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Chapter 7: Estimation: Single Population 7.1 a. Check for nonnormality The distribution shows no significant evidence of nonnormality. b. Point estimate of the population mean that is unbiased, efficient and consistent. X i 560 = 7.0 Unbiased point estimator is the sample mean: X n 8 c. The unbiased point estimate of the variance of the sample mean: s 2 (2.268) 2 Var ( X ) .64298 n 8 Chapter 7: Estimation: Single Population 7.2 173 a. There appears to be no evidence of non-normality, as shown by the normal probability plot shown next. b. Assuming normality, the unbiased and most efficient point estimator for the population mean is the sample mean, X . X i 2411 301.375 thousand dollars X n 8 c. Assuming normality, the sample mean is an unbiased estimator of the population mean with variance Var ( X ) 2 . Also, assuming normality, the unbiased and most n efficient point estimator for the population variance is the sample variance, s 2 . s 2 8373.125 Var ( X ) 1046.64 n 8 d. The unbiased and most efficient point estimator for a proportion is the sample proportion, p̂ . x 3 pˆ 0.375 n 8 174 7.3 Statistics for Business & Economics, 7th edition n = 10 economists forecast for percentage growth in real GDP in the next year Descriptive Statistics: RGDP_Ex7.3 Variable RGDP_Ex7.3 N 10 N* 0 Variable RGDP_Ex7.3 Minimum 2.2000 Mean 2.5700 Q1 2.4000 SE Mean 0.0716 TrMean 2.5625 StDev 0.2263 Median 2.5500 Q3 2.7250 Maximum 3.0000 Variance 0.0512 Range 0.8000 CoefVar 8.81 Sum 25.7000 IQR 0.3250 a. Unbiased point estimator of the population mean is the sample mean: X i 2.57 X n b. The unbiased point estimate of the population variance: s 2 .0512 c. Unbiased point estimate of the variance of the sample mean s 2 .0512 Var ( X ) .00512 n 10 x 7 d. Unbiased estimate of the population proportion: pˆ .70 n 10 e. Unbiased estimate of the variance of the sample proportion: pˆ (1 pˆ ) .7(1 .7) Var ( pˆ ) .021 n 10 7.4 n = 12 employees. Number of hours of overtime worked in the last month: a. Unbiased point estimator of the population mean is the sample mean: X i 24.42 X n b. The unbiased point estimate of the population variance: s 2 85.72 c. Unbiased point estimate of the variance of the sample mean s 2 85.72 Var ( X ) 7.1433 n 12 x 3 d. Unbiased estimate of the population proportion: pˆ .25 n 12 e. Unbiased estimate of the variance of the sample proportion: pˆ (1 pˆ ) .25(1 .25) Var ( pˆ ) .015625 n 12 Chapter 7: Estimation: Single Population a. Check each variable for normal distribution: Normal Probability Plot .999 .99 Probability .95 .80 .50 .20 .05 .01 .001 45 50 55 Meals Average: 50.1 StDev: 2.46842 N: 30 Anderson-Darling Normality Test A-Squared: 0.413 P-Value: 0.318 Normal Probability Plot .999 .99 .95 Probability 7.5 .80 .50 .20 .05 .01 .001 15 20 25 Attendance Average: 21.24 StDev: 2.50466 N: 25 Anderson-Darling Normality Test A-Squared: 0.377 P-Value: 0.383 175 176 Statistics for Business & Economics, 7th edition Normal Probability Plot .999 .99 Probability .95 .80 .50 .20 .05 .01 .001 9 10 11 12 13 14 15 16 Ages Average: 12.24 StDev: 1.96384 N: 25 Anderson-Darling Normality Test A-Squared: 0.569 P-Value: 0.126 No evidence of non-normality in Meals, Attendance or Ages b. Unbiased estimates of population mean and variance: Descriptive Statistics: Meals, Attendance, Ages Variable Meals Attendan Ages Variable Meals Attendan Ages N 30 25 25 Minimum 45.000 15.000 9.000 Mean 50.100 21.240 12.240 Maximum 55.000 25.000 16.000 Median 50.000 21.000 12.000 Q1 48.000 19.500 10.000 TrMean 50.192 21.348 12.217 StDev 2.468 2.505 1.964 Q3 52.000 23.500 14.000 Variable Unbiased estimate of mean Unbiased estimate of variance (s2) Meals 50.100 (2.468)2 = 6.0910 Attendance 21.240 (2.505)2 = 6.2750 Ages 12.240 (1.964)2 = 3.8573 7.6 1 1 E ( X1 ) E ( X 2 ) 2 2 2 2 1 3 3 E (Y ) E ( X 1 ) E ( X 2 ) 4 4 4 4 1 2 2 E (Z ) E ( X1 ) E ( X 2 ) 3 3 3 3 a. E ( X ) SE Mean 0.451 0.501 0.393 Chapter 7: Estimation: Single Population 1 1 12 2 Var ( X 1 ) Var ( X 2 ) n 4 4 2 8 4 2 1 9 5 Var (Y ) Var ( X 1 ) Var ( X 2 ) 16 16 8 2 1 4 5 Var ( Z ) Var ( X 1 ) Var ( X 2 ) 9 9 9 is most efficient since Var ( X ) Var (Y ) Var ( Z ) X Var (Y ) 5 c. Relative efficiency between Y and X : 2.5 Var ( X ) 2 Var ( Z ) 20 Relative efficiency between Z and X : 2.222 Var ( X ) 9 b. Var ( X ) a. Evidence of non-normality? Normal Probability Plot for Leak Rates ( ML Estimates 99 ML Estimates 95 Mean 0.0515 StDev 0.0216428 90 Goodness of Fit 80 Percent 7.7 2 AD* 70 60 50 40 30 20 10 5 1 0.00 0.05 0.10 Data No evidence of nonnormality exists. b. The minimum variance unbiased point estimator of the population X i .0515 mean is the sample mean: X n c. The unbiased point estimate of the variance of the sample mean: s 2 x (.0216428)2 .0004684 Var ( X ) 2 n ˆ (X ) ; Var s 2 .0004684 .00000937 n 50 0.596 177 178 7.8 Statistics for Business & Economics, 7th edition a. Evidence of non-normality? No evidence of the data distribution coming from a non-normal population b. The minimum variance unbiased point estimator of the population X i 3.8079 mean is the sample mean: X n Descriptive Statistics: Volumes Variable Volumes Variable Volumes N 75 Minimum 3.5700 Mean 3.8079 Median 3.7900 Maximum 4.1100 Q1 3.7400 TrMean 3.8054 StDev 0.1024 Q3 3.8700 c. Minimum variance unbiased point estimate of the population variance is the sample variance s2 = .0105 7.9 Reliability factor for each of the following: a. 96% confidence level: z 2 = +/- 2.05 b. 88% confidence level: z 2 = +/- 1.56 c. 85% confidence level: z 2 = +/- 1.44 d. .07 z 2 = +/- 1.81 e. 2 = .07 z 2 = +/- 1.48 7.10 Calculate the margin of error to estimate the population mean a. 98% confidence level; n = 64, variance = 144 = 3.495 ME z 2 = 2.33 12 n 64 b. 99% confidence interval, n=120; standard deviation = 100 = 23.552 = 2.58 100 ME z 2 n 120 7.11 Calculate the width to estimate the population mean, for a. 90% confidence level, n = 100, variance = 169 = 4.277 = 2 1.645 13 width = 2ME = 2 z 2 100 n b. 95% confidence interval, n = 120, standard deviation = 25 = 8.9461 = 2 1.96 25 width = 2ME = 2 z 2 120 n SE Mean 0.0118 Chapter 7: Estimation: Single Population 7.12 Calculate the LCL and UCL = 40.2 to 59.8 = 50 1.96 40 n 64 = 81.56 to 88.44 b. x z 2 = 85 2.58 20 n 225 = 506.2652 to 513.73478 c. x z 2 = 510 1.645 50 n 485 a. x z 2 7.13 a. n 9, x 187.9, 32.4, z.10 1.28 187.9 1.28(32.4/3) = 174.076 up to 201.724 b. 210.0 – 187.9 = 22.1 = z / 2 (32.4 / 3), z / 2 2.05 2[1 Fz (2.05)] .0404 100(1-.0404)% = 95.96% 7.14 a. Find the reliability factor for 92% confidence level: z 2 = +/- 1.75 b. Calculate the standard error of the mean = .63246 6 n 90 c. Calculate the width of the 92% confidence interval for the population mean = 2.2136 = 2 1.75 6 width = 2ME = 2 z 2 90 n 7.15 a. n 25, x 2.90, .45, z.025 1.96 = 2.90 1.96(.45/5) = 2.7236 up to 3.0764 x z n b. 2.99 – 2.90 = .09 = z / 2 (.45 / 5), z / 2 1 2[1 Fz (1)] .3174 100(1-.3174)% = 68.26% 7.16 a. n 16, x 4.07, .12, z.005 2.58 4.07 2.58(.12/4) = 3.9926 up to 4.1474 b. narrower since the z score for a 95% confidence interval is smaller than the z score for the 99% confidence interval c. narrower due to the smaller standard error d. wider due to the larger standard error 7.17 Find the reliability factor tv , a. b. c. d. 2 n = 20, 90% confidence level; t = 1.729 n = 7, 98% confidence level; t = 3.143 n = 16, 95% confidence level; t = 2.131 n = 23, 99% confidence level; t = 2.819 179 180 Statistics for Business & Economics, 7th edition 7.18 Find the ME a. n = 20, 90% confidence level, s = 36 = 13.9182 = ME 1.729 36 ME t 2 s n 120 b. n = 7, 98% confidence level, s = 16 = ME 3.143 16 = 19.007 ME t 2 s n 7 c. n = 16, 99% confidence level, x1 = 15, x2 = 17, x3 = 13, x4 = 11, x5 = 14 = 7.5407 x 14, s 2.58199 ME 5.841 2.58199 4 7.19 Time spent driving to work for n = 5 people Descriptive Statistics: Driving_Ex7.19 Variable Driving_Ex7.19 Variable Driving_Ex7.19 N 5 N* 0 Mean 38.40 Minimum 30.00 SE Mean 2.66 Q1 32.50 TrMean * Median 40.00 StDev 5.94 Q3 43.50 Variance 35.30 Maximum 45.00 CoefVar 15.47 Range 15.00 Sum 192.00 IQR 11.00 a. Calculate the standard error s 2.6571 5.94138 n 5 b. Find the value of t for the 95% confidence interval tv , 2 = 2.776 c. Calculate the width for a 95% confidence interval for the population mean ME t 2 s = ME 2.776 5.94138 = 14.752 n 5 7.20 Find the LCL and UCL for each of the following: a. alpha = .05, n = 25, sample mean = 560, s = 45 = 560 2.064 45 = 541.424 to 578.576 x t 2 s n 25 b. alpha/2 = .05, n = 9, sample mean = 160, sample variance = 36 = 160 1.860 6 = 156.28 to 163.72 x t 2 s n 9 c. 1 = .98, n = 22, sample mean = 58, s = 15 = 58 2.518 15 = 49.9474 to 66.0526 x t 2 s n 22 Chapter 7: Estimation: Single Population 7.21 Calculate the margin of error to estimate the population mean for: a. 98% confidence level; n = 64, sample variance = 144 = 3.585 = ME 2.390 12 ME t 2 s n 64 b. 99% confidence level; n = 120, sample standard deviation = 100 = 24.2824 = ME 2.660 100 ME t 2 s n 120 c. 95% confidence level; n = 200, sample standard deviation = 40 = 5.65685 = ME 2.000 40 ME t 2 s n 200 7.22 Calculate the width for each of the following: a. alpha = 0.05, n = 6, s = 40 w 2ME 2t 2 s 2 2.571 40 2(41.98425) 83.9685 n 6 b. alpha = 0.01, n = 22, sample variance = 400 2(12.07142) 24.1428 w 2ME 2t 2 s 2 2.831 20 n 22 c. alpha = 0.10, n = 25, s = 50 2(17.11) 34.22 w 2ME 2t 2 s 2 1.711 50 n 25 7.23 181 a. 95% confidence interval: Results for: TOC.xls One-Sample T: Leak Rates (cc/sec.) Variable Leak Rates ( N 50 Mean 0.05150 StDev 0.02186 SE Mean 0.00309 95.0% CI ( 0.04529, 0.05771) b. 98% confidence interval: One-Sample T: Leak Rates (cc/sec.) Variable Leak Rates ( 7.24 N 50 Mean 0.05150 StDev 0.02186 SE Mean 0.00309 98.0% CI ( 0.04406, 0.05894) a. Results for: Sugar.xls Descriptive Statistics: Weights Variable Weights Variable Weights N 100 Mean 520.95 Minimum 504.70 Maximum 544.80 Median 518.75 TrMean 520.52 Q1 513.80 StDev 9.45 SE Mean 0.95 Q3 527.28 90% confidence interval: Results for: Sugar.xls One-Sample T: Weights Variable Weights N 100 Mean 520.948 StDev 9.451 SE Mean 0.945 90.0% CI ( 519.379, 522.517) 182 Statistics for Business & Economics, 7th edition b. narrower since a smaller value of z will be used in generating the 80% confidence interval. 7.25 n 9, x 157.82, s 38.89, t8,.025 2.306 margin of error: 2.306(38.89/3) = 29.8934 7.26 n 7, x 74.7143, s 6.3957, t6,.025 2.447 margin of error: 2.447(6.3957/ 7 ) = 5.9152 7.27 a. n 10, x 15.90, s 5.30, t9,.005 3.25 15.90 3.25(5.30/ 10 ) = 10.453 up to 21.347 b. narrower since the t-score will be smaller for a 90% confidence interval than for a 99% confidence interval 7.28 n 25, x 42, 740, s 4, 780, t24,.05 1.711 42,740 1.711(4780/5) = $41,104.28 up to $44,375.72 7.29 n 9, x 16.222, s 4.790, t8,.10 1.86 We must assume a normally distributed population 16.222 1.86(4.790/3) = 13.252 up to 19.192 7.30 Find the standard error of the proportion for pˆ (1 pˆ ) .3(.7) a. n = 250, p̂ = 0.3 = .02898 n 250 pˆ (1 pˆ ) .45(.55) b. n = 175, p̂ = 0.45 = .03761 n 175 pˆ (1 pˆ ) .05(.95) c. n = 400, p̂ = 0.05 = .010897 n 400 7.31 Find the margin of error for a. n = 250, p̂ = 0.3, α = .05 z 2 b. n = 175, p̂ = 0.45, α = .08 z 2 c. n = 400, p̂ = 0.05, α = .04 z 2 pˆ (1 pˆ ) .3(.7) = .056806 1.96 n 250 pˆ (1 pˆ ) .45(.55) = .05810 1.75 n 175 pˆ (1 pˆ ) .05(.95) = .02234 2.05 n 400 Chapter 7: Estimation: Single Population 7.32 Find the confidence level for estimating the population proportion for a. 92.5% confidence level; n = 650, p̂ = .10 pˆ (1 pˆ ) .10(1 .10) = .10 1.78 = .079055 to .120945 n 650 b. 99% confidence level; n = 140, p̂ = .01 pˆ z 2 pˆ (1 pˆ ) .01(1 .01) = .01 2.58 = 0.0 to .031696 n 140 c. alpha = .09; n = 365, p̂ = .50 pˆ z 2 pˆ z 2 7.33 7.34 .5(.5) pˆ (1 pˆ ) = .50 1.70 = .4555 to .5445 650 n n = 142, 87 answered GMAT or GRE is ‘very important’. 95% confidence interval for the population proportion: pˆ (1 pˆ ) .61268(1 .61268) = .61268 1.96 = .53255 to .6928 pˆ z 2 n 142 n 95, pˆ 67 / 95 .7053, z.005 2.58 pˆ (1 pˆ ) .7053(.2947) = .7053 (2.58) = n 95 99% confidence interval: .5846 up to .8260 pˆ z / 2 7.35 a. unbiased point estimate of proportion: Tally for Discrete Variables: Adequate Variety Adequate Variety 1 2 N= Count CumCnt 135 135 221 356 356 Percent CumPct 37.92 37.92 62.08 100.00 x 135 .3792 n 356 b. 90% confidence interval: n 356, pˆ 135 / 356 .3792, z.05 1.645 pˆ pˆ (1 pˆ ) = .3792 (1.645) .3792(.6208) / 356 = n .3369 up to .4215 pˆ z / 2 7.36 n 320, pˆ 80 / 320 .25, z.025 1.96 pˆ (1 pˆ ) = .25 (1.96) .25(.75) / 320 = n 95% confidence interval: .2026 up to .2974 pˆ z / 2 183 184 7.37 Statistics for Business & Economics, 7th edition n 400, pˆ 320 / 400 .80, z.01 2.326 pˆ (1 pˆ ) .80(1 .80) = .80 2.326 = .75348 n 400 b. width of a 90% confidence interval pˆ (1 pˆ ) .8(1 .8) w 2ME 2 z 2 2 1.645 2(.0329) .0658 n 400 a. LCL = pˆ z / 2 7.38 width = .545-.445 = .100; ME = 0.05 pˆ 0.495 pˆ (1 pˆ ) .495(.505) .0355 n 198 .05 = z / 2 (.0355), z / 2 1.41 2[1 Fz (1.41)] .0793 100(1-.1586)% = 84.14% 7.39 n 420, pˆ 223/ 420 .5310, z.025 1.96 pˆ (1 pˆ ) = .5310 (1.96) .5310(.4690) / 420 = n 95% confidence interval: .4833 up to .5787 The margin of error is .0477 pˆ z / 2 7.40 n 246, pˆ 40 / 246 .1626, z.01 2.326 pˆ (1 pˆ ) = .1626 (2.326) .1626(.8374) / 246 = n 98% confidence interval: .1079 up to .2173 pˆ z / 2 7.41 a. n 246, pˆ 232 / 246 .9431, z.01 2.326 pˆ (1 pˆ ) = .9431 (2.326) .9431(.0569) / 246 = n 98% confidence interval: .9087 up to .9775 b. n 246, pˆ 10 / 246 .0407, z.01 2.326 pˆ z / 2 pˆ (1 pˆ ) = .0407 (2.326) .0407(.9593) / 246 = n 98% confidence interval: .0114 up to .0699 pˆ z / 2 Chapter 7: Estimation: Single Population 7.42 a. n 21, s 2 16, taking .05, 220,.975 9.59, 220,.025 34.17 (n 1) s 2 2 n 1, / 2 2 (n 1)s 2 2 = n 1,1 / 2 21(16) 21(16) 2 = 9.8332 < 2 34.17 9.59 < 35.036 b. n 16, s 8, 215,.975 6.26, 215,.025 27.49 (n 1) s 2 2 n1, / 2 2 (n 1)s 2 2 n1,1 / 2 15(8)2 15(8)2 2 = = 34.9218 < 2 27.49 6.26 < 153.3546 c. n 28, s 15, 227,.995 11.81, 227,.005 49.64 (n 1) s 2 2 n1, / 2 2 (n 1)s 2 2 n1,1 / 2 = 28(15)2 28(15)2 = 126.9138 < 2 < 2 49.64 11.81 533.446 7.43 a. n 21, s 2 16, taking .05, 220,.975 9.59, 220,.025 34.17 (n 1) s 2 2 n 1, / 2 2 (n 1)s 2 2 = n 1,1 / 2 21(16) 21(16) 2 = 9.8332 < 2 34.17 9.59 < 35.036 b. n 16, s 8, 215,.975 6.26, 215,.025 27.49 (n 1) s 2 2 n1, / 2 2 (n 1)s 2 2 n1,1 / 2 = 15(8)2 15(8)2 = 34.9218 < 2 2 27.49 6.26 < 153.3546 c. n 28, s 15, 227,.995 11.81, 227,.005 49.64 (n 1) s 2 2 n1, / 2 2 (n 1)s 2 2 n1,1 / 2 = 28(15)2 28(15)2 = 126.9138 < 2 < 2 49.64 11.81 533.446 7.44 Random sample from a normal population a. Find the 90% confidence interval for the population variance Descriptive Statistics: Ex7.44 Variable Ex7.44 Mean 11.00 SE Mean 1.41 StDev 3.16 Variance 10.00 Minimum 8.00 Maximum 16.00 n 5, s 2 10, 24,.95 .711, 24,.05 9.49 (n 1) s 2 2 n 1, / 2 56.25879 2 (n 1)s 2 2 n 1,1 / 2 = 4(10) 4(10) 2 = 9.49 .711 4.21496 < 2 < 185 186 Statistics for Business & Economics, 7th edition b. Find the 95% confidence interval for the population variance n 5, s 2 10, 24,.975 .484, 24,.025 11.14 (n 1) s 2 2 n 1, / 2 2 (n 1)s 2 2 = n 1,1 / 2 4(10) 4(10) 2 = 3.59066 < 2 11.14 .484 < 82.6446 7.45 No evidence of non-nomality Probability Plot of Leak Rates (cc/sec.) Normal 99 Mean StDev N AD P-Value 95 90 0.0515 0.02186 50 0.359 0.437 Percent 80 70 60 50 40 30 20 10 5 1 0.00 0.02 0.04 0.06 0.08 Leak Rates (cc/sec.) 0.10 0.12 n = 50, s 2 0.000478 Since df = 49 is not in the Chi-Square Table in the Appendix, we will approximate the interval using df = 50. 49(0.000478) 49(0.000478) (n 1) s 2 (n 1)s 2 2 2 2 2 71.42 32.36 n1, / 2 n1,1 / 2 = 3.279E-4 < 2 < 7.238E-4 7.46 n 10, s 2 28.898, 29,.05 16.92, 29,.95 3.33 (n 1) s 2 2 n1, / 2 2 (n 1)s 2 2 n1,1 / 2 = 9(28.898) 9(28.898) 2 = 15.3713 up to 78.1027 16.92 3.33 7.47 n 20, s 2 6.62, 219,.025 32.85, 219,.975 8.91 (n 1) s 2 (n 1)s 2 19(6.62) 19(6.62) 2 32.85 8.91 n1, / 2 n1,1 / 2 = 3.8289 up to 14.1167. Assume that the population is normally distributed. 2 2 2 = Chapter 7: Estimation: Single Population 7.48 n 18, s 2 108.16, 217,.05 27.59, 217,.95 8.67 (n 1) s 2 (n 1)s 2 17(108.16) 17(108.16) 2 27.59 8.67 n1, / 2 n1,1 / 2 = 66.6444 up to 212.0784. Assume that the population is normally distributed. 2 2 2 = 7.49 a. n 15, s 2 (2.36)2 5.5696 14(5.5696) 14(5.5696) 2 = 2.9852 up to 13.8498 26.12 5.63 b. wider since the chi-square statistic for a 99% confidence interval is larger than for a 95% confidence interval 7.50 n 9, s 2 .7875, 28,.05 15.51, 28,.95 2.73 8(.7875) 8(.7875) 2 = .4062 up to 2.3077 15.51 2.73 7.51 a. ˆ x2 7.52 a. The confidence interval is x t n 1, / 2ˆ x x t n 1, / 2ˆ x . Use 102 1200 80 1.1676 80 1200 1 64 1425 90 0.6667 b. ˆ x2 90 1425 1 129 3200 200 0.6049 c. ˆ x2 200 3200 1 t79,0.025 1.990, and ˆ x2 1.1676 from exercise 7.51 part a. x t n 1, / 2ˆ x x t n 1, / 2ˆ x 142 (1.990)(1.1676) 142 (1.990)(1.1676) or (139.68, 144.32) b. The confidence interval is x t n 1, / 2ˆ x x t n 1, / 2ˆ x . Use t89,0.025 1.987, and ˆ x2 0.6667 from exercise 7.51 part b. x t n 1, / 2ˆ x x t n 1, / 2ˆ x 232.4 (1.987)(0.6667) 232.4 (1.987)(0.6667) or (231.08, 233.72) 187 188 Statistics for Business & Economics, 7th edition c. The confidence interval is x t n 1, / 2ˆ x x t n 1, / 2ˆ x . Use t199,0.025 1.972, and ˆ x2 0.6049 from exercise 7.51 part c. x t n 1, / 2ˆ x x t n 1, / 2ˆ x 59.3 (1.972)(0.6049) 59.3 (1.972)(0.6049) or (58.11, 60.49) 7.53 a. A 100(1 )% confidence interval for the population total is obtained from the following formula. Nx t n 1, / 2 Nˆ x N Nx t n 1, / 2 Nˆ x As stated, N 1325, n 121, s 20, and x 182. For a 95% confidence level, note that tn 1, / 2 t120,0.025 1.98. Next calculate N̂ x . Nˆ x Ns n N n (1325)( 20) 1325 121 2297.325 N 1 1325 1 121 So the 95% confidence interval is Nx t n 1, / 2 Nˆ x N Nx t n 1, / 2 Nˆ x or (1325)(182) (1.98)( 2297.325) N (1325)(182) (1.98)( 2297.325) or (236601, 245699) b. As stated, N 2100, n 144, s 50, and x 1325. For a 98% confidence level, note that tn 1, / 2 t143,0.01 2.35. Next calculate N̂ x . Nˆ x Ns n N n (2100)(50) 2100 144 8446.684 N 1 2100 1 144 So the 98% confidence interval is Nx t n 1, / 2 Nˆ x N Nx t n 1, / 2 Nˆ x or (2100)(1325) (2.35)(8446.684) N (2100)(1325) (2.35)(8446.684) or (2762650, 2802350) Chapter 7: Estimation: Single Population 7.54 a 189 A 100(1 )% confidence interval for the population total is obtained from the following formula. pˆ z / 2ˆ pˆ P pˆ z / 2ˆ pˆ As stated, N 1058, n 160, and x 40. For a 95% confidence level, note that z / 2 z0.025 1.96. Next calculate p̂ and ˆ p̂ . x 40 0.25 n 160 pˆ (1 pˆ ) ( N n) 0.25(1 0.25) 1058 160 ˆ p2ˆ 0.0009956 n 1 ( N 1) 160 1 1058 1 pˆ ˆ pˆ 0.0009956 0.03155 So the 95% confidence interval is pˆ z / 2ˆ pˆ P pˆ z / 2ˆ pˆ or 0.25 (1.96)(0.03155) P 0.25 (1.96)(0.03155) or (0.1882, 0.3118) b. As stated, N 854, n 81, and x 50. For a 99% confidence level, note that z / 2 z0.005 2.576. Next calculate p̂ and ˆ p̂ . x 50 0.6173 n 80 pˆ (1 pˆ ) ( N n) 0.6173(1 0.6173) 854 81 ˆ p2ˆ 0.002643 n 1 ( N 1) 81 1 854 1 pˆ ˆ pˆ 0.002643 0.05141 So the 99% confidence interval is pˆ z / 2ˆ pˆ P pˆ z / 2ˆ pˆ or 0.6173 (2.576)(0.05141) P 0.6173 (2.576)(0.05141) or (0.4849, 0.7497) 190 Statistics for Business & Economics, 7th edition 7.55 Within Minitab, go to Calc Make Patterned Data… in order to generate a simple set of numbers of size ‘n’ or ‘N’. Enter first value as 1, last value as 984 which is the total number of pages in the text. Go to Calc Random Data Sample from Columns… in order to generate a simple random sample of size ‘n’. “Sample ____ rows from column(s):” Enter 50 as the number of rows to sample from. The results will be the observation numbers in the list to include in the sample. 7.56 a. x 9.7, s 6.2 , ˆ x ( s)2 N n (6.2) 2 139 = .7519 n N 50 189 9.7 1.96 (.7519) (8.2262, 11.1738) b. 99% confidence interval: Nx Z / 2 Nˆ x N Nx Z / 2 Nˆ x where, Nx (189)(9.7) 1833.30 s2 (6.2)2 N ( N n) 189(189 50) = 142.1167 n 50 1833.30 2.58(142.1167) 1466.6390 < N < 2199.9610 Nˆ x 7.57 a. x 127.43 s 2 N n (43.27)2 760 2 b. ˆ x = 28.9216 n N 60 820 c. 127.43 1.645 ( 28.9216 ) (118.5834, 136.2766) d. [137.43 – 117.43]/2 = 10 = z / 2 28.9216 , solving for z: 1.86 tabled value of .9686 yields a confidence level of 93.72% or an of .0628 s 2 N n (43.27)2 760 e. 95% confidence interval: using ˆ x 2 n N 60 820 Nx Z / 2 Nˆ x N Nx Z / 2 Nˆ x , where Nx (820)(127.43) 104, 492.6 s2 (43.27)2 N ( N n) 820(820 60) = 4,409.8619 n 60 104,492.6 1.96(4409.8619) 95,849.2706 < N < 113,135.9294 Nˆ x Chapter 7: Estimation: Single Population 7.58 (5.32) 2 85 = .6936 40 125 7.28 2.58 (.6936) (5.4904, 9.0696) b. 90% confidence interval: Nx Z / 2 Nˆ x N Nx Z / 2 Nˆ x , where Nx (125)(7.28) 910 a. ˆ x s2 (5.32)2 N ( N n) 125(125 40) = 86.7054 n 40 910 1.645(86.7054) 767.3696 < N < 1,052.6304 Nˆ x 7.59 a. false: as n increases, the confidence interval becomes narrower for a given N and s2 b. true c. true: the finite population correction factor is larger to account for the fact that a smaller proportion of the population is represented as N increases relative to n. d. true 7.60 x = 143/35 = 4.0857 90% confidence interval: Nx Z / 2 Nˆ x N Nx Z / 2 Nˆ x , where Nx (120)(4.0857) 490.2857 s2 (3.1) 2 Nˆ x N ( N n) 120(120 35) = 52.9210 n 35 490.2857 1.645(52.9210) 403.2307 < N < 577.3407 7.61 p̂ = x/n = 39/400 = .0975 pˆ [ pˆ (1 pˆ ) /(n 1)][( N n) / N ] [(.0975)(.9025) /(399)][(1395 400) /1395] = .0125 95% confidence interval: .0975 1.96(.0125): .073 up to .1220 7.62 p̂ = 56/100 = .56 pˆ [ pˆ (1 pˆ ) /(n 1)][( N n) / N ] [(.56)(.44) / 99][(420 100) / 420] = .0435 90% confidence interval: .56 1.645(.0435): .4884 up to .6316 191 192 7.63 Statistics for Business & Economics, 7th edition p̂ = 37/120 = .3083 pˆ [ pˆ (1 pˆ ) /(n 1)][( N n) / N ] [(.3083)(.6917) /(119)][(257 120) / 257] = .0309 95% confidence interval: .3083 1.96(.0309): .2477 up to .3689 7.64 p̂ = 31/80 = .3875 pˆ [ pˆ (1 pˆ ) /(n 1)][( N n) / N ] [(.3875)(.6125) /(79)][(420 80) / 420] = .0493 90% confidence interval: .3875 1.645(.0493): .3064 up to .4686 128.688 < Np < 196.812 or between 129 and 197 students intend to take the final. a. n 10, x 257, s 37.2, t9,.05 1.833 7.65 = 257 1.833(37.2/ 10 ) = 235.4318 up to 278.5628 x t 2 s n assume that the population is normally distributed b. 95% and 98% confidence intervals: = 257 2.262(37.2/ 10 ) = 230.39 up to 283.61 [95%]: x t 2 s n = 257 2.821(37.2/ 10 ) = 223.815 up to [98%]: x t 2 s n 290.185 n 16, x 150, s 12, t15,.025 2.131 7.66 = 150 2.131(12/4) = 143.607 up to 156.393 x t 2 s n It is recommended that he stock 157 gallons. n 50, x 30, s 4.2, z.05 1.645 7.67 = 30 1.645(4.2/ 50 ) = 29.0229 up to 30.9771 7.68 Results from Minitab: Descriptive Statistics: Passengers7_68 Variable Passenge Variable Passenge N 50 Minimum 86.00 Mean 136.22 Maximum 180.00 One-Sample T: Passengers7_68 Variable Passengers8_ N 50 Mean 136.22 Median 141.00 TrMean 136.75 Q1 118.50 StDev 24.44 SE Mean 3.46 Q3 152.00 SE Mean 3.46 95% confidence interval: 129.27 up to 143.17 StDev 24.44 ( 95.0% CI 129.27, 143.17) Chapter 7: Estimation: Single Population 7.69 a. Use a 5% risk. Incorrect labels = 8/48 = 0.1667 0.1667(0.8333) 0.1667 1.96 0.1667 0.1054 = 0.0613 up to 0.2721 48 b. For a 90% confidence interval, 0.1667(0.8333) 0.1667 1.645 0.1667 0.0885 = 0.0782 up to 48 0.2552 7.70 a. The minimum variance unbiased point estimator of the population X i 27 3.375. The unbiased point mean is the sample mean: X n 8 2 2 x nx 94.62 8(3.375) 2 i 2 .4993 estimate of the variance: s n 1 7 x 3 b. pˆ .375 n 8 7.71 3.69 – 3.59 = 0.10 = z / 2 (1.045/ 457), z / 2 2.05 2[1 Fz (2.05)] .0404 100(1-.0404)% = 95.96% 7.72 n 174, x 6.06, s 1.43 6.16 – 6.06 = .1 = z / 2 (1.43/ 174), z / 2 .922 2[1 Fz (.92)] .3576 100(1-.3576)% = 64.24% 7.73 n = 33 accounting students who recorded study time a. An unbiased, consistent, and efficient estimator of the population mean is the sample mean x = 8.545 b. Find the sampling error for a 95% confidence interval; Using degrees of freedom = 30, = 1.3568 ME 2.042 3.817 33 7.74 n = 25 patient records – the average length of stay is 6 days with a standard deviation of 1.8 days a. find the reliability factor for a 95% interval estimate t / 2 2.064 b. Find the LCL for a 99% confidence interval estimate of the population mean . The LCL = 5.257 ME t 2 s = 6 2.064 1.8 n 25 193 194 Statistics for Business & Economics, 7th edition 7.75 n = 250, x = 100 a. Find the standard error to estimate population proportion of first timers pˆ (1 pˆ ) .4(1 .4) = .03098 n 250 b. Find the sampling error. Since no confidence level is specified, we find the sampling error (Margin of Error) for a 95% confidence interval. ME = 1.96 (0.03098) = 0.0607 c. For a 92% confidence interval, ME = 1.75 (0.03098) = 0.05422 0.40 .05422 giving 0.3457 up to 0.4542 a. 90% confidence interval reliability factor = t / 2 1.729 b. Find the LCL for a 99% confidence interval LCL = 60.75 – 2.861 21.83159 = 46.78 or approximately 47 20 passengers. 7.76 7.77 a. Find a 95% confidence interval estimate for the population proportion of students who would like supplements in their smoothies. Tally for Discrete Variables: Supplements, Health Consciousness Supplements No 0 Yes 1 N= n 113, Count 42 71 113 Percent 37.17 62.83 Health Consciousness Very 1 Moderately 2 Slight 3 Not Very 4 N= Count 29 55 20 9 113 Percent 25.66 48.67 17.70 7.96 pˆ 71/113 .62832, z.05 1.96 pˆ (1 pˆ ) .62832(1 .62832) = .62832 1.96 n 113 0.0891 = 0.5392 up to 0.71742. pˆ z / 2 = 0.62832 ± b. pˆ 29 / 113 0.2566 For 98% confidence level, pˆ 2.33 (0.2566)(1 0.2566) 0.2566 0.09573 113 or 0.1609 up to 0.3523 c. pˆ 77 / 113 0.6814 0.6814 1.645 0.7535 (0.6814)(1 0.6814) 0.6814 0.0721 or 0.6093 up to 113 Chapter 7: Estimation: Single Population 7.78 n = 100 students at a small university. a. Estimate the population grade point average with 95% confidence level One-Sample T: GPA Variable GPA N 100 Mean 3.12800 StDev 0.36184 SE Mean 0.03618 95% CI (3.05620, 3.19980) b. Estimate the population proportion of students who were very dissatisfied (code 1) or moderately dissatisfied (code 2) with parking. Use a 90% confidence level. Tally for Discrete Variables: Parking Parking 1 2 3 4 5 N= Count 19 26 18 18 19 100 n 100, Percent 19.00 26.00 18.00 18.00 19.00 pˆ 45 /100 .45, z.05 1.645 , pˆ z / 2 .45 1.645 pˆ (1 pˆ ) n .45(1 .45) = .45 ± .08184 = .368162 up to .53184 100 c. Estimate the proportion of students who were at least moderately satisfied (codes 4 and 5) with on-campus food service Tally for Discrete Variables: Dining Dining 1 2 3 4 5 N= Count 14 26 21 20 19 100 Percent 14.00 26.00 21.00 20.00 19.00 n 100, pˆ 39 /100 .39, z.05 1.645 pˆ z / 2 pˆ (1 pˆ ) .39(1 .39) = .39 1.645 = .39 ± .08023 = .30977 up n 100 to .47023. 7.79 a. Estimate the average age of the store’s customers by the sample mean xi 6310 50.48 x n 125 To find a confidence interval estimate we will assume a 95% confidence level: 13.06 50.48 1.96 50.48 2.29 ; 48.19 up to 52.77 years 125 195 196 Statistics for Business & Economics, 7th edition b. Estimate the population proportion of customers dissatisfied with the delivery system Tally for Discrete Variables: Dissatisfied with Delivery Dissatisfied with Delivery 1 2 N= Count 9 116 125 Percent 7.20 92.80 pˆ 9 /125 .072 ; Assuming a 95% confidence level, we find: 0.072 1.96 (0.072)(1 0.072) 0.072 0.0453 or 0.0267 up to 0.1173 125 c. Estimate the population mean amount charged to a Visa credit card Descriptive Statistics: Cost of Flowers Variable Cost of Flowers Method of Payment American Express Cash Master Card Other Visa Mean 52.99 51.34 54.58 53.42 52.65 SE Mean 2.23 4.05 3.11 2.99 2.04 TrMean 52.83 51.46 54.43 53.72 52.58 StDev 10.68 16.19 15.25 14.33 12.71 The population mean can be estimated by the sample mean amount charged to a Visa credit card = $52.65. 7.80 n = 500 motor vehicle registrations, 200 were mailed, 160 paid in person, remainder paid on-line. a. Estimate the population proportion to pay for vehicle registration renewals in person, use a 90% confidence level. Test and CI for One Proportion Test of p = 0.5 vs p not = 0.5 Sample X N Sample p 90% CI 1 160 500 0.320000 (0.285686, 0.354314) The 90% confidence interval is from 28.56856% up to 35.4314% b. Estimate the population proportion of on-line renewals, use 95% confidence. Test and CI for One Proportion Test of p = 0.5 vs p not = 0.5 Sample X N Sample p 95% CI 1 140 500 0.280000 (0.240644, 0.319356) The 95% confidence interval is from 24.0644% up to 31.9356% Median 50.55 50.55 55.49 54.85 50.65 Chapter 7: Estimation: Single Population 7.81 From the data in 7.80, find the confidence level if the interval extends from 0.34 up to 0.46. ME = ½ the width of the confidence interval. 0.46 – 0.34 = 0.12 / 2 = 0.06 pˆ (1 pˆ ) or ME z 2 n (0.4)(0.6) 0.06 z 2 500 Solving for z: z 2 2.74 Area from the z-table = .4969 x 2 = .9938. The confidence level is 99.38% 7.82 From the data in 7.80, find the confidence level if the interval extends from 23.7% up to 32.3%. ME = ½ the width of the confidence interval. .323 – .237 = .086 / 2 = .043 and pˆ .28 pˆ (1 pˆ ) .28(1 .28) solving for z: z 2 2.14 .043 z 2 n 500 Area from the z-table = .4838 x 2 = .9676. The confidence level is 96.76% ME z 2 7.83 a. What is the margin of error for a 99% confidence interval pˆ (1 pˆ ) pˆ x 250 .7143 , ME z 2 n 350 n .7143(1 .7143) ME 2.58 350 ME = .0623 = b. Is the margin of error for a 95% confidence larger, smaller or the same as the 99% confidence level? The margin of error will be smaller (more precise) for a lower confidence level. The difference in the equation is the value for z which would drop from 2.58 down to 1.96. 7.84 Compute the 98% confidence interval of the mean age of on-line renewal users. n= 460, sample mean = 42.6, s = 5.4. = 42.6 ± .58664 = 42.0134 up to x t 2 s = 42.6 2.33 5.4 n 460 43.18664 747 (11.44)2 90 10 74.7 , s = 11.44, ˆ 2 x 11.633 10 10 90 90% confidence interval: 74.7 1.645 11.633 : 69.089 up to 80.311 b. The interval would be wider; the z-score would increase to 1.96 7.85 a. x 197 198 Statistics for Business & Economics, 7th edition Chapter 7: Estimation: Single Population (149.92)2 272 50 366.888 50 272 99% confidence interval: 492.36 2.58 366.888 : 442.9139 up to 541.7501 b. 95% confidence interval: 492.36 1.96 366.888 : 454.8175 up to 529.9025 c. The 90% interval is narrower; the z-score would decline to 1.645 7.86 a. ˆ 2 x 7.87 pˆ 36 .6(.4) 148 60 .6 , ˆ pˆ .0024 60 60 148 95% confidence interval: .6 1.96 .0024 : .504 up to .696 199