商务统计学(第四版)课后习题答案第八章

CHAPTER 88.1 X ±Z ⋅σn= 85±1.96⋅864 83.04 ≤µ≤ 86.968.2 X ±Z ⋅σn= 125±2.58⋅2436114.68 ≤µ≤ 135.328.3 If all possible samples of the same size n are taken, 95% of them include the true population average monthly sales of the product within the interval developed. Thus you are 95 percent confident that this sample is one that does correctly estimate the true average amount. 8.4Since the results of only one sample are used to indicate whether something has gone wrong in the production process, the manufacturer can never know with 100% certainty that the specific interval obtained from the sample includes the true population mean. In order to have 100% confidence, the entire population (sample size N ) would have to be selected.8.5To the extent that the sampling distribution of sample means is approximately normal, it is true that approximately 95% of all possible sample means taken from samples of that same size will fall within 1.96 times the standard error away from the true population mean. But the population mean is not known with certainty. Since the manufacturer estimated the mean would fall between 10.99408 and 11.00192 inches based on a single sample, it is not necessarily true that 95% of all sample means will fall within those same bounds. 8.6 Approximately 5% of the intervals will not include the true population. Since the true population mean is not known, we do not know for certain whether it is contained in the interval (between 10.99408 and 11.00192 inches) that we have developed.8.7 (a) X ±Z ⋅σn=0.995±2.58⋅0.02500.9877≤µ≤1.0023(b) Since the value of 1.0 is included in the interval, there is no reason to believe that the mean is different from 1.0 gallon.(c) No. Since σ is known and n = 50, from the Central Limit Theorem, we may assume that the sampling distribution of X is approximately normal.(d) The reduced confidence level narrows the width of the confidence interval.X ±Z ⋅σn=0.995±1.96⋅0.02500.9895≤µ≤1.0005(b)Since the value of 1.0 is still included in the interval, there is no reason to believe that the mean is different from 1.0 gallon.8.8 (a)350 1.96X Z ±=± 325.5≤µ≤374.50(b) No. The manufacturer cannot support a claim that the bulbs last an average 400 hours. Based on the data from the sample, a mean of 400 hours would represent a distance of 4 standard deviations above the sample mean of 350 hours.(c)No. Since σ is known and n = 64, from the Central Limit Theorem, we may assume that the sampling distribution of X is approximately normal.(d) The confidence interval is narrower based on a process standard deviation of 80hours rather than the original assumption of 100 hours.(a)X ±Z ⋅σn=350±1.96⋅8064330.4≤µ≤369.6(b)Based on the smaller standard deviation, a mean of 400 hours wouldrepresent a distance of 5 standard deviations above the sample mean of 350 hours. No, the manufacturer cannot support a claim that the bulbs have a mean life of 400 hours.8.9 (a) X ±Z ⋅σn=1.99±1.96⋅0.051001.9802≤µ≤1.9998 (b) No. Since σ is known and n = 100, from the central limit theorem, we may assume thatthe sampling distribution of X is approximately normal.(c) An individual value of 2.02 is only 0.60 standard deviation above the sample mean of 1.99. The confidence interval represents bounds on the estimate of the average of a sample of 100, not an individual value.(d)A shift of 0.02 units in the sample average shifts the confidence interval by the same distance without affecting the width of the resulting interval.(a)X ±Z ⋅σn=1.97±1.96⋅0.051001.9602≤µ≤1.97988.10 (a) t 9 = 2.2622 (b) t 9 = 3.2498 (c) t 31 = 2.0395 (d) t 64 = 1.9977 (e) t 15 = 1.75318.11 75 2.0301X t ±=± 66.8796 ≤µ≤ 83.1204 8.12 50 2.9467X t ±=± 38.9499 ≤µ≤ 61.05018.13 Set 1: 4.5±2.3646⋅3.74178 1.3719 ≤µ≤ 7.6281 Set 2: 4.5±2.3646⋅2.449582.4522 ≤µ≤ 6.5478The data sets have different confidence interval widths because they have different values forthe standard deviation.8.14 Original data:5.8571±2.4469⋅6.46607 – 0.1229 ≤µ≤ 11.8371 Altered data: 4.00±2.4469⋅2.160272.0022 ≤µ≤ 5.9978The presence of an outlier in the original data increases the value of the sample mean and greatly inflates the sample standard deviation.8.15 (a) 1.67 2.0930X t ±=± $1.52 ≤µ≤ $1.82(b)The store owner can be 95% confident that the population mean retail value ofgreeting cards that the store has in its inventory is somewhere in between $1.52 and $1.82. The store owner could multiply the ends of the confidence interval by the number of cards to estimate the total value of her inventory.8.16 (a) 32 2.0096X t ±=± 29.44 ≤µ≤ 34.56 (b) The quality improvement team can be 95% confident that the population meanturnaround time is somewhere in between 29.44 hours and 34.56 hours.(c)The project was a success because the initial turnaround time of 68 hours does not fall inside the 95% confidence interval.8.17 (a) 195.3 2.1098X t ±=±≤µ≤ 205.9419 (b)No, a grade of 200 is in the interval.(c) It is not unusual. A tread-wear index of 210 for a particular tire is only 0.69 standard deviation above the sample mean of 195.3.8.18 (a) 23 2.0739X t ±=±≤µ≤ $24.99 (b) You can be 95% confident that the mean bounced check fee for the population issomewhere between $21.01 and $24.99.8.19 (a)7.1538 2.0595X t ±=±≤µ≤ $8.39 (b)You can be 95% confident that the population mean monthly service fee, if acustomer’s account falls below the minimum required balance, is somewhere between $5.92 and $8.39.8.20 (a)43.04 2.0096X t ±=±≤µ≤ 54.96 (b) The population distribution needs to be normally distribution.(c)Normal Probability Plot-3-2-10123Z ValueDa y sBox-and-whisker Plot050100150Both the normal probability plot and the box-and-whisker plot suggest that the distribution is skewed to the right.(d)Even though the population distribution is not normally distributed, with a sample of 50, the t distribution can still be used due to the Central Limit Theorem.8.21 (a)43.8889 2.0555X t ±=± 33.89 ≤µ≤ 53.89 (b) The population distribution needs to be normally distributed.(c)Box-and-whisker Plot1030507090Normal Probability Plot0102030405060708090100-2-1.5-1-0.50.511.52Z ValueT i m eBoth the normal probability plot and the box-and-whisker show that the population distribution is not normally distributed and is skewed to the right.8.22 (a)182.4 2.0930X t ±=± $161.68 ≤µ≤ $203.12(b) 45 2.0930X t ±=± $40.31 ≤µ≤ $49.69(c)The population distribution needs to be normally distributed.8.22 (d) cont.Normal Probability Plot050100150200250300-2-1.5-1-0.50.511.52Z ValueH o t e lBox-and-whisker Plot110160210260310Both the normal probability plot and the box-and-whisker show that the population distribution for hotel cost is not normally distributed and is skewed to the right.8.22 (d) cont.Normal Probability Plot010********607080-2-1.5-1-0.500.511.52Z ValueC a r sBox-and-whisker Plot3040506070Both the normal probability plot and the box-and-whisker show that the population distribution for rental car cost is not normally distributed and is skewed to the right.8.23 (a)0.00023 1.9842X t ±=−± –0.000566 ≤µ≤ 0.000106(b)The population distribution needs to be normally distributed. However, with a sampleof 100, the t distribution can still be used as a result of the Central Limit Theorem even if the population distribution is not normal.(c)Both the normal probability plot and the box-and-whisker plot suggest that the distribution is skewed to the right.(d)We are 95% confident that the mean difference between the actual length of the steel part and the specified length of the steel part is between -0.000566 and 0.000106 inches, which is narrower than the plus or minus 0.005 inches requirement. The steel mill is doing a good job at meeting the requirement. This is consistent with the finding in Problem 2.23.Normal Probability Plot-0.004-0.003-0.002-0.00100.0010.0020.0030.0040.0050.006Z ValueE r r orBox-and-whisker Plot-0.006-0.004-0.00200.0020.0040.0068.2450200X p n === 0.25 0.25p Z ±=±0.19 π≤≤ 0.318.2525400X p n === 0.0625 0.0625p Z ±=± 0.0313 0.09378.26 (a)135500X p n === 0.27 0.27p Z ±=±0.22π≤≤ 0.32(b)The manager in charge of promotional programs concerning residential customers can infer that the proportion of households that would purchase an additionaltelephone line if it were made available at a substantially reduced installation cost is between 0.22 and 0.32 with a 99% level of confidence.8.27 (a) 330500X p n === 0.66 0.66p Z ±=± 0.62 π≤≤ 0.70(b) You can be 95% confident that the population proportion of highly educated womenwho left careers for family reasons that want to return to work is between 0.62 and 0.70.8.28 (a) p = 0.77 0.77p Z ±=±0.74 π≤≤ 0.80(b) p = 0.77 0.77p Z ±=± 0.75 π≤≤ 0.79(c) The 95% confidence interval is wider. The loss in precision reflected as a widerconfidence interval is the price you have to pay to achieve a higher level of confidence.8.29 (a) p = 0.27 0.27p Z ±=±0.25 π≤≤ 0.29 (b)You can be 95% confident that the population proportion of older consumers that don’t think they have enough time to be good money managers is somewhere between 0.25 and 0.29.8.30 (a)0.46Xp n== 0.46p Z ±=±0.41630.5037π<<(b) 0.10Xp n== 0.10p Z ±=±0.07370.1263π<<8.31 (a) p = 0.2697 0.2697p Z ±=±0.24 π≤≤ 0.30(b) p = 0.2697 0.2697p Z ±=±0.23π≤≤ 0.31 (c)The 99% confidence interval is wider. The loss in precision reflected as a widerconfidence interval is the price one has to pay to achieve a higher level of confidence.8.32 (a) 4500.451000X p n ===0.45p Z ±=±0.41920.4808π<<(b) You are 95% confidence that the proportion of all working women in North America who believe that companies should hold positions for those on maternity leave for more than six months is between 0.4192 and 0.4808.298 Chapter 8: Confidence Interval Estimation8.33 (a)1260.21600X p n === 0.21p Z ±=±0.18π≤≤ 0.24(b)1260.21600X p n === 0.21p Z ±=±0.17π≤≤ 0.25 (c) You are 95% confidence that the population proportion of employers who have amandatory mail order program in place or are adopting one by the end of 2004 is between 0.18 and 0.24.You are 99% confidence that the population proportion of employers who have amandatory mail order program in place or are adopting one by the end of 2004 is between 0.17 and 0.25.(d)When the level of confidence is increased, the confidence interval becomes wider. The loss in precision reflected as a wider confidence interval is the price youhave to pay to achieve a higher level of confidence.8.34 n =Z 2σ2e 2=1.962⋅15252= 34.57 Use n = 358.35 n =Z 2σ2e 2=2.582⋅1002202= 166.41 Use n = 1678.36 2222(1–) 2.58(0.5)(0.5)(0.04)Z n e ππ== = 1,040.06Use n = 1,0418.37 2222(1–) 1.96(0.4)(0.6)(0.02)Z n e ππ== = 2,304.96Use n = 2,3058.38 (a) n =Z 2σ2e 2=1.962⋅4002502= 245.86 Use n = 246 (b) 2222221.9640025Z n eσ⋅== = 983.41 Use n = 9848.39 n =Z 2σ2e 2=1.962⋅(0.02)2(0.004)2= 96.04 Use n = 978.40 n =Z 2σ2e 2=1.962⋅(100)2(20)2= 96.04 Use n = 978.41 n =Z 2σ2e 2=1.962⋅(0.05)2(0.01)2= 96.04 Use n = 97Solutions to End-of-Section and Chapter Review Problems 2998.42 (a) n =Z 2σ2e 2=2.582⋅25252= 166.41 Use n = 167 (b) n =Z 2σ2e 2=1.962⋅25252= 96.04 Use n = 978.43 (a) n =Z 2σ2e 2=1.6452⋅45252= 219.19 Use n = 220 (b) n =Z 2σ2e 2=2.582⋅45252= 539.17 Use n = 5408.44 n =Z 2σ2e 2=1.962⋅20252= 61.47 Use n = 628.45 (a) 2222(1–) 1.96(0.1577)(10.1577)(0.06)Z n e ππ−== = 141.74 Use n = 142 (b) 2222(1–) 1.96(0.1577)(10.1577)(0.04)Z n e ππ−== = 318.91 Use n = 319 (c) 2222(1–) 1.96(0.1577)(10.1577)(0.02)Z n eππ−== = 1275.66 Use n = 12768.46 (a) 2222(1) 1.96(0.45)(0.55)(0.02)Z n e ππ−== = 2376.8956 Use n = 2377 (b) 2222(1) 1.96(0.29)(0.71)(0.02)Z n eππ−== = 1977.3851 Use n = 1978(c) The sample sizes differ because the estimated population proportions are different.(d)Since purchasing groceries at wholesale clubs and purchasing groceries at convenience stores are not necessary mutually exclusive events, it is appropriate to use one sample and ask the respondents both questions. However, drawing two separate samples is also appropriate for the setup.8.47 (a) 2222(1) 1.96(0.35)(10.35)546.2068(0.04)Z n eππ−−=== Use n = 547 (b) ()22222.5758(0.35)(10.35)(1)943.4009(0.04)Z n e ππ−−=== Use n = 944(c)2222(1) 1.96(0.35)(10.35)2184.8270(0.02)Z n e ππ−−=== Use n = 2185 (d) ()22222.5758(0.35)(10.35)(1)3773.6034(0.02)Z n e ππ−−=== Use n = 3774(e)Holding everything else constant, the higher the confidence level desired or the lowerthe acceptable sampling error, the larger the sample size needed.300 Chapter 8: Confidence Interval Estimation8.48 (a) 3030.9294326X p n ===0.9294 1.96p Z ±=±0.90170.9572π≤≤(b)You are 95% confident that the population proportion of business men and women whohave their presentations disturbed by cell phones is between 0.9017 and 0.9572.(c) 2222(1) 1.96(0.9294)(10.9294)157.5372(0.04)Z n e ππ−−=== Use n = 158(d)2222(1) 2.5758(0.9294)(10.9294)272.0960(0.04)Z n e ππ−−=== Use n = 2738.49 (a) 0.52 1.96p Z ±=±0.50530.5347π≤≤(b)You are 95% confident that the proportion of families that held stocks in 2001 isbetween 0.5053 and 0.5347.(c) 2222(1) 1.96(0.52)(10.52)9588.2695(0.01)Z n e ππ−−=== Use n = 95898.50The only way to have 100% confidence is to obtain the parameter of interest, rather than a sample statistic. From another perspective, the range of the normal and t distribution is infinite, so a Z or t value that contains 100% of the area cannot be obtained.8.51 The t distribution is used for obtaining a confidence interval for the mean when σ isunknown. 8.52 If the confidence level is increased, a greater area under the normal or t distribution needs to be included. This leads to an increased value of Z or t , and thus a wider interval.8.53When estimating the rate of noncompliance, it is commonplace to use a one-sided confidence interval instead of a two-sided confidence interval since only the upper bound on the rate of noncompliance is of interest.8.54 In some applications such as auditing, interest is primarily on the total amount of a variable rather than the mean amount.8.55 Difference estimation involves determining the difference between two amounts, rather than a single amount.Solutions to End-of-Section and Chapter Review Problems 3018.56 (a) The population from which this sample was drawn was the collection of all the people who visited the magazine's web site.(b) The sample is not a random sample from this population. The sample consisted of only those who visited the magazine's web site and chose to fill out the survey. (c)This is not a statistically valid study. There was selection bias since only those who visited the magazine's web site and chose to answer the survey were represented. There was possibly nonresponse bias as well. Visitors to the web site who chose to fill out the survey might not answer all questions and there was no way for the magazine to get back to them to follow-up on the nonresponses if this was an anonymous survey.(d)To avoid the above potential pitfalls, the magazine could have drawn a random sample from t he list of all subscribers to the magazine and offer them the option of filling out the survey over the Internet or on the survey form that is mailed to the subscribers. The magazine should also keep track of the subscribers who are invited to fill out the survey and follow up on the nonresponses after a specified period of time with mail or telephone to encourage them to participate in the survey.The sample size needed is 2222(1) 1.96(0.6195)(10.6195)2264(0.02)Z n e ππ⋅⋅−⋅⋅−===8.57 (a) 0.44 1.96p Z ±=± 0.39650.4835π<<(b)Since the range that covers 0.5 and above is not contained in the 95% confidenceinterval, we can conclude that less than half of all applications contain inaccuracies with past employers.(c)0.44 1.96p Z ±=± 0.37120.5088π<<(d) Since the 95% confidence interval contains 0.5, it is incorrect to conclude that lessthan half of all applications contain inaccuracies with past employers.(e)The smaller sample size in (c) results in a wider confidence interval and, hence, results in a loss of precision in the interval estimate.8.58 (a) 0.58 1.96p Z ±=±0.51160.6484π<<(b) 0.50 1.96p Z ±=±0.43070.5693π<<(c) 0.22 1.96p Z ±=±0.16260.2774π<<(d)0.19 1.96p Z ±=±0.13560.2444π<<(e) 2222(1) 1.96(0.5)(0.5)2400.92401(0.02)Z n eππ⋅⋅−⋅⋅===≅302 Chapter 8: Confidence Interval Estimation8.59 Cheat at golf:0.8204 1.96p Z ±=± 0.78290.8580π<<Hate others who cheat at golf:0.8204 1.96p Z ±=± 0.78290.8580π<<Believe business and golf behavior parallel:0.7207 1.96p Z ±=± 0.67680.7646π<<Would let a client win to get business:0.1995 1.96p Z ±=± 0.16040.2386π<<Would call in sick to play golf:0.0998 1.96p Z ±=±0.07040.1291π<<From these confidence intervals, we can conclude with 95% level of confidence that more thanhalf of the CEOs will cheat at golf, hate others who cheat at golf and believe business and golf behavior parallel and less than half of the CEOs will let a client win to get business or call in sick to play golf.8.60 (a) 15.3 2.0227X t ±=± 14.085 ≤µ≤ 16.515(b)0.675 1.96p Z ±=± 0.530 π≤≤ 0.820 (c) n =Z 2⋅σ2e 2=1.962⋅5222= 24.01 Use n = 25(d) 2222(1–) 1.96(0.5)(0.5)(0.035)Z n eππ⋅⋅⋅⋅== = 784 Use n = 784(e)If a single sample were to be selected for both purposes, the larger of the two samplesizes (n = 784) should be used.8.61 (a) 1759 2.6490X t ±=± 1,638.69 ≤µ≤ 1,879.31(b)0.60 1.96p Z ±=±π≤≤ 0.715Solutions to End-of-Section and Chapter Review Problems 3038.62 (a) 9.7 2.0639X t ±=± 8.049 ≤µ≤ 11.351(b)0.48 1.96p Z ±=±π≤≤ 0.676 (c) n =Z ⋅e 2=1.962⋅4.521.52= 34.57 Use n = 35 (d) 2222(1–) 1.645(0.5)(0.5)(0.075)Z n e ππ⋅⋅⋅⋅== = 120.268 Use n = 121 (e)If a single sample were to be selected for both purposes, the larger of the two samplesizes (n = 121) should be used.8.63 (a) n =Z 2⋅σ2e 2=2.582⋅18252= 86.27Use n = 87Note : If the Z -value used is carried out to 2.5758, the value of n is 85.986 and only 86 women would need to be sampled.(b)2222(1–) 1.645(0.5)(0.5)(0.045)Z n e ππ⋅⋅⋅⋅== = 334.07 Use n = 335(c)If a single sample were to be selected for both purposes, the larger of the two samplesizes (n = 335) should be used.8.64 (a) $28.52 1.9949X t ±=± $25.80 ≤µ≤ $31.24(b)0.40 1.645p Z ±=±π≤≤ 0.4963 (c) n =Z ⋅σe 2=1.962⋅10222= 96.04 Use n = 97 (d) 2222(1–) 1.645(0.5)(0.5)(0.04)Z n e ππ⋅⋅⋅⋅== = 422.82 Use n = 423 (e) If a single sample were to be selected for both purposes, the larger of the two samplesizes (n = 423) should be used.8.65 (a) $21.34 1.9949X t ±=± $19.14 ≤µ≤ $23.54(b) 0.3714 1.645p Z ±=±0.2764 π≤≤ 0.4664(c)n =Z 2⋅σ2e 2=1.962⋅1021.52= 170.74 Use n = 171 (d)2222(1–) 1.645(0.5)(0.5)(0.045)Z n e ππ⋅⋅⋅⋅== = 334.08 Use n = 335304 Chapter 8: Confidence Interval Estimation8.65 (e) If a single sample were to be selected for both purposes, the larger of the two sample cont. sizes (n = 335) should be used.8.66 (a)$38.54 2.0010X t ±=±≤µ≤ $40.42(b) 0.30 1.645p Z ±=±π≤≤ 0.3973 (c) n =Z 2⋅σ2e 2=1.962⋅821.52= 109.27 Use n = 110 (d) 2222(1–) 1.645(0.5)(0.5)(0.04)Z n e ππ⋅⋅⋅⋅== = 422.82 Use n = 423 (e)If a single sample were to be selected for both purposes, the larger of the two samplesizes (n = 423) should be used.8.67 (a) 2222(1) 1.96(0.5)(0.5)(0.05)Z n eππ⋅⋅−⋅⋅== = 384.16 Use n = 385If we assume that the population proportion is only 0.50, then a sample of 385 wouldbe required. If the population proportion is 0.90, the sample size required is cut to 103.(b) 0.84 1.96p Z ±=± 0.7384≤ π ≤ 0.9416 (c) The representative can be 95% confidence that the actual proportion of bags that will dothe job is between 74.5% and 93.5%. He/she can accordingly perform a cost-benefit analysis to decide if he/she wants to sell the Ice Melt product.8.68 (a) 8.4209 2.0106X t ±=± 8.418.43µ≤≤(b) With 95% confidence, the population mean width of troughs is somewhere between 8.41and 8.43 inches. Hence, the company's requirement of troughs being between 8.31 and 8.61 is being met with a 95% level of confidence.8.69 (a) 5.5014 2.6800X t ±=± 5.46 5.54µ≤≤(b) Since 5.5 grams is within the 99% confidence interval, the company can claim that themean weight of tea in a bag is 5.5 grams with a 99% level of confidence.Solutions to End-of-Section and Chapter Review Problems 3058.70 (a)0.2641 1.9741X t ±=±0.24250.2856µ≤≤(b)0.218 1.9772X t ±=±0.19750.2385µ≤≤(c)Normal Probability Plot00.10.20.30.40.50.60.70.80.9-3-2-1123Z ValueV e r m o n tNormal Probability Plot00.20.40.60.811.2-3-2-1123Z ValueB o s t o nThe amount of granule loss for both brands are skewed to the right.(d)Since the two confidence intervals do not overlap, we can conclude that the mean granule loss of Boston shingles is higher than that of Vermont Shingles.306 Chapter 8: Confidence Interval Estimation8.71 (a)3124.2147 1.9665X t ±=±3120.663127.77µ≤≤(b)3704.0424 1.9672X t ±=± 3698.98<3709.10µ<(c)Normal Probability Plot3000305031003150320032503300-4-3-2-11234Z ValueB o s t o nNormal Probability Plot35503600365037003750380038503900-4-3-2-11234Z ValueV e r m o n tThe weight for Boston shingles is slightly skewed to the right while the weight for Vermont shingles appears to be slightly skewed to the left.(d)Since the two confidence intervals do not overlap, the mean weight of Vermont shingles is greater than the mean weight of Boston shingles.Solutions to End-of-Section and Chapter Review Problems 3078.72(a)NY, Food: 20.1 2.0096X t ±=± 19.5120.69µ≤≤ LI, 20.54 2.0096X t ±=± 19.7321.35µ≤≤ NY, 17.12 2.0096X t ±=± 16.3517.89µ≤≤ LI, 17.64 2.0096X t ±=± 16.6518.63µ≤≤ NY, 18.4 2.0096X t ±=± 17.7419.06µ≤≤ LI, 19.04 2.0096X t ±=±18.3719.71µ≤≤NY, 39.74 2.0096X t ±=± 37.0042.48µ≤≤ LI, 33.7400 2.0096X t ±=± 31.5535.93µ≤≤(b)With 95% confidence you can conclude the mean price per person in New York City isgreater than the mean in Long Island. With 95% confidence you can conclude that there are no differences in the ratings for food, décor and service between the two cities.。