answer key of chapter 7-1

answer key of chapter 7-1 - 7.6 7.12 I E_fi 1.0_H H_The...

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Unformatted text preview: 7.6 7.12 _ I E __ _fi. 1.0. .__H _. H _The test statistic is-z = I; (15-5. 25-6) 19 2 2 01 a 9 16 __= +_: _ _=./_25=.5 “Ii-x2 —1 :22 100 + 100 The sampling distribution of 351 — E2 is approximately normal by the Central Limit Theorem since n1 2 30 and n2 2 30. nil—32 = #1 - #2 =10 fl — I: =15.5 ~ 26.6 = —11.1 w-11.1 contradicts the null hypothesis H9: til — p2 = 10. Yes, it appears that $1 — I; The rejection region requires all = .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::“-1_r”2#10 =' —-42..2'.f' - .5 _ _ . - meanness is :2 '<'" '—'i.9é'olr'%">' 1:96; (Refer have) Since the observed value of the test statistic falls in the rejection region (2 __= 5-422 < — 1.96), H9 is rejected. There is sufficient evidence to indibate the differénce in the 9013mm normal-to 10 3:0: = ' - -' - ' " ' ' The form of confidence interval is: .. .-Forconfidence..c0efficient .953: = 1 — .95 =".os and-eta? .0512 = .025. From Table IV, Appendix E; -z "= 1:96. Theeonfidence interval is:: . —-26..6' . ..._- _ - .--11-.' 2+. .—12.0 -,—- .12 (15.5 )i_l--9'6|100'+_100 1.... 28=f( “Sr 10 ). 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 5114de who have access to lecture notes and-the second-population is the setof'responses for-all business students not having access to lecture notes; _ ' To determine if there is a difference in the mean response of the two groups, we test: _- as. r.» .. Ha; _p._1_ .0' 7.18 The test statistic is_z._= = (8'48. -- 3780) O = J9 . ' “The rejection region and": .le2 .005 of theiz'edistribntion. From “ Table IV, Appendix 5 27.58; ' The rejection region is z .<. , —2.58_ 0rsz > 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 mean sf the. felt-s: .=3'. '01:. Brzms =. 2.535. ’ITseI-mfidence'interv‘al-is: " ‘ ooriooafioom .99,_,o e .01 age-on '='--.0u2*= .005. From Table Iv, Appendix +' =5 (3.43... 1-30) i 253. 2.99 n: 86 35 - '-".='>'.68'i'.801'-'==v'('-.12i-,2'l.481) _- '- We are 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 intei'val: of-zms _=- 2.58. To determine if the mean annual percentage turnover for U.S. plants exceeds that for Japanese" plants, we test: H0: pl—ng=0 H3: n1—n2>0 The test statistic is r = 4.46 (from printout). The rejection region requires or = .05 in the upper tail of the r~disrribution with df 2 321 + it: h 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 .0031l2 = .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 3‘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 = 3113 -— 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 A stern-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 r-test procedure works pretty well even if the data are not exactly normal. 7.36 First, find the sample sizes needed for width 5, or bound 2.5. For confidence coefficient .9, o: : 1 — .9 = .1 and (#2 = .1f2 = .05. From Table IV, Appendix B, 2.05 : 2'2 2 Z ll 0 + 0' 2 7 2 n1: r12 = 1 2) 2 (L645) (10- +10) = 86.59 a: 87 32 2.52 Thus, the necessary sample size from each pepulation is 87 . Therefore, sufficient funds have not been allocated to meet the specifications since :11 = n2 = 100 are large enough samples. 7.46 a. To determine if the variance for population 2 is greater than that for population 1, we test: Ho: 0i = 0% Iii: 0% ‘z 0% 52 92 The test statistic is F = .3“. = 2‘972 = 4.29 512 1.43592 The rejection region requires a = .05 in the upper tail of the F-distribution with v1 = n2 - l = 5 -— I = 4and v2 = r11 — 1 = 6 — 1 = 5. From Table IX, Appendix B. Fm = 5.19. The rejection region is F > 5.19. . Since the observed value of the test statistic does not fall in the rejection region (F = 4.29 r 5.19), H0 is not rejected. There is insufficient evidence to indicate the variance for population 2 is greater than that for p0pulation l 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 or < .05 but there is evidence to reject H0 for o: = .10. E51; 3 Let a? = variance of the solid- I 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: 7.76 02 1 H0: 1—1 Er2 2 H- U1¢1 a‘_2' 02 The test statistic is F = 7.03 (from printout). The p-value is p = 0.0800 (from printout). Since the p-value is not less than or (p = .0800 a .05). H0 is not rejected. There is insufficient evidence to indicate the variances for the two groups differ at a = .05. From the test, there is insufficient evidence to indicate the variances are different at a z .05. However, the p-value of .0800 is quite close to or = .05. If we conclude that the variances are equal, we have a chance of committing a Type II error. Without specifying a fixed alternate value for the ratio of the variances, we cannot compute the probability of committing a Type II error, .6. If we would inflate the value of or, (say to the value of .25 or higher), then we know that the value of .8 will decrease. If we still do not reject H0 with the inflated value of a, then we can have confidence that the variances are equai. In this problem. however, the p-value is quite small (p I .0800). Thus, the assumption of equal variances is questionable and the two-sample t-test might not be appropriate. df for Error is 41 - 6 = 35 SSE = SS(Total) — SST = 46.5 — 17.5 = 29.0 _ SST 17.5 _ SSE 290 MST _ = _ — 2.9167 MSE = = _.__- = P_1 6 n_p 35 .8236 MST 2.9167 F = _ = = 3. MSE .8286 52 The ANOVA table is: Source df SS MS F Ireatment 6 “.5 2.9167 3.5? Error 35 29.0 .8286 iofii 31 36.5 The number of treatments is p. We knowP — 1 = 6 =>p = 7. To determine if there is a difference among the population means, we test: Ho: #1=#2=”‘=F/7 Ha: At least one of the population means differs from the rest The test statistic is F = 3.52. The rejection region requires a = .10 in the upper tail of the F-distribution with numerator df = p — 1 = 6 and denominator df = n — p = 35. From Table VIII, Appendix B, F10 = 1.98. The rejecti0n region is F > 1.98. ’ Since the observed vaiue of the test statistic falls in the rejection region (F = 3.52 > 1.98}, H0 is rejected. There is sufficient evidence to indicate a difference among the population means at [X = .10. 7.78 b. The observed significance level is P(F a 3.52). With numerator df = 6 and denominator df = 35, and Table XI, P(F a 3.52) < .01. H03 #1 = #2 Ha: ,ul 95 pig 3 —E _ The test statistic isr = 1—2 = 3'7 4'1 = -316 1 1 MSE i + i -3286[g + 3] I11 "2 The rejection region requires (:12 = .1072 = .05 in each tail of the I-distribution with df = n — p = 35. From Table VI, Appendix B, {05 3 1.697. The rejection region is r < —1.697 or r > 1.697. Since the observed value of the test statistic does not fall in the rejection region (I = u .76 t *1.697), H0 is not rejected. There is insufficient evidence to indicate that it] and p2 differ at a = .10. For confidence coefficient .90, o: = .10 and 0272 = .05. From Table VI, Appendix B, with df = 35, {05 2 1.697. The coufidence interval is: = (3.7 — 4.1) i 1.692[.8286[l + l] 6 6 = —.4 i .892 = (1.292, .492) The confidence interval is: ii i rosthSEm n 3.7 i 1.697V‘.8286!6 n 3.7 i .631 = (3.069, 4.331) a. For diagram 3, -_Ex1~7+8+9+9+10+11 54 Il— —....._.—_. :F___=9 n 6 6 J—Ezzg’e:12+13+14+14+15+nszsftfl4 n 6 6 For diagram b, 51:20=5+5+7+11+13+13=fi=9 rt 6 6 J—C2=}:’~’2=10+10+12+16+13+18aria“4 n 6 6 For diagram 3, 2 SST = Zfli-(EI- — if = 6(9 _ }1_5)2 + 6(14 _11-5)2 = 75 i=1 (E = L" = _.._54 + 34 = 11.5) n 12 For diagram 1), 2 SST = Ema — if = 6(9 —11.s::2 + 604 —11.5>2 = 75 i=1 I For diagram a, '2 x 2 2102-121) 496-34. 92: nl = =2 1 n—l 6—1 1 ' '2 x 2 £122 ‘ (E 2') lam-EL 2_ n2 _ 6 =2 Sz—"fiE—T-TT 2 SSE = (In — 1):? + (:22 -1)32={6—1)2 + (6 w 02 = 20 For diagram 13, . x 2 Ex? — 1} 558 - 5i s§=_wn_l‘ 2—6_l =14.4 1 _ I x 2 2 Ex; " (EH 2) 124-8 - E1 2_ 2 _ 6 _ 52 — —_ — .—..—.—.—-—_ — HZ 1 6 1 SSE 2021* U521 +012 — 0:2 = {6 — U144 + {6 a 014.4 =1441 For diagram 3, SS(T0tal) = SST + SSE = 75 + 20 = 95 ssr is “ESL x 100% = g x 100% = 73.95% ofSS(T0tal) SS(Total) For diagram 1), SS(TOtal) = SST + SSE = 75 + 144 = 219 SST is i x 100% = x 100% : 34.25% of SS(Total) SS(Total} 219 For diagram 3. MST = fl = i = 75 p — 1 2 v 1 MSE= SSE = 20 =2 'F=__MST=3§=375 n — p 12 — 2 SE 2 For diagram b. MST = fl = i = 75 p — 1 2 — 1 MSE= SSE = 144 =14.4 F=flsl=i=521 n ‘ p 12 — 2 MSE 14.4 The rajection region for both diagrams requires a: = .05 in the upper tail of the F-distributinn with 012;) +1: 2 —1=1and v2 = n —p =12 - 2 =10. From Table 1X. Appendix 8, F05 = 4.96. The rcjccdon region is F > 4.96. 7.86 180 a. For diagram 3, the observed value of the test statistic falls in the rejection region (F = 3?.5 b 4.96). Thus, H0 is rejected. There is sufficient evidence to indicate the samples were drawn from populations with different means at o: = .05. For diagram 1), the observed value of the test statistic falls in the rejection region (F = 5.21 > 4.96}. Thus. H0 is rejected. There is sufficient evidence to indicate the samples were drawn from populations with different means at o = .05. g. We must assume both populations are normally distributed with women variances. Refer to Exercise 7.78, the ANOVA table is: For diagram 3: Source df SS MS F Treatment 1 75 75 7 . 5 Error 1 0 20 2 Total 1 l 95. For diagram b: Source df SS MS F Treatment 1 75 75 5 . 21 Error 10 144 14 .4 Total 1 1 21 9 To determine if differences exist among the mean sorption rates of the three organic solvent types, we test: Ho: #1 = M = #3 Ha: At least two treatment means differ where ,ui represents the mean sorption rate for solvent type t'. The test statistic is F = 24.51. The rejection region requires or = .10 in the upper tail of the F—distribution with v1 = p —- 1 = 3 —1=2andv2 = n —p = 32 — 3 = 29. UsingTableWIl,AppendixB. F110 = 2.50. The rejection region is F > 2.50. Since the observed value of the test statistic falls inthe rejection region (F = 24.51 > 2.50), H0 is rejected. There is sufficient evidence to indicate that differences exist among the mean sorption rates of the three organic solvent types at a = .10. The assumptions required for the analysis to be valid are: l. The samples are selected randome and independently. 2. The probability distributions of sorption rates for each type are normal. 3. The variances of the sorption rate probability distributions for each type are equai. To check the assumptions, we will look at the stem—and-leaf displays and the dot plots. The stem-and-leaf displays are: Stem-and—leaf of Sorption Type Leaf Unit = 0.10 II —> 2 Ii 0 2 0 67 (3) 0 889 4 1 0011 Stem~and-leaf of Sorption Type Leaf Unit = 0.10 II N 2 II 0: NmbbNN D m 0 Stem-and-Leaf of Sorption Type = 3 N = 15 Leaf Unit = 0.10 6 0 000111 (2) 0 23 7 0 4&555 2 0 66 These stem-and-leaf displays do not look particularly mound-shaped. Thus the assumption that each of the samples is selected from a normal distribution may not be valid. The dot plots for the three types are: Dolpr for Sorption Sorplion The spread of the sorptiou rates for type 2 appears to be greater than the spreads for the other two types Thus‘ the assumption of equal variances may not be met. ...
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answer key of chapter 7-1 - 7.6 7.12 I E_fi 1.0_H H_The...

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