商务与经济统计习题答案（第8版中文版）SBE8-SM12.doc

精品资料商务与经济统计习题答案〔第8版，中文版〕SBE8-SM12Chapter 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 sle 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 .2 11.29 Red 0.20 79 .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 pany 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 parison 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 Expe。