商务与经济统计习题答案（第8版中文版）SBE8-SM11.doc

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1、Inferences About Population VariancesChapter 11Inferences About Population VariancesLearning Objectives1.Understand the importance of variance in a decision-making situation.2Understand the role of statistical inference in developing conclusions about the variance of a single population.3.Know the s

2、ampling distribution of (n - 1) s2/s 2 has a chi-square distribution and be able to use this result to develop a confidence interval estimate ofs 2.4.Know how to test hypotheses involving s 2.5.Understand the role of statistical inference in developing conclusions about the variances of two populati

3、ons.6.Know that the sampling distribution of has an F distribution and be able to use this result to test hypotheses involving the variances of two populations. Solutions:1.a.11.0705b.27.4884c.9.59083d.23.2093e.9.390462.s2 = 25a.With 19 degrees of freedom = 30.1435 and = 10.117015.76 s 2 46.95b.With

4、 19 degrees of freedom = 32.8523 and = 8.9065514.46 s 2 53.33c.3.8 s 2 7.33.With 15 degrees of freedom = 24.9958Reject H0 if 24.9958Do not reject H0 4.a.n = 18s2 = .36 = 27.5871= 8.67176 (17 degrees of freedom).22 s 2 .71b.47 s .845.a. b. = 16.0128 = 1.6898713.58 s 2 128.71c.3.69 s 11.346.a. b. = 11

5、.1433 = 0.48441963.52 s 2 1461.217.97 s 38.237.a.b. = 16.0128 = 1.689871.14 s 2 10.85c.1.07 s 3.298.a.b.c.11 degrees of freedom = 21.92 = 3.81575.0042 s 2 .0244d.0651 s .15619.H0: s 2 .0004Ha: s 2 .0004n = 30 = 42.5569 (29 degrees of freedom)Do not reject H0; the product specification does not appea

6、r to be violated.10.H0: s 2 .75Ha: s 2 .75 = 42.5569 (29 degrees of freedom)Since= 206.22 42.5569, reject H0The standard deviation for television sets is greater than the standard deviation for VCRs.11.19 degrees of freedom = 8.90655 = 32.8523Reject H0 if 32.8523Do not reject H0. Cannot conclude the

7、 variance in interest rates has changed.12.H0: s 2 = .94Ha: s 2 .94With 11 degrees of freedom, reject if= 21.92.Since = 9.49 is not in the rejection region, we cannot reject H0.13.a.F.05 = 2.91b.F.025 = 2.76c.F.01 = 4.50d.Remember to reverse the degrees of freedom in the F.025 above.14.F.05,15,19 =

8、2.23Reject H0 if F 2.23Reject H0: conclude 15.We recommend placing the larger sample variance in the numerator. With a = .05,F.025,20,24 = 2.33. Reject if F 2.33.F = 8.2/4.0 = 2.05 Do not reject H0Or if we had the lower tail F value,F = 4.0/8.2 = .49 F .41 Do not reject H016.F.01,24,29 = 2.49 Reject

9、 H0 if F 2.49 Reject H0; Conclude adults have a greater variance in online times than teens.17.a.Let= variance in repair costs (4 year old automobiles)= variance in repair costs (2 year old automobiles)b.= (170)2 = 28,900= (100) 2 = 10,000F.01,24,24 = 2.66 Reject H0; conclude that 4 year old automob

10、iles have a larger variance in annual repair costs compared to 2 year old automobiles. This is expected due to the fact that older automobiles are more likely to have some very expensive repairs which lead to greater variance in the annual repair costs.18.Fa/2 = F.025,9,6 = 5.52 Do not reject H0; Ca

11、nnot conclude any difference between variances of the two industries.19.F.025 = 2.37 (Degrees of freedom are 24 numerator, 21 denominator)Using Minitab,Machine 1:n1 = 25s1 = .2211=3.328Machine 1:n1 = 22s1 = .0768=3.278Reject H0; the process variances are significantly different. Machine 1 offers the

12、 best opportunity for process quality improvements.Note that the sample means are similar with the mean bag weights of approximately 3.3 grams. However, the process variances are significantly different.20.F.025 = 2.37 (Degrees of freedom are 24 numerator, 24 denominator)With 11.1 the larger sample

13、variance, we haveF = 11.1/2.1 = 5.29Reject H0; the variances are not equal for seniors and managers.21.a.= 9663.57= 19,237.73b.F.05,9,9 = 3.18Since F = 1.99 2.40, reject H0 and conclude that there is greater variability in stopping distances on wet pavement.b.Drive carefully on wet pavement because

14、of the uncertainty in stopping distances.23.a.s2 = (30) 2 = 900b. = 30.1435 and = 10.1170 (19 degrees of freedom)567.29 s 2 1690.22c.23.82 s 41.1124.With 12 degrees of freedom,= 23.3367 = 4.40379114.93 s 2 609.0310.72 s 24.6825.a.b.c. = 32.8523 = 8.906552889.87 s 2 10,659.4553.76 s 103.2426.a.H0: s

15、2 .0001Ha: s 2 .0001 = 21.0642 (14 degrees of freedom)Reject H0; s 2 exceeds maximum variance requirement.b. = 23.6848 and = 6.57063 (14 degrees of freedom).00012 s 2 .0004227.H0: s 2 .02Ha: s 2 .02= 55.7585 (40 degrees of freedom)Do not reject H0; the variance does not appear to be exceeding the st

16、andard.28.n = 22 s2 = 1.5H0: s 2 1Ha: s 2 1= 29.6151 (21 degrees of freedom)Reject H0; conclude that s 2 1.29.H0: s 2 = 10Ha: s 2 10With 8 degrees of freedom, reject if = 15.5073Since = 10.16 is not in the rejection region, we cannot reject H0.30.a.Try n = 15= 26.1190 = 5.62872 (14 degrees of freedo

17、m)34.30 s 2 159.185.86 s 12.62 A sample size of 15 was used.b.n = 25; expected the width of the interval to be smaller. = 39.3641 = 12.4011 (24 degrees of freedom)39.02 s 2 126.866.25 s 11.1331.Fa/2 = F.05,9,9 = 3.18 Reject H0. Conclude the variances differ with NASDAQ stocks showing the greater var

18、iance.32.F.025 = 1.466Do not reject H0; We are not able to conclude students who complete the course and students who drop out have different variances of grade point averages.33.n1 = 16= 5.4n2 = 16= 2.3F.05 = 2.40 (Degrees of freedom are 15 numerator, 15 denominator)Do not reject H0; data does not indicate a difference between the population variances.34.F.05 = 1.94 (30 numerator and 24 denominator degrees of freedom)Reject H0; conclude that the variances of assembly times are not equal.11 - 11