/South-Western 10 Computer Z Test Figure 12.4 (a)

11. ©2006 Thomson/South-Western 11 Computer Z Test Figure 12.4 (b)

12. ©2006 Thomson/South-Western 12 Comparing Two Population Proportions (Large, Independent Samples) Standard error s p -p = + 12 ^^ p 1 (1 - p 1 ) n 1 - 1 ^^ p 2 (1 - p 2 ) n 2 - 1 ^^ Confidence interval (p 1 - p 2 ) - Z  /2 ^^ p 1 (1 - p 1 ) n 1 - 1 ^^ p 2 (1 - p 2 ) n 2 - 1 ^^ + to (p 1 - p 2 ) + Z  /2 ^^ p 1 (1 - p 1 ) n 1 - 1 ^^ p 2 (1 - p 2 ) n 2 - 1 ^^ +

13. ©2006 Thomson/South-Western 13 Hypothesis Test for Two Population Proportions Two-Tailed Test H o : p 1 = p 2 H a : p 1 ≠ p 2 reject H o if |Z| > Z  /2 For test statistic Z =Z =Z =Z = p(1 - p) n 1 − − p(1 - p) n 2 −− + p 1 - p 2 ^^

14. ©2006 Thomson/South-Western 14 Hypothesis Test for Two Population Proportions For test statistic One-Tailed Test H o : p 1 ≤ p 2 H a : p 1 > p 2 reject H o if Z > Z  H o : p 1 ≥ p 2 H a : p 1 < p 2 reject H o if Z < -Z  Z =Z =Z =Z = p(1 - p) n 1 − − p(1 - p) n 2 −− + p 1 - p 2 ^^

15. ©2006 Thomson/South-Western 15 Z Curve Showing p-Value Figure 12.5 Z * = −.68 Z p-value= 2(area) = 2(.5 −.2517) =.4966

16. ©2006 Thomson/South-Western 16 Z Curve Showing p-Value Z * = − 3.29 Z p-value= area =.5 −.4995 =.0005 Figure 12.6

17. ©2006 Thomson/South-Western 17 Computer Solution - Z Test Figure 12.7 (a)

18. ©2006 Thomson/South-Western 18 Computer Solution - Z Test Figure 12.7 (b)

19. ©2006 Thomson/South-Western 19 Chi-Square Distribution 2222  a, df 2 Area = a Figure 12.8

20. ©2006 Thomson/South-Western 20 Chi-Square Distribution 2222 18.5493 Area =.1 Figure 12.9

21. ©2006 Thomson/South-Western 21 The Multinomial Situation Assumptions 1.The experiment consists of n independent repetitions (trials) 2.Each trial outcome falls in exactly one of k categories 3.The probabilities of the k outcomes are denoted by p 1, p 2,..., p k and remain the same on each trial. Further: p 1 + p 2, +...+ p k = 1

22. ©2006 Thomson/South-Western 22 Hypothesis Testing for the Multinomial Situation 1. The summation is across all categories (outcomes) 2. The O’s are the observed frequencies in each category using the sample 3. The E’s are the expected frequencies in each category if H o is true 4. The df for the chi-square statistic are k-1, where k is the number of categories  2 = ∑ (O - E) 2 Ewhere:

23. ©2006 Thomson/South-Western 23 Chi-Square Test of Independence Null and Alternative Hypothesis H o : the classifications are independent H a : the classifications are dependent Estimating the Expected Frequencies E =E =E =E = ^ (row total for this cell)(column total for this cell) n

24. ©2006 Thomson/South-Western 24 Expected Frequencies Figure 12.10 Classification 1 Classification 2 1 2 3 4c r 123 C1C1C1C1 C2C2C2C2 C3C3C3C3 C4C4C4C4 CcCcCcCc R1R1R1R1 R2R2R2R2 R3R3R3R3 RrRrRrRr E =E =E =E = ^ R2C3R2C3nnR2C3R2C3nnn

25. ©2006 Thomson/South-Western 25 Chi-Square Test of Independence The Testing Procedure H o :the row and column classifications are independent H a :the row and column classifications are dependent reject H o if  2 >  2 ,df where df = (r-1)(c-1)  2 = ∑ (O - E) 2 E

26. ©2006 Thomson/South-Western 26 Chi-Square Test of Independence 1.The summation is over all cells of the contingency table consisting of r rows and c columns 4.The degrees of freedom are df = (r-1)(c-1) 2.O is the observed frequency 3.E is the expected frequency ^ E =E =E =E = ^ (total of all cells) total of row in which the cell lies total of column in which the cell lies