Chapter 12 Tests of Goodness of Fit and Independence Learning Objectives 1. Know how to conduct a goodness of fit test. 2. Know how to use sample data to test for independence of two variables. 3. Understand the role of the chi-square distribution in conducting tests of goodness of fit and independence. 4. Be able to conduct a goodness of fit test for cases where the population is hypothesized to have either a multinomial, a Poisson, or a normal probability distribution. 5. For a test of independence, be able to set up a contingency table, determine the observed and expected frequencies, and determine if the two variables are independent. Solutions: 1. Expected frequencies: e1 = 200 (.40) = 80, e2 = 200 (.40) = 80 e3 = 200 (.20) = 40 Actual frequencies: f1 = 60, f2 = 120, f3 = 20 = 9.21034 with k - 1 = 3 - 1 = 2 degrees of freedom Since = 35 > 9.21034 reject the null hypothesis. The population proportions are not as stated in the null hypothesis. 2. Expected frequencies: e1 = 300 (.25) = 75, e2 = 300 (.25) = 75 e3 = 300 (.25) = 75, e4 = 300 (.25) = 75 Actual frequencies: f1 = 85, f2 = 95, f3 = 50, f4 = 70 = 7.81473 with k - 1 = 4 - 1 = 3 degrees of freedom Since c2 = 15.33 > 7.81473 reject H0 We conclude that the proportions are not all equal. 3. H0 = pABC = .29, pCBS = .28, pNBC = .25, pIND = .18 Ha = The proportions are not pABC = .29, pCBS = .28, pNBC = .25, pIND = .18 Expected frequencies: 300 (.29) = 87, 300 (.28) = 84 300 (.25) = 75, 300 (.18) = 54 e1 = 87, e2 = 84, e3 = 75, e4 = 54 Actual frequencies: f1 = 95, f2 = 70, f3 = 89, f4 = 46 = 7.81 (3 degrees of freedom) Do not reject H0; there is no significant change in the viewing audience proportions. 4. Observed Expected Hypothesized Frequency Frequency Category Proportion (fi) (ei) (fi - ei)2 / ei Brown 0.30 177 151.8 4.18 Yellow 0.20 135 101.2 11.29 Red 0.20 79 101.2 4.87 Orange 0.10 41 50.6 1.82 Green 0.10 36 50.6 4.21 Blue 0.10 38 50.6 3.14 Totals: 506 29.51 = 11.07 (5 degrees of freedom) Since 29.51 > 11.07, we conclude that the percentage figures reported by the company have changed. 5. Observed Expected Hypothesized Frequency Frequency Category Proportion (fi) (ei) (fi - ei)2 / ei Full Service 1/3 264 249.33 0.86 Discount 1/3 255 249.33 0.13 Both 1/3 229 249.33 1.66 Totals: 748 2.65 = 4.61 (2 degrees of freedom) Since 2.65 < 4.61, there is no significant difference in preference among the three service choices. 6. Observed Expected Hypothesized Frequency Frequency Category Proportion (fi) (ei) (fi - ei)2 / ei News and Opinion 1/6 20 19.17 .04 General Editorial 1/6 15 19.17 .91 Family Oriented 1/6 30 19.17 6.12 Business/Financial 1/6 22 19.17 .42 Female Oriented 1/6 16 19.17 .52 African-American 1/6 12 19.17 2.68 Totals: 115 10.69 = 9.24 (5 degrees of freedom) Since 10.69 > 9.24, we conclude that there is a difference in the proportion of ads with guilt appeals among the six types of magazines. 7. Expected frequencies: ei = (1 / 3) (135) = 45 With 2 degrees of freedom, = 5.99 Do not reject H0; there is no justification for concluding a difference in preference exists. 8. H0: p1 = .03, p2 = .28, p3 = .45, p4 = .24 df = 3 = 11.34 Reject H0 if c2 > 11.34 Rating Observed Expected (fi - ei)2 / ei Excellent 24 .03(400) = 12 12.00 Good 124 .28(400) = 112 1.29 Fair 172 .45(400) = 180 .36 Poor 80 .24(400) = 96 2.67 400 400 c2 = 16.31 Reject H0; conclude that the ratings differ. A comparison of observed and expected frequencies show telephone service is slightly better with more excellent and good ratings. 9. H0 = The column variable is independent of the row variable Ha = The column variable is not independent of the row variable Expected Frequencies: A B C P 28.5 39.9 45.6 Q 21.5 30.1 34.4 = 7.37776 with (2 - 1) (3 - 1)= 2 degrees of freedom Since c2 = 7.86 > 7.37776 Reject H0 Conclude that the column variable is not independent of the row variable. 10. H0 = The column variable is independent of the row variable Ha = The column variable is not independent of the row variable Expected Frequencies: A B C P 17.5000 30.6250 21.8750 Q 28.7500 50.3125 35.9375 R 13.7500 24.0625 17.1875 = 9.48773 with (3 - 1) (3 - 1)= 4 degrees of freedom Since c2 = 19.78 > 9.48773 Reject H0 Conclude that the column variable is not independent of f the row variable. 11. H0 : Type of ticket purchased is independent of the type of flight Ha: Type of ticket purchased is not independent of the type of flight. Expected Frequencies: e11 = 35.59 e12 = 15.41 e21 = 150.73 e22 = 65.27 e31 = 455.68 e32 = 197.32 Observed Expected Frequency Frequency Ticket Flight (fi) (ei) (fi - ei)2 / ei First Domestic 29 35.59 1.22 First International 22 15.41 2.82 Business Domestic 95 150.73 20.61 Business International 121 65.27 47.59 Full Fare Domestic 518 455.68 8.52 Full Fare International 135 197.32 19.68 Totals: 920 100.43 = 5.99 with (3 - 1)(2 - 1) = 2 degrees of freedom Since 100.43 > 5.99, we conclude that the type of ticket purchased is not independent of the type of flight. 12. a. Observed Frequency (fij) Domestic European Asian Total Same 125 55 68 248 Different 140 105 107 352 Total 265 160 175 600 Expected Frequency (eij) Domestic European Asian Total Same 109.53 66.13 72.33 248 Different 155.47 93.87 102.67 352 Total 265 160 175 600 Chi Square (fij - eij)2 / eij Domestic European Asian Total Same 2.18 1.87 0.26 4.32 Different 1.54 1.32 0.18 3.04 c2 = 7.36 Degrees of freedom = 2 = 5.99 Reject H0; conclude brand loyalty is not independent of manufacturer. b. Brand Loyalty Domestic 125/265 = .472 (47.2%) ¬ Highest European 55/160 = .344 (34.4%) Asian 68/175 = .389 (38.9%) 13. Industry Major Oil Chemical Electrical Computer Business 30 22.5 17.5 30 Engineering 30 22.5 17.5 30 Note: Values shown above are the expected frequencies. = 11.3449 (3 degrees of freedom: 1 x 3 = 3) c2 = 12.39 Reject H0; conclude that major and industry not independent. 14. Expected Frequencies: e11 = 31.0 e12 = 31.0 e21 = 29.5 e22 = 29.5 e31 = 13.0 e32 = 13.0 e41 = 5.5 e42 = 5.5 e51 = 7.0 e52 = 7.0 e61 = 14.0 e62 = 14.0 Observed Expected Frequency Frequency Most Difficult Gender (fi) (ei) (fi - ei)2 / ei Spouse Men 37 31.0 1.16 Spouse Women 25 31.0 1.16 Parents Men 28 29.5 0.08 Parents Women 31 29.5 0.08 Children Men 7 13.0 2.77 Children Women 19 13.0 2.77 Siblings Men 8 5.5 1.14 Siblings Women 3 5.5 1.14 In-Laws Men 4 7.0 1.29 In-Laws Women 10 7.0 1.29 Other Relatives Men 16 14.0 0.29 Other Relatives Women 12 14.0 0.29 Totals: 200 13.43 = 11.0705 with (6 - 1) (2 - 1) = 5 degrees of freedom Since 13.43 > 11.0705. we conclude that gender is not independent of the most difficult person to buy for. 15. Expected Frequencies: e11 = 17.16 e12 = 12.84 e21 = 14.88 e22 = 11.12 e31 = 28.03 e32 = 20.97 e41 = 22.31 e42 = 16.69 e51 = 17.16 e52 = 12.84 e61 = 15.45 e62 = 11.55 Observed Expected Frequency Frequency Magazine Appeal (fi) (ei) (fi - ei)2 / ei News Guilt 20 17.16 0.47 News Fear 10 12.84 0.63 General Guilt 15 14.88 0.00 General Fear 11 11.12 0.00 Family Guilt 30 28.03 0.14 Family Fear 19 20.97 0.18 Business Guilt 22 22.31 0.00 Business Fear 17 16.69 0.01 Female Guilt 16 17.16 0.08 Female Fear 14 12.84 0.11 African-American Guilt 12 15.45 0.77 African-American Fear 15 11.55 1.03 Totals: 201 3.41 = 15.09 with (6 - 1) (2 - 1) = 5 degrees of freedom Since 3.41 < 15.09, the hypothesis of independence cannot be rejected. 34. a. Observed Frequency (fij) Pharm Consumer Computer Telecom Total Correct 207 136 151 178 672 Incorrect 3 4 9 12 28 Total 210 140 160 190 700 Expected Frequency (eij) Pharm Consumer Computer Telecom Total Correct 201.6 134.4 153.6 182.4 672 Incorrect 8.4 5.6 6.4 7.6 28 Total 210 140 160 190 700 Chi Square (fij - eij)2 / eij Pharm Consumer Computer Telecom Total Correct .14 .02 .04 .11 .31 Incorrect 3.47 .46 1.06 2.55 7.53 c2 = 7.85 Degrees of freedom = 3 = 7.81473 Do not reject H0; conclude order fulfillment is not independent of industry. b. The pharmaceutical industry is doing the best with 207 of 210 (98.6%) correctly filled orders. 17. Expected Frequencies: Part Quality Supplier Good Minor Defect Major Defect A 88.76 6.07 5.14 B 173.09 11.83 10.08 C 133.15 9.10 7.75 c2 = 7.96 = 9.48773 (4 degrees of freedom: 2 x 2 = 4) Do not reject H0; conclude that the assumption of independence cannot be rejected 18. Expected Frequencies: Party Affiliation Education Level Democratic Republican Independent Did not complete high school 28 28 14 High school degree 32 32 16 College degree 40 40 20 c2 = 13.42 = 13.2767 (4 degrees of freedom: 2 x 2 = 4) Reject H0; conclude that party affiliation is not independent of education level. 19. Expected Frequencies: e11 = 11.81 e12 = 8.44 e13 = 24.75 e21 = 8.40 e22 = 6.00 e23 = 17.60 e31 = 21.79 e32 = 15.56 e33 = 45.65 Observed Expected Frequency Frequency Siskel Ebert (fi) (ei) (fi - ei)2 / ei Con Con 24 11.81 12.57 Con Mixed 8 8.44 0.02 Con Pro 13 24.75 5.58 Mixed Con 8 8.40 0.02 Mixed Mixed 13 6.00 8.17 Mixed Pro 11 17.60 2.48 Pro Con 10 21.79 6.38 Pro Mixed 9 15.56 2.77 Pro Pro 64 45.65 7.38 Totals: 160 45.36 = 13.28 with (3 - 1) (3 - 1) = 4 degrees of freedom Since 45.36 > 13.28, we conclude that the ratings are not independent. 20. First estimate m from the sample data. Sample size = 120. Therefore, we use Poisson probabilities with m = 1.3 to compute expected frequencies. x Observed Frequency Poisson Probability Expected Frequency Difference (fi - ei) 0 39 .2725 32.700 6.300 1 30 .3543 42.516 -12.516 2 30 .2303 27.636 2.364 3 18 .0998 11.976 6.024 4 or more 3 .0430 5.160 - 2.160 = 7.81473 with 5 - 1 - 1 = 3 degrees of freedom Since c2 = 9.0348 > 7.81473 Reject H0 Conclude that the data do not follow a Poisson probability distribution. 21. With n = 30 we will use six classes with 16 2/3% of the probability associated with each class. = 22.80 s = 6.2665 The z values that create 6 intervals, each with probability .1667 are -.98, -.43, 0, .43, .98 z Cut off value of x -.98 22.8 - .98 (6.2665) = 16.66 -.43 22.8 - .43 (6.2665) = 20.11 0 22.8 + 0 (6.2665) = 22.80 .43 22.8 + .43 (6.2665) = 25.49 .98 22.8 + .98 (6.2665) = 28.94 Interval Observed Frequency Expected Frequency Difference less than 16.66 3 5 -2 16.66 - 20.11 7 5 2 20.11 - 22.80 5 5 0 22.80 - 25.49 7 5 2 25.49- 28.94 3 5 -2 28.94 and up 5 5 0 = 9.34840 with 6 - 2 - 1 = 3 degrees of freedom Since c2 = 3.20 £ 9.34840 Do not reject H0 The claim that the data comes from a normal distribution cannot be rejected. 22. Use Poisson probabilities with m = 1. c2 = 4.30 = 5.99147 (2 degrees of freedom) Do not reject H0; the assumption of a Poisson distribution cannot be rejected. 23. x Observed Poisson Probabilities Expected 0 15 .1353 13.53 1 31 .2707 27.07 2 20 .2707 27.07 3 15 .1804 18.04 4 13 .0902 9.02 5 or more 6 .0527 5.27 c2 = 4.98 = 7.77944 (4 degrees of freedom) Do not reject H0; the assumption of a Poisson distribution cannot be rejected. 24. = 24.5 s = 3 n = 30 Use 6 classes Interval Observed Frequency Expected Frequency less than 21.56 5 5 21.56 - 23.21 4 5 23.21 - 24.50 3 5 24.50 - 25.79 7 5 25.79 - 27.44 7 5 27.41 up 4 5 c2 = 2.8 = 6.25139 (3 degrees of freedom: 6 - 2 - 1 = 3) Do not reject H0; the assumption of a normal distribution cannot be rejected. 25. = 71 s = 17 n = 25 Use 5 classes Interval Observed Frequency Expected Frequency less than 56.7 7 5 56.7 - 66.5 7 5 66.5 - 74.6 1 5 74.6 - 84.5 1 5 84.5 up 9 5 c2 = 11.2 = 9.21034 (2 degrees of freedom) Reject H0; conclude the distribution is not a normal distribution. 26. Observed 60 45 59 36 Expected 50 50 50 50 c2 = 8.04 = 7.81473 (3 degrees of freedom) Reject H0; conclude that the order potentials are not the same in each sales territory. 27. Observed 48 323 79 16 63 Expected 37.03 306.82 126.96 21.16 37.03 Since 41.69 > 13.2767, reject H0. Mutual fund investors＇ attitudes toward corporate bonds differ from their attitudes toward corporate stock. 28. Observed 20 20 40 60 Expected 35 35 35 35 Since 31.43 > 7.81473, reject H0. The park manager should not plan on the same number attending each day. Plan on a larger staff for Sundays and holidays. 29. Observed 13 16 28 17 16 Expected 18 18 18 18 18 c2 = 7.44 = 9.48773 Do not reject H0; the assumption that the number of riders is uniformly distributed cannot be rejected. 30. Observed Expected Hypothesized Frequency Frequency Category Proportion (fi) (ei) (fi - ei)2 / ei Very Satisfied 0.28 105 140 8.75 Somewhat Satisfied 0.46 235 230 0.11 Neither 0.12 55 60 0.42 Somewhat Dissatisfied 0.10 90 50 32.00 Very Dissatisfied 0.04 15 20 1.25 Totals: 500 42.53 = 9.49 (4 degrees of freedom) Since 42.53 > 9.49, we conclude that the job satisfaction for computer programmers is different than the job satisfaction for IS managers. 31. Expected Frequencies: Quality Shift Good Defective 1st 368.44 31.56 2nd 276.33 23.67 3rd 184.22 15.78 c2 = 8.11 = 5.99147 (2 degrees of freedom) Reject H0; conclude that shift and quality are not independent. 32. Expected Frequencies: e11 = 1046.19 e12 = 632.81 e21 = 28.66 e22 = 17.34 e31 = 258.59 e32 = 156.41 e41 = 516.55 e42 = 312.45 Observed Expected Frequency Frequency Employment Region (fi) (ei) (fi - ei)2 / ei Full-Time Eastern 1105 1046.19 3.31 Full-time Western 574 632.81 5.46 Part-Time Eastern 31 28.66 0.19 Part-Time Western 15 17.34 0.32 Self-Employed Eastern 229 258.59 3.39 Self-Employed Western 186 156.41 5.60 Not Employed Eastern 485 516.55 1.93 Not Employed Western 344 312.45 3.19 Totals: 2969 23.37 = 7.81 with (4 - 1) (2 - 1) = 3 degrees of freedom Since 23.37 > 7.81, we conclude that employment status is not independent of region. 33. Expected frequencies: Loan Approval Decision Loan Offices Approved Rejected Miller 24.86 15.14 McMahon 18.64 11.36 Games 31.07 18.93 Runk 12.43 7.57 c2 = 2.21 = 7.81473 (3 degrees of freedom) Do not reject H0; the loan decision does not appear to be dependent on the officer. 34. a. Observed Frequency (fij) Never Married Married Divorced Total Men 234 106 10 350 Women 216 168 16 400 Total 450 274 26 750 Expected Frequency (eij) Never Married Married Divorced Total Men 210 127.87 12.13 350 Women 240 146.13 13.87 400 Total 450 274 26 750 Chi Square (fij - eij)2 / eij Never Married Married Divorced Total Men 2.74 3.74 .38 6.86 Women 2.40 3.27 .33 6.00 c2 = 12.86 Degrees of freedom = 2 = 9.21 Reject H0; conclude martial status is not independent of gender. b. Martial Status Never Married Married Divorced Men 66.9% 30.3% 2.9% Women 54.0% 42.0% 4.0% Men 100 - 66.9 = 33.1% have been married Women 100 - 54.0 = 46.0% have been married 35. Expected Frequencies: = 9.48773 (4 degrees of freedom) Since 9.76 < 9.48773, reject H0. Banking tends to have lower P/E ratios. We can conclude that industry type and P/E ratio are related. 36. Expected Frequencies: Days of the Week County Sun Mon Tues Wed Thur Fri Sat Total Urban 56.7 47.6 55.1 56.7 60.1 72.6 44.2 393 Rural 11.3 9.4 10.9 11.3 11.9 14.4 8.8 78 Total 68 57 66 68 72 87 53 471 c2 = 6.20 = 12.5916 (6 degrees of freedom) Do not reject H0; the assumption of independence cannot be rejected. 37. = 76.83 s = 12.43 Interval Observed Frequency Expected Frequency less than 62.54 5 5 62.54 - 68.50 3 5 68.50 - 72.85 6 5 72.85 - 76.83 5 5 76.83 - 80.81 5 5 80.81 - 85.16 7 5 85.16 - 91.12 4 5 91.12 up 5 5 c2 = 2 = 11.0705 (5 degrees of freedom) Do not reject H0; the assumption of a normal distribution cannot be rejected. 38. Expected Frequencies: Los Angeles San Diego San Francisco San Jose Total Occupied 165.7 124.3 186.4 165.7 642 Vacant 34.3 25.7 38.6 34.3 133 Total 200.0 150.0 225.0 200.0 775 = 7.81473 with 3 degrees of freedom Since c2 = 7.78 £ 7.81473 Do not reject H0. We cannot conclude that office vacancies are dependent on metropolitan area, but it is close: the p-value is slightly larger than .05. 39. a. x Observed Frequencies Binomial Prob. n = 4, p = .30 Expected Frequencies 0 30 .2401 24.01 1 32 .4116 41.16 2 25 .2646 26.46 3 10 .0756 7.56 4 3 .0081 .81 100 100.00 The expected frequency of x = 4 is .81. Combine x = 3 and x = 4 into one category so that all expected frequencies are 5 or more. x Observed Frequencies Expected Frequencies 0 30 24.01 1 32 41.16 2 25 26.46 3 or 4 13 8.37 100 100.00 b. With 3 degrees of freedom, = 7.81473. Reject H0 if c2 > 7.81473. Do not reject H0; conclude that the assumption of a binomial distribution cannot be rejected.
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