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# answer key of chapter 7_first - 7.6 7.12 2 2 C’1 “2 9...

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Unformatted text preview: 7.6 7.12 2 2 _ C’1 “2 _ 9 16 =./25 = 5 ”Erin" J's—1 a m *no' ' The sampling distribution of 351 — E2 is approximately normal by the Central Limit Theorem since n1 2 30 and n2 2 30. \$1.32 = :11 — #2 = 10 x1-x2--11.1 113-22 -10 ’71-"2 r. — :2 =15.5 ~ 26.6 = —11.1 Yes. it appears that E1 — x2 w-11.1 contradicts the null hypothesis H9: #1 — p2 = 10. The rejection region requires (IQ = .025 = .052 in each tail of the z-distribution. From Table IV, Appendix B, 2:025 = 1.96. The rejection region is z < -1.96 or z > 1.96. Ha:i“-1_r”2¢10 _ . .5 31.1.0. _ . . _The test statistic is-z = £‘_x2_)__.;£1_55__2.5‘W—_ = —-42..2.= E "+_' .___2__--. smashiregios is :3 '<'" 41950.}; 1.96. (Refer "minds Since the observed value of the test statistic falls in the rejection region (_2= -_-42_. 2 < —l. 96) H9 is rejected There' is sufﬁcient evidence to indicate the difference in the populanonmeansisnoteqnaltomata— ..05_- " - The form of the conﬁdence interval is: . For confidence. .coefﬁCient 95 n: = 1 — .95 = .05 and (Liz. = .052’2 = .025. From Table IV AppendixB',-z_1023— 1:96. Theconﬁdenceintervalis: -___|+ 9 16 — 26.6 .6 - --11-+_-. —-,1.20 — ._12 We are 95% conﬁdent that the difference in the two means is between “12.08 and —10.12. The conﬁdence interval gives more information. The ﬁrst pepulation .is the set of responses for all business students who have access to lecture notes and the second- population is the set .of responses for all business students not having access to lecture notes: To determine if there is a difference in the menu response of the two groups, we test: Hem =...0. n2¢0 7.18 The test statistic is_z._= -(i—l—i)—_ = (8'48. -- 3780) O = J9 . ' "The rejection region requires and": .01f2.= .005 ineaeh tail of theiz'edistribntion. From :' Table IV, Appendix 33-53005 5 27.58; ' The rejection region is z .<. . —2.58_ or__'z" > 2.58. Since the observed value of the test statistic does not fall in the rejection region (2: = 2.19 a 2.58);“HD is not rejected: There is insufﬁcient evidence to indicate a difference in the mean response sf that“ was its: .=3'. '01:. Brzms =. 2.535. the".conndence-.iaterva1-is:. " ‘ resonance coefﬁcwm .994: ep-01aeci-ai2'ezoiir2'e .005. From Table Iv, Appendix —— '=" .4 '-- 7.‘ i 2. 8' ' .- + n: (8 8 80) 5 86 + 35 ='>'.68' a 1801219”('—.I2t-,-:'l.-481) _- '- We are 99% conﬁdent that the difference in the mean response between the trio groups is between —.121 and 11.481. A 95 % conﬁdence interval would be smaller than the 99% conﬁdence intervat. The 2: value used in the 95% conﬁdence interval is {025 = 1.96 compared with the z value used in the - 99 % conﬁdence interval: of-zms _=- 2.58. To determine if the mean annual percentage turnover for U.S. plants exceeds that for Japanese" plants, we test: H0: #1‘HQ=0 H3: ul—n2>0 The test statistic is r = 4.46 (from printout). The rejection region requires or = .05 in the upper tail of the r~distribution with df 2 321 + it; _ 2 = 5 + 5 — 2 = 8. From Table VI, Appendix B, {05 = 1.860. The rejection region is t > 1.860. Since the observed value of the test statistic falls in the rejection region (I = 4.46 > 1.860), H0 is rejected. There is sufﬁcient evidence to indicate the mean annual percentage turnover for U.S. plants exceeds that for Japanese plants at o: = .05. The observed signiﬁcance is .0031f2 = .00155. Since the p-value is so small, there is evidence to reject H0 for o: > .005. The necessary assumptions are: 1. Both sampled populations are approximately normal. 2. The population variances are equal. 3. The samples are randomly and independently sampled. There is no indication that the populations are not normal. Both sample variances are similar, so there is no evidence the population variances are unequal. There is no indication the assumptions are not valid. 7.22 a. 13- #D=#1-p2 c. For conﬁdence coefﬁcient .95, o: = .05 and 0:10 = .025. From Table VI, Appendix B, with df = nD — 1 = 6 - 1 = 5, 11025 = 2.571. The conﬁdence interval is: S — o It) i tar: = 2 i 2.571é =9 2 i 1.484 = (.516, 3.484) a: rt (1. H0: #1) = 0 Ha: [in 9'5 0 . . . ID 2 _ The test statistic ts t z = — 3.46 EDIE «EH/6. The rejection region requires orfz = .05t’2 = .025 in each tail of the z-distribution with df = ”D -— 1 = 6 - 1 = 5. From Table VI, Appendix B, {025 = 2.571. The rejection region is I < —2.5710rr > 2.571. Since the observed value of the test statistic falls in the rejection region (3.46 > 2.571), H0 is rejected. There is sufﬁcient evidence to indicate that the mean difference is different from 0 at o: = .05. 7.28 a. To determine whether male students’ attitudes toward their fathers differ from their attitudes toward their mothers, on average, we test: in. From the printout. the value of the test statistic is I = ﬁ1.08. The p-value is p = .3033. Since the p-value is not less than tr = .05, H0 is not rejected. There is insufﬁcient evidence to indicate male students’ attitudes toward their fathers differ from their attitudes toward their mothers. on average at o: = .05. c. In order for the above test to be valid, we must assume that I. The population of differences is normal 2. The differences are randomly selected 7.46 351’; 3 Let a? = variance of the solid- A stem-and-leaf display of the differences is: Character Stem-and-Leaf Display Stem-and-leaf of Diff N Leaf Unit = 0.10 13 1 -2 0 (6) -1 000000 6 0 00 4 1 0000 We know that the data are not normal, because there are only 4 different values for the differences. Although the stem-and-leaf display does not look particularly normal, it is rather mound-shaped. The Heat procedure works pretty well even if the data are not exactly normal. First, find the sample sizes needed for width 5, or bound 2.5. For conﬁdence coefﬁcient .9, o: : 1 — .9 = .1 and M2 = .1f2 = .05. From Table IV, Appendix B, 2.05 : 1.645. , 2 (1.6459003 + 102) 32 2.52 . 2 2' — W = 86.59 .._.. 87 Thus, the necessary sample size from each p0pulation is 87 . Therefore, sufﬁcient funds have not been allocated to meet the speciﬁcations since :11 = n2 = 100 are large enough samples. a. To determine if the variance for population 2 is greater than that for population 1, we test: Ho: 0i = 0% Iii: 0% <1 0% 52 92 The test statistic is F = .31. = 2‘972 = 4.29 512 1.43592 The rejection region requires a = .05 in the upper tail of the F-distribution with r1 = n2 - l = 5 -— I = 4and r2 = r11 — 1 = 6 — 1 = 5. From Table IX, Appendix B. F.05 = 5.19. The rejection region is F > 5.19. Since the observed value of the test statistic does not fail in the rejection region (F = 4.29 :l 5.19), H0 is not rejected. There is insufﬁcient evidence to indicate the variance for population 2 is greater than that for population 1 at or = .05. b. The p-value is P(F a 4.29). From Tables VIII-and IX, with v1 = 4 and v2 = 5, .05 < P(F 2 4.29) < .10 There is no evidence to reject H0 for o: < .05 but there is evidence to reject H0 for o: = .10. waste generation rates for industrialized countries and 0% = variance of the solid-waste generation rates for middle~income countries. To determine if the variances differ, we test: Hg: —E=1 ”a s H- ”I 3. _;—e1 2 ”a The test statistic is F = \$.03 (freni printeut). The p-value is p = 0.0300 {frern printeut). Since the p-value is net less than n: (p = .0300 r. .05}. Hg is net rejected. There is insufﬁcient evidence te indicate the variances fer the twe greups differ at ct = .05. Frent the test, there is insufﬁcient evidence te indicate the variances are different at er = .05. Hewever, the p-value ef .0800 is quite clese te e: = .05. If we cenclude that the variances are equal, we have a chance ef cernrnitting a Type II errer. Witheut specifying a fit-ted alternate value fer the ratie ef the variances, we cannet centpute the prehahilityr ef cennnirting a Type II errer, 5. if we weuld inﬂate the value ef er. (say te the value cf .25 er higher). then we knew that the value ef B will decrease. If we still de net reject He with the inﬂated value ef es. then we can have cenﬁdence that the variances are equal. In this prehletn. hewever, the p~value is quite small (p I .0800). Thus. the assutnptien ef equal variances is questienahle and the twe-santple t-test might net be apprepriate. ...
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answer key of chapter 7_first - 7.6 7.12 2 2 C’1 “2 9...

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