288 Chapter 8: Confidence Interval Estimation CHAPTER 8 8.1 8.2 8.3 8.4 X ±Z⋅X ±Z⋅σn = 85±1.96⋅8 6424 3683.04 ≤µ≤ 86.96 σn = 125±2.58⋅114.68 ≤µ≤ 135.32 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. Since 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. To 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. 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. 8.5 8.6 8.7 (a) X ±Z⋅ (b) (c) (d) σn=0.995±2.58⋅0.02 0.9877≤µ≤1.0023 50Since 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. 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. The reduced confidence level narrows the width of the confidence interval. X ±Z⋅(b) σn=0.995±1.96⋅0.02 0.9895≤µ≤1.0005 50Since 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. Solutions to End-of-Section and Chapter Review Problems 289 8.8 (a) (b) (c) (d) 8.9 (a) (b) (c) (d) 8.10 (a) (b) (c) (d) (e) X±Z⋅σn=350±1.96⋅10064 325.5≤µ≤374.50 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. 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. The confidence interval is narrower based on a process standard deviation of 80 hours rather than the original assumption of 100 hours. (a) X ±Z⋅σn=350±1.96⋅8064 330.4≤µ≤369.6 (b) Based on the smaller standard deviation, a mean of 400 hours would represent 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. X ±Z⋅σ0.05n=1.99±1.96⋅100 1.9802≤µ≤1.9998 No. Since σ is known and n = 100, from the central limit theorem, we may assume that the sampling distribution of X is approximately normal. 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. 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⋅0.05100 1.9602≤µ≤1.9798 t9 = 2.2622 t9 = 3.2498 t31 = 2.0395 t64 = 1.9977 t15 = 1.7531 290 Chapter 8: Confidence Interval Estimation 8.11 8.12 X±t⋅24S 66.8796 =75±2.0301⋅≤µ≤ 83.1204 36n15S 38.9499 ≤µ≤ 61.0501 =50±2.9467⋅16nX±t⋅3.7417 1.3719 ≤µ≤ 7.6281 82.4495 Set 2: 4.5±2.3646⋅ 2.4522 ≤µ≤ 6.5478 88.13 Set 1: 4.5±2.3646⋅The data sets have different confidence interval widths because they have different values for the standard deviation. 5.8571±2.4469⋅8.14 Original data: 6.4660 – 0.1229 ≤µ≤ 11.8371 72.1602 Altered data: 2.0022 4.00±2.4469⋅≤µ≤ 5.9978 7The 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) X±t⋅ (b) 0.32S $1.52 ≤µ≤ $1.82 =1.67±2.0930⋅20nThe store owner can be 95% confident that the population mean retail value of greeting 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) X±t⋅ 8.17 (a) X±t⋅ 8.18 (a) X±t⋅ (b) (b) (c) (b) (c) 9S 29.44 ≤µ≤ 34.56 =32±2.0096⋅50nThe quality improvement team can be 95% confident that the population mean turnaround time is somewhere in between 29.44 hours and 34.56 hours. The project was a success because the initial turnaround time of 68 hours does not fall inside the 95% confidence interval. 21.4S 184.6581 =195.3±2.1098⋅≤µ≤ 205.9419 18nNo, a grade of 200 is in the interval. 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. 4.6024S $21.01 ≤µ≤ $24.99 =23±2.0739⋅23nYou can be 95% confident that the mean bounced check fee for the population is somewhere between $21.01 and $24.99. Solutions to End-of-Section and Chapter Review Problems 291 8.19 (a) X±t⋅ 8.20 (a) X±t⋅(b) 3.0554S $5.92 ≤µ≤ $8.39 =7.1538±2.0595⋅26nYou can be 95% confident that the population mean monthly service fee, if a customer’s account falls below the minimum required balance, is somewhere between $5.92 and $8.39. 41.9261S 31.12 ≤µ≤ 54.96 =43.04±2.0096⋅50n (b) The population distribution needs to be normally distribution. (c) Normal Probability Plot180160140120Days-3100806040200-2-10123 Z ValueBox-and-whisker Plot Days (d) Both the normal probability plot and the box-and-whisker plot suggest that the distribution is skewed to the right. 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. 050100150 292 Chapter 8: Confidence Interval Estimation 8.21 (a) X±t⋅25.2835S 33.89 =43.8889±2.0555⋅≤µ≤ 53.89 27n (b) The population distribution needs to be normally distributed. (c) 8.22 (a) (b) (c) Box-and-whisker PlotTime 1030507090 Normal Probability Plot100908070e60mi50T403020100-2-1.5-1-0.500.511.52Z ValueBoth the normal probability plot and the box-and-whisker show that the population distribution is not normally distributed and is skewed to the right. X±t⋅Sn=182.4±2.0930⋅44.270020 $161.68 ≤µ≤ $203.12 X±t⋅S10.0263n=45±2.0930⋅20 $40.31 ≤µ≤ $49.69 The population distribution needs to be normally distributed. Solutions to End-of-Section and Chapter Review Problems 293 8.22 (d) cont. Normal Probability Plot300250200Hotel150100500-2-1.5-1-0.500.511.52 Z Value Box-and-whisker PlotHotel 110160210260310Both 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. 294 Chapter 8: Confidence Interval Estimation 8.22 (d) cont. Normal Probability Plot80706050Cars403020100-2-1.5-1-0.50Z Value0.511.52 Box-and-whisker PlotCars 3040506070 Both 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. Solutions to End-of-Section and Chapter Review Problems 295 8.23 (a) X±t⋅ (b) 0.0017S –0.000566 ≤µ≤ 0.000106 =−0.00023±1.9842⋅100nThe population distribution needs to be normally distributed. However, with a sample of 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) Normal Probability Plot0.0060.0050.0040.003Error0.0020.0010-0.001-3-0.002-0.003-0.004-2-10123Z ValueBox-and-whisker PlotError-0.006-0.004-0.00200.0020.0040.006 (d) Both the normal probability plot and the box-and-whisker plot suggest that the distribution is skewed to the right. 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. 296 Chapter 8: Confidence Interval Estimation 8.24 p=X50= 0.25 =n200p±Z⋅p(1–p)0.25(0.75) =0.25±1.96n2000.19 ≤π≤ 0.31 8.25 p=X25== 0.0625 n400 8.26 (a) 0.0625(0.9375)p(1−p) =0.0625±2.58400n 0.0313 ≤π≤ 0.0937 p±Z⋅ (b) 8.27 (a) (b) 8.28 (a) (b) (c) p=Xp(1–p)−0.27)n=135500= 0.27 p±Z⋅0.27(1n=0.27±2.5758500 0.22 ≤π≤ 0.32 The manager in charge of promotional programs concerning residential customers can infer that the proportion of households that would purchase an additional telephone 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. p=X330(1–p)n=500= 0.66 p±Z⋅pn=0.66±1.960.66(0.37)500 0.62 ≤π≤ 0.70 You can be 95% confident that the population proportion of highly educated women who left careers for family reasons that want to return to work is between 0.62 and 0.70. p= 0.77 p±Z⋅p(1–p)0.77(0.23)n=0.77±1.961000 0.74 ≤π≤ 0.80 p= 0.77 p±Z⋅p(1–p)0.77(0.23)n=0.77±1.6451000 0.75 ≤π≤ 0.79 The 95% confidence interval is wider. The loss in precision reflected as a wider confidence interval is the price you have to pay to achieve a higher level of confidence. Solutions to End-of-Section and Chapter Review Problems 297 8.29 (a) p= 0.27 p±Z⋅0.27(0.73)p(1–p) =0.27±1.961900n 0.25 ≤π≤ 0.29 (b) 8.30 (a) (b) 8.31 (a) (b) (c) 8.32 (a) (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. p=Xp(1−p)n=0.46 p±Z⋅n=0.46±1.960.46(1−0.46)500 0.4163<π<0.5037 p=Xp(1−p)n=0.10 p±Z⋅n=0.10±1.960.10(1−0.10)500 0.0737<π<0.1263 p= 0.2697 p±Z⋅p(1–p)0.2697(1n=0.2697±1.96−0.2697)723 0.24 ≤π≤ 0.30 p= 0.2697 p±Z⋅p(1–p)n=0.2697±2.57580.2697(1−0.2697)723 0.23 ≤π≤ 0.31 The 99% confidence interval is wider. The loss in precision reflected as a wider confidence interval is the price one has to pay to achieve a higher level of confidence. p=X450n=1000=0.45 p±Z⋅p(1−p)n=0.45±1.960.45(1−0.45)1000 0.4192<π<0.4808 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 Estimation 8.33 (a) p=X126==0.21 n600 p±Z⋅p(1−p)0.21(1−0.21)=0.21±1.96 600n ≤π≤ 0.24 0.18 p±Z⋅p(1−p)0.21(1−0.21)=0.21±2.5758 600np= (b) X126==0.21 n600 ≤π≤ 0.25 0.17 (c) You are 95% confidence that the population proportion of employers who have a mandatory 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 a mandatory 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 you have to pay to achieve a higher level of confidence. Z2σ21.962⋅152n=2= = 34.57 8.34 2e5 Use n = 35 Z2σ22.582⋅10028.35 n=2= = 166.41 2e20 Use n = 167 Z2π(1–π)2.582(0.5)(0.5)8.36 n== = 1,040.06 (0.04)2e2 Use n = 1,041 Z2π(1–π)1.962(0.4)(0.6)8.37 n== = 2,304.96 (0.02)2e2 Use n = 2,305 Z2σ21.962⋅40028.38 (a) n== = 245.86 22e50Z2σ21.962⋅4002 = 983.41 (b) n=2=225e 8.39 8.40 8.41 Use n = 246 Use n = 984 Z2σ21.962⋅(0.02)2n=2= = 96.04 2e(0.004)Z2σ21.962⋅(100)2n=2= = 96.04 2e(20)Z2σ21.962⋅(0.05)2n=2= = 96.04 2e(0.01)Use n = 97 Use n = 97 Use n = 97 Solutions to End-of-Section and Chapter Review Problems 299 Z2σ22.582⋅2528.42 (a) n== = 166.41 22e5Z2σ21.962⋅252 (b) n=2= = 96.04 2e5 Use n = 167 Use n = 97 Z2σ21.6452⋅4528.43 (a) n== = 219.19 22e5Z2σ22.582⋅452 (b) n=2= = 539.17 2e5 8.44 Use n = 220 Use n = 540 Z2σ21.962⋅202n=2= = 61.47 2e5Use n = 62 Z2π(1–π)1.962(0.1577)(1−0.1577)8.45 (a) n= = 141.74 =22(0.06)eZ2π(1–π)1.962(0.1577)(1−0.1577) = 318.91 (b) n==22(0.04)eZ2π(1–π)1.962(0.1577)(1−0.1577)n== (c) = 1275.66 22(0.02)e Use n = 142 Use n = 319 Use n = 1276 Z2π(1−π)1.962(0.45)(0.55)n== = 2376.8956 8.46 (a) 22(0.02)eZ2π(1−π)1.962(0.29)(0.71) = 1977.3851 (b) n==22(0.02)e (c) (d) Use n = 2377 Use n = 1978 The sample sizes differ because the estimated population proportions are different. 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. Z2π(1−π)1.962(0.35)(1−0.35)n = 547 ==546.2068 Use 8.47 (a) n=22(0.04)eZ2π(1−π)(2.5758)(0.35)(1−0.35) (b) n===943.4009 Use n = 944 e2(0.04)2Z2π(1−π)1.962(0.35)(1−0.35)n===2184.8270 Use n = 2185 (c) e2(0.02)2Z2π(1−π)(2.5758)(0.35)(1−0.35)n===3773.6034 (d) Use n = 3774 e2(0.02)2 (e) Holding everything else constant, the higher the confidence level desired or the lower the acceptable sampling error, the larger the sample size needed. 22 300 Chapter 8: Confidence Interval Estimation 8.48 (a) p= (b) X303==0.9294 n326p(1−p)0.9294(1−0.9294)p±Z⋅=0.9294±1.96⋅ n3260.9017≤π≤0.9572 You are 95% confident that the population proportion of business men and women who have their presentations disturbed by cell phones is between 0.9017 and 0.9572. Z2π(1−π)1.962(0.9294)(1−0.9294)n= (c) ==157.5372 Use n = 158 e2(0.04)2Z2π(1−π)2.57582(0.9294)(1−0.9294)n===272.0960 Use n = 273 (d) e2(0.04)2 8.49 (a) p±Z⋅ (b) p(1−p)0.52(1−0.52)=0.52±1.96⋅ n44490.5053≤π≤0.5347 You are 95% confident that the proportion of families that held stocks in 2001 is between 0.5053 and 0.5347. Z2π(1−π)1.962(0.52)(1−0.52)n===9588.2695 Use (c) n = 9589 22e(0.01) 8.50 The 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 is unknown. 8.52 8.53 When 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. In some applications such as auditing, interest is primarily on the total amount of a variable rather than the mean amount. Difference estimation involves determining the difference between two amounts, rather than a single amount. 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. 8.54 8.55 8.56 (a) (b) (c) (d) 8.57 (a) (b) (c) (d) (e) 8.58 (a) (b) (c) (d) (e) Solutions to End-of-Section and Chapter Review Problems 301 The population from which this sample was drawn was the collection of all the people who visited the magazine's web site. 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. 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. To avoid the above potential pitfalls, the magazine could have drawn a random sample from the 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 n=Z2⋅π⋅(1−π)e2=1.962⋅(0.6195)⋅(1−0.6195)(0.02)2=2264 p±Z⋅p(1−p)n=0.44±1.96⋅0.44(0.56)500 0.3965<π<0.4835 Since the range that covers 0.5 and above is not contained in the 95% confidence interval, we can conclude that less than half of all applications contain inaccuracies with past employers. p±Z⋅p(1−p)0.44(0.56)n=0.44±1.96⋅200 0.3712<π<0.5088 Since the 95% confidence interval contains 0.5, it is incorrect to conclude that less than half of all applications contain inaccuracies with past employers. The smaller sample size in (c) results in a wider confidence interval and, hence, results in a loss of precision in the interval estimate. p±Z⋅p(1−p)n=0.58±1.96⋅0.58(1−0.58)200 0.5116<π<0.6484 p±Z⋅p(1−p)n=0.50±1.96⋅0.50(1−0.50)200 0.4307<π<0.5693 p±Z⋅p(1−p)n=0.22±1.96⋅0.22(1−0.22)200 0.1626<π<0.2774 p±Z⋅p(1−p)n=0.19±1.96⋅0.19(1−0.19)200 0.1356<π<0.2444 Z2⋅π⋅(1−π)1.962n=e2=⋅(0.5)⋅(0.5)(0.02)2=2400.9≅2401 302 Chapter 8: Confidence Interval Estimation 8.59 Cheat at golf: p±Z⋅p(1−p)0.8204(0.1796)=0.8204±1.96⋅ n401p(1−p)0.8204(0.1796) =0.8204±1.96⋅n401p(1−p)0.7207(0.2793)=0.7207±1.96⋅ n401p(1−p)0.1995(0.8005)=0.1995±1.96⋅ n401p(1−p)0.0998(0.9002)=0.0998±1.96⋅ n4010.7829<π<0.8580 Hate others who cheat at golf: p±Z⋅0.7829<π<0.8580 Believe business and golf behavior parallel: p±Z⋅0.6768<π<0.7646 Would let a client win to get business: p±Z⋅0.1604<π<0.2386 Would call in sick to play golf: p±Z⋅0.0704<π<0.1291 From these confidence intervals, we can conclude with 95% level of confidence that more than half 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) X±t⋅S3.8=15.3±2.0227⋅ n4014.085 ≤µ≤ 16.515 p±Z⋅ (b) p(1–p)0.675(0.325)=0.675±1.96⋅ 0.530 ≤π≤ 0.820 n40Z2⋅σ21.962⋅52n== = 24.01 Use n = 25 (c) 22e2 Use n = 784 Z2⋅π⋅(1–π)1.962⋅(0.5)⋅(0.5)n== = 784 (d) 22(0.035)e 8.61 (a) X±t⋅(e) If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 784) should be used. S380=1759±2.6490⋅ 1,638.69 ≤µ≤ 1,879.31 n70p(1–p)0.60(0.40)=0.60±1.96⋅ 0.485 ≤π≤ 0.715 n70p±Z⋅ (b) Solutions to End-of-Section and Chapter Review Problems 303 8.62 (a) X±t⋅S4=9.7±2.0639⋅ 8.049 ≤µ≤ 11.351 n25p(1–p)0.48(0.52)=0.48±1.96⋅ 0.284 ≤π≤ 0.676 n252222Z⋅σ1.96⋅4.5n== = 34.57 Use n = 35 (c) 22e1.5Z2⋅π⋅(1–π)1.6452⋅(0.5)⋅(0.5)n== = 120.268 Use n = 121 (d) 22p±Z⋅ (b) (e) 8.63 (a) (b) (c) 8.64 (a) (b) (c) (d) (e) 8.65 (a) (b) (c) (d) e(0.075)If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 121) should be used. n=Z2⋅σ2e2=2.582⋅18252 = 86.27 Use n = 87 Note: 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. n=Z2⋅π⋅(1–π)1.6452⋅(0.5)⋅(0.5)e2=(0.045)2 = 334.07 Use n = 335 If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 335) should be used. X±t⋅Sn=$28.52±1.9949⋅$11.3970 $25.80 ≤µ≤ $31.24 p±Z⋅p(1–p)n=0.40±1.645⋅0.40(0.60)70 0.3037 ≤π≤ 0.4963 n=Z2⋅σ222e2=1.96⋅1022 = 96.04 Use n = 97 n=Z2⋅π⋅(1–π)1.6452⋅(0.5)⋅(0.5)e2=(0.04)2 = 422.82 Use n = 423 If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 423) should be used. X±t⋅Sn=$21.34±1.9949⋅$9.2270 $19.14 ≤µ≤ $23.54 p±Z⋅p(1–p)n=0.3714±1.645⋅0.3714(0.6286)70 0.2764 ≤π≤ 0.4664 n=Z2⋅σ21.962⋅102e2=1.52 = 170.74 Use n = 171 n=Z2⋅π⋅(1–π)1.6452⋅(0.5)⋅(0.5)e2=(0.045)2 = 334.08 Use n = 335 304 Chapter 8: Confidence Interval Estimation 8.65 (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. X±t⋅8.66 (a) (b) p±Z⋅S$7.26=$38.54±2.0010⋅ $36.66 ≤µ≤ $40.42 n60p(1–p)0.30(0.70)=0.30±1.645⋅ 0.2027 ≤π≤ 0.3973 n60Z2⋅σ21.962⋅82n== = 109.27 Use n = 110 (c) 22 (d) (e) 8.67 (a) (b) (c) 8.68 (a) (b) 8.69 (a) (b) e1.5n=Z2⋅π⋅(1–π)1.6452⋅(0.5)⋅(0.5)e2=(0.04)2 = 422.82 Use n = 423 If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 423) should be used. n=Z2⋅π⋅(1−π)1.962⋅(0.5)⋅(0.5)e2=(0.05)2 = 384.16 Use n = 385 If we assume that the population proportion is only 0.50, then a sample of 385 would be required. If the population proportion is 0.90, the sample size required is cut to 103. p±Z⋅p(1−p)n=0.84±1.96⋅0.84(0.16)50 0.7384 ≤ π ≤ 0.9416 The representative can be 95% confidence that the actual proportion of bags that will do the 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. X±t⋅Sn=8.4209±2.0106⋅0.046149 8.41≤µ≤8.43 With 95% confidence, the population mean width of troughs is somewhere between 8.41 and 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. X±t⋅Sn=5.5014±2.6800⋅0.105850 5.46≤µ≤5.54 Since 5.5 grams is within the 99% confidence interval, the company can claim that the mean 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 305 S0.1424=0.2641±1.9741⋅ n170S0.1227X±t⋅=0.218±1.9772⋅ (b) n1408.70 (a) X±t⋅ (c) Normal Probability Plot0.90.80.70.60.50.40.30.20.10-3-2-101230.2425≤µ≤0.2856 0.1975≤µ≤0.2385 Vermont 1.21Z Value Normal Probability PlotBoston0.80.60.40.20-3-2-10123 (d) Z ValueThe amount of granule loss for both brands are skewed to the right. 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 Estimation S34.713=3124.2147±1.9665⋅ n368S46.7443X±t⋅=3704.0424±1.9672⋅ (b) n3308.71 (a) X±t⋅ (c) Normal Probability Plot330032503120.66≤µ≤3127.77 3698.98<µ<3709.10 Boston32003150310030503000-4-3-2-101234 390038503800Z Value Normal Probability PlotVermont37503700365036003550-4-3-2-101234 (d) The weight for Boston shingles is slightly skewed to the right while the weight for Vermont shingles appears to be slightly skewed to the left. Since the two confidence intervals do not overlap, the mean weight of Vermont shingles is greater than the mean weight of Boston shingles. Z Value Solutions to End-of-Section and Chapter Review Problems 307 S2.0825=20.1±2.0096⋅ n50S2.8586X±t⋅=20.54±2.0096⋅ LI, Food: n50S2.6927X±t⋅=17.12±2.0096⋅ NY, Décor: n50S3.4862X±t⋅=17.64±2.0096⋅ LI, Décor: n50S2.3123X±t⋅=18.4±2.0096⋅ NY, Services: n50S2.3730X±t⋅=19.04±2.0096⋅ LI, Services: n508.72 (a) NY, Food: X±t⋅ 19.51≤µ≤20.69 19.73≤µ≤21.35 16.35≤µ≤17.89 16.65≤µ≤18.63 17.74≤µ≤19.06 18.37≤µ≤19.71 S9.6528=39.74±2.0096⋅ n50S7.7191X±t⋅=33.7400±2.0096⋅ LI, Price: n50X±t⋅ NY, Price: (b) 37.00≤µ≤42.48 31.55≤µ≤35.93 With 95% confidence you can conclude the mean price per person in New York City is greater 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.
