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150 Chapter 8: Confidence Interval Estimation CHAPTER 8 8 8.1 X Z = 85 1.96 83.04 86.96 n 64 24 8.2 X Z = 125 2.58 114.68 135.32 n 36 8.3 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 we are 95 percent confident that this sample is one that does correctly estimate the true average amount. 8.4 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 every possible interval estimate of the true mean, the entire population (sample size N) would have to be selected. 8.5 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. 8.6 Approximately 5% of the intervals will not include the true population mean somewhere in the interval. Since the true population mean is not known, we do not know for certain whether it is in the one interval we have developed, between 10.99408 and 11.00192 inches. 0.02 8.7 (a) X Z 0.995 2.58 0.9877 1.0023 n 50 (b) Since the value of 1.0 is included in the interval, there is no reason to believe that the average is different from 1.0 gallon. (c) No. Since is known and n = 50, from the central limit theorem, we may assume that the sampling distribution of X is approximately normal. (d) An individual can with an observed value of 0.98 gallon of paint yields a Z-value of – 1.0. The confidence interval represents bounds on the estimate of the average of a sample of 50 cans of paint, not an individual value. (e) The reduced confidence level narrows the width of the confidence interval. 0.02 (a) X Z 0.995 1.96 0.9895 1.0005 n 50 (b) Since the value of 1.0 is still included in the interval, there is no reason to believe that the average is different from 1.0 gallon. Solutions to End-of-Section and Chapter Review Problems 151 8.8 (a) (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 is known and n = 64, from the central limit theorem, we may assume that the sampling distribution of X is approximately normal. (d) An individual value of 320 is only 0.30 standard deviations below the sample mean of 350. The confidence interval represents bounds on the estimate of a sample of 64, not an individual value. (e) The confidence interval is narrower based on a process standard deviation of 80 hours rather than the original assumption of 100 hours. 80 (a) X Z 350 1.96 330.4 369.6 n 64 (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 last an average of 400 hours. 0.05 8.9 (a) X Z 1.99 1.96 1.9802 1.9998 n 100 (b) No. Since is known and n = 100, from the central limit theorem, we may assume that the 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 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. 0.05 (a) X Z 1.97 1.96 1.9602 1.9798 n 100 8.10 (a) t9 = 2.2622 (b) t9 = 3.2498 (c) t31 = 2.0395 (d) t64 = 1.9977 (e) t15 = 1.7531 152 Chapter 8: Confidence Interval Estimation s 24 8.11 X t 75 2.0301 66.8796 83.1204 n 36 s 15 8.12 X t 50 2.9467 38.9499 61.0501 n 16 3.7417 8.13 Set 1: 4.5 2.3646 1.3719 7.6281 8 2.4495 Set 2: 4.5 2.3646 2.4522 6.5478 8 The data sets have different confidence interval widths because they have different values for the standard deviation. 6.4660 8.14 Original data: 5.8571 2.4469 – 0.1229 11.8371 7 2.1602 Altered data: 4.00 2.4469 2.0022 5.9978 7 The presence of an outlier in the original data increases the value of the sample mean and greatly inflates the sample standard deviation. s 21.4 8.15 (a) X t 195.3 2.1098 184.6581 205.9419 n 18 (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. s 0.32 8.16 (a) X t 1.67 2.0930 $1.52 $1.82 n 20 (b) The store owner can be 95% confidence that the population mean retail value of greeting cards that it has in its inventory is somewhere in between $1.52 and $1.82. 8.17 (a) (b) The population of dental expenses must be approximately normally distributed. Solutions to End-of-Section and Chapter Review Problems 153 s 138.80 8.17 (d) X t 261.40 2.2622 $162.11 $360.69 n 10 cont. (e) The additional $500 in dental expenses, divided across the sample of 10, raises the mean by $50 and increases the standard deviation by nearly $20. The interval half- width increases over $11 in the process. The new interval is: s 157.056 X t 311.40 1.8331 $220.36 $402.22 n 10 s 194645.95 8.18 (a) X t 2153666.667 2.145 107575.24< 323158.09 n 15 (b) The number of shares traded is approximately normally distributed. s 1.8154 8.19 (a) X t 3.8415 2.0227 3.26< 4.42 n 40 s 1.8154 (b) X t 3.8415 2.7079 3.06< 4.62 n 40 (c) The 99% confidence interval is wider because a higher percentage of all the intervals that could be constructed will have to include the true population mean. s 5.9005 8.20 (a) X t 4.3171 2.0322 2.29< 6.34 n 35 s 16.9253 (b) X t 11.4 2.0322 5.59< 17.21 n 35 s 0.6143 (c) X t 0.6954 2.0322 0.48< 0.91 n 35 s 53.5766 8.21 (a) X t 164.85 2.0930 139.78< 189.92 n 20 s 8.8572 (b) X t 45.65 2.0930 41.50< 49.80 n 20 (c) We need to assume that the population has a normal distribution since the sample size of 20 is not large enough to invoke the central limit theorem for distributions that are not symmetric. X 50 ps (1 – ps ) 0.25(0.75) 8.22 ps = 0.25 ps Z 0.25 1.96 n 200 n 200 0.19 p 0.31 X 25 ps (1 – ps ) 0.0625(0.9375) 8.23 ps = 0.0625 ps Z 0.25 2.58 n 400 n 400 0.0313 p 0.0937 154 Chapter 8: Confidence Interval Estimation ps (1 – ps ) 0.41(0.59) 8.24 (a) ps = 0.41 ps Z 0.41 1.96 n 200 0.3418 p 0.4782 (b) The dealer can infer that the proportion of all customers who still own the cars they purchased at the dealership 5 years earlier is somewhere between 03418 and 0.4782 with a 95% level of confidence. 8.25 (a) (b) 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 somewhere between 0.22 and 0.32 with a 99% level of confidence. X 453 ps (1 – ps ) 0.53(1 0.53) 8.26 (a) ps 0.5304 ps Z 0.53 1.96 n 854 n 854 0.497 p 0.564 X 453 (b) ps 0.5304 n 854 ps (1 – ps ) 0.53(1 0.53) ps Z 0.53 2.5758 0.486 p 0.574 n 854 (c) The 99% confidence interval is wider because a higher percentage of all the intervals that could be constructed will have to include the true population proportion of personnel officials that believe job seekers sometimes falsify past salaries. Solutions to End-of-Section and Chapter Review Problems 155 X 229 8.27 (a) ps 0.3001 n 763 ps (1 – ps ) 0.3001(1 0.3001) ps Z 0.3001 1.6449 n 763 0.2728 p 0.3274 X 99 (b) ps 0.1298 n 763 ps (1 – ps ) 0.1298(1 0.1298) ps Z 0.1298 1.6449 n 763 0.1097 p 0.1498 X 1123 8.28 (a) ps 0.9027 n 1244 ps (1 – ps ) 0.9027(1 0.9027) ps Z 0.9027 1.96 n 1244 0.8863 p 0.9192 X (1244 684) (b) ps 0.4502 n 1244 ps (1 – ps ) 0.4502(1 0.4502) ps Z 0.4502 1.96 n 1244 0.4225 p 0.4778 ps (1 – ps ) 0.66(0.34) 8.29 (a) ps Z 0.66 1.96 0.6311 p 0.6896 n 1007 ps (1 – ps ) 0.301(1 0.301) (b) ps Z 0.301 1.96 0.266 p 0.336 n 665 (c) With this information, the Tupperware Corporation can better predict the demand for their lunch boxes. ps (1 – ps ) 0.67(0.33) 8.30 (a) ps Z 0.67 1.96 0.611 p 0.735 n 220 With 95% confidence, between 61.1% and 73.5% of all child-care providers have at least one safety hazard. ps (1 – ps ) 0.3818(1 0.3818) (b) ps Z 0.3818 1.96 0.3176 p 0.4460 n 220 With 95% confidence, between 31.76% and 44.6% of all child-care providers have children wearing clothing with neck drawstrings. 156 Chapter 8: Confidence Interval Estimation ps (1 – ps ) 0.191(1 0.191) 8.30 (c) ps Z 0.191 1.96 0.139 p 0.243 n 220 cont. With 95% confidence, between 13.9% and 24.3% of all child-care providers have cribs with soft bedding that can lead to suffocation. ps (1 – ps ) 0.383(1 0.383) 8.31 (a) ps Z 0.383 1.96 0.365 p 0.401 n 2943 With 95% confidence, between 36.5% and 40.1% of the population of U.S. Internet households plan to buy a computer during the year. ps (1 – ps ) 0.383(1 0.383) (b) ps Z 0.383 2.576 0.360 p 0.406 n 2943 With 99% confidence, between 36.0% and 40.6% of the population of U.S. Internet households plan to buy a computer during the year. ps (1 – ps ) 0.231(1 0.231) (c) ps Z 0.231 1.96 0.216 p 0.246 n 2943 With 95% confidence, between 21.6% and 24.6% of the population of U.S. Internet households plan to buy a new car or truck during the year. ps (1 – ps ) 0.231(1 0.231) (d) ps Z 0.231 2.576 0.211 p 0.251 n 2943 With 99% confidence, between 21.1% and 25.1% of the population of U.S. Internet households plan to buy a new car or truck during the year. Z 2 2 1.962 152 8.32 n 2 2 = 34.57 Use n = 35 e 5 Z 2 2 2.582 100 2 8.33 n 2 2 = 166.41 Use n = 167 e 20 Z 2 p(1– p) 2.582 (0.5)(0.5) 8.34 n 2 2 = 1,040.06 Use n = 1,041 e (0.04) Z 2 p(1– p) 1.962 (0.4)(0.6) 8.35 n 2 2 = 2,304.96 Use n = 2,305 e (0.02) Z 2 2 1.962 4002 8.36 (a) n 2 2 = 245.86 Use n = 246 e 50 Z 2 2 1.962 4002 (b) n 2 = 983.41 Use n = 984 e 252 Z 2 2 1.962 (0.02)2 8.37 n 2 2 = 96.04 Use n = 97 e (0.004) Solutions to End-of-Section and Chapter Review Problems 157 Z 2 2 1.962 (100)2 8.38 n 2 2 = 96.04 Use n = 97 e (20) Z 2 2 1.962 (0.05) 2 8.39 n 2 2 = 96.04 Use n = 97 e (0.01) Z 2 2 2.582 252 8.40 (a) n 2 2 = 166.41 Use n = 167 e 5 Z 2 2 1.962 252 (b) n 2 2 = 96.04 Use n = 97 e 5 Z 2 2 1.6452 45 2 8.41 (a) n 2 2 = 219.19 Use n = 220 e 5 Z 2 2 2.582 45 2 (b) n 2 2 = 539.17 Use n = 540 e 5 Z 2 2 1.962 202 8.42 n 2 2 = 61.47 Use n = 62 e 5 s 108947.59 8.43 (a) X t 134088.89 2.110 79910.45 188267.33 n 18 Z 2 2 1.96 108947.6 2 2 (b) n 2 114 e 200002 (c) The sample size is computed using the sample standard deviation of the shares traded for the 18 companies as an estimate for the population standard deviation. This is not likely to be an accurate estimate and, hence, the result will not likely by valid if the study is performed today. Z 2 p(1– p) 1.6452 (0.5)(0.5) 8.44 (a) n 2 2 = 422.82 Use n = 423 e (0.04) Z 2 p(1– p) 1.962 (0.5)(0.5) (b) n 2 2 = 600.25 Use n = 601 e (0.04) Z 2 p (1 – p ) 1.96 2 (0.5)(0.5) (c) n = 1067.07 Use n = 1068 e2 (0.03) 2 (d) In general, larger sample size is needed if a higher level of confidence is required holding everything else fixed. This increase in sample size reflects the higher price for the increased in level of confidence. If a higher precision is required, which is reflected in lower acceptable sampling error, the sample size will need to be raised to reflect the higher associated cost holding everything else fixed. 158 Chapter 8: Confidence Interval Estimation Z 2 p(1– p) 2.582 (0.10)(0.90) 8.45 (a) n 2 2 = 122.26 Use n = 123 e (0.07) Z 2 p(1– p) 1.962 (0.10)(0.90) (b) n 2 2 = 96.04 Use n = 97 e (0.06) X 94 8.46 (a) ps 0.1604 n 586 ps (1 – ps ) 0.1604(1 0.1604) ps Z 0.1604 1.96 0.1307 p 0.1901 n 586 Z 2 p (1 – p ) 1.96 2 (0.1604)(1 0.1604) (b) n 1293.34 Use n = 1294 e2 (0.02) 2 Z 2 p (1 – p ) 2.57582 (0.1604)(1 0.1604) (c) n 2233.84 Use n = 2234 e2 (0.02) 2 X 707 8.47 (a) ps 0.6805 n 1039 ps (1 – ps ) 0.6805(1 0.6805) ps Z 0.6805 1.96 n 1039 0.6521 p 0.7088 Z 2 p (1 – p ) 1.96 2 (0.6805)(1 0.6805) (b) n 2088.02 Use n = 2089 e2 (0.02) 2 Z 2 p (1 – p ) 1.96 2 (0.6805)(1 0.6805) (c) n 8352.07 Use n = 8353 e2 (0.01) 2 Z 2 p (1 – p ) 1.962 (0.50)(0.50) 8.48 n 2400.91 Use n = 2401 e2 (0.02) 2 ps (1 – ps ) 0.3(0.7) 8.49 (a) ps Z 0.3 1.96 0.273 p 0.327 n 1100 Z p (1 – p ) 2.576 (0.30)(0.70) 2 2 (b) n 13933.34 Use n = 13934 e2 (0.01) 2 s N –n 7.8 500 – 25 8.50 (b) N X N t 500 25.7 500 2.7969 n N –1 25 500 –1 $10,721.53 Population Total $14,978.47 Solutions to End-of-Section and Chapter Review Problems 159 8.51 Using Excel, Di (Di – D )2 13.76 164.719256 42.87 1759.3243 34.65 1137.32841 11.09 103.312994 14.54 185.349164 22.87 481.552302 25.52 604.879592 9.81 78.9307865 10.03 82.8882785 15.49 209.868823 Mean = 1.00315 Sum, first 10 items = 4780.82023 191*(0 – 1.00315)2 = 192.205195 (Di – D )2 4973.025425 St Dev = 4.99900765 sD N –n 4.9990 10, 000 – 200 N D N t 10, 000 1.00315 10, 000 1.96 n N –1 200 10, 000 –1 3,172.53 Total Difference in the Population 16,890.47 ps (1 – ps ) 0.04(1 0.04) 8.52 (a) ps Z 0.04+1.2816 p 0.0545 n 300 ps (1 – ps ) 0.04(1 0.04) (b) ps Z 0.04+1.645 p 0.0586 n 300 ps (1 – ps ) 0.04(1 0.04) (c) ps Z 0.04+2.3263 p 0.0663 n 300 s N –n $0.32 300 – 20 8.53 N X N t 300 $1.67 300 2.0930 n N –1 20 300 –1 $457.52 Population Total $544.48 s N –n $138.8046 3000 –10 8.54 N X N t 3000 $261.40 3000 1.8331 n N –1 10 3000 –1 $543,176.96 Population Total $1,025,223.04 s N –n $93.67 1546 – 50 8.55 N X N t 1546 $252.28 1546 2.0096 n N –1 50 1546 –1 $349,526.64 Population Total $430,523.12 160 Chapter 8: Confidence Interval Estimation sD N –n $29.5523 4000 –150 8.56 N D N t 4000 $7.45907 4000 2.6092 n N –1 150 4000 –1 $5,125.99 Total Difference in the Population $54,546.57 Note: The t-value of 2.6092 for 95% confidence and df = 149 was derived on Excel. sD N –n $25.2792 1200 –120 8.57 N D N t 1200 (–$0.9583) 1200 1.98 n N –1 120 1200 –1 – $6,354.02 Total Difference in the Population $4,054.10 ps (1 – ps ) 0.0367(1 0.0367) 8.58 (a) ps Z 0.0367+1.645 p 0.0545 n 300 (b) Since the upper bound is higher than the tolerable exception rate of 0.04, the auditor should request a larger sample. ps (1 – ps ) 0.024(1 0.024) 8.59 (a) ps Z 0.024+1.645 p 0.0353 n 500 (b) With 95% level of confidence, the auditor can conclude that the rate of noncompliance is less than 0.0353, which is less than the 0.05 tolerable exception rate for internal control, and, hence, the internal control compliance is adequate. 8 60 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. 8.61 The t distribution is used for obtaining a confidence interval for the mean when is unknown. 8.62 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. 8.63 The confidence interval will be narrower and the sample size will be smaller. 8.64 In some applications such as auditing, interest is primarily on the total amount of a variable rather than the average amount. 8.65 Difference estimation involves determining the difference between two amounts, rather than a single amount. Solutions to End-of-Section and Chapter Review Problems 161 ps (1 – ps ) 0.43(0.57) 8.66 (a) ps Z 0.43 1.96 0.416 p 0.444 n 5000 The sampling error is 0.014. ps (1 – ps ) 0.159(1 0.159) (b) ps Z 0.159 1.96 0.149 p 0.169 n 5000 The sampling error is 0.010 (c) The sampling error is the largest when ps 0.5 . Hence, the sampling error of (a) with ps 0.43 is larger than that of (b) when ps 0.159 8.67 (a) The population from which this sample was drawn was the collection of all subscribers to the Redbook magazine who visited the magazine's web site. (b) The sample is not a random sample from this population. The sample consisted of only those subscribers to the magazine who visited the magazine's web site and chose to fill up 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 up 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 drawn a random sample from the list of all subscribers to the magazine and offer them the option of filling up 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 in 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. Z 2 p (1 – p ) 1.96 2 (0.6195) (0.6195) The sample size needed is n 2264 e2 (0.02) 2 ps (1 – ps ) 0.657(1 0.657) 8.68 (a) ps Z 0.657 1.96 0.608 p 0.705 n 364 Z p (1 – p ) 1.96 (0.657) (0.657) 2 2 (b) n 2165 e2 (0.02) 2 ps (1 – ps ) 0.870(1 0.870) (c) ps Z 0.870 1.96 0.828 p 0.913 n 239 ps (1 – ps ) 0.46(1 0.460) (d) ps Z 0.46 1.96 0.397 p 0.523 n 239 162 Chapter 8: Confidence Interval Estimation s 0.1058 8.69 (a) X t 5.5014 2.6800 5.46< 5.54 n 50 (b) Since 5.5 gram is within the 99% confidence interval, the company can claim that the average weight of tea in a bag is 5.5 gram with a 99% level of confidence. s 25.2835 8.70 (a) X t 43.8889 2.0555 33.89< 53.89 n 27 (b) With 95% confidence, the average approval process takes somewhere between 33.89 to 53.89 days. s 0.0461 8.71 (a) X t 8.4209 2.0106 8.41< 8.43 n 49 (b) With 95% confidence, the population average 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. s 89.5508 8.72 (a) X t 1723.4 2.0452 1689.96< 1756.84 n 30 (b) With 95% confidence, the population average force is somewhere between 1689.96 and 1756.84 pounds. Hence, the company's requirement of strength of the insulators being at least 1500 pounds is being met with a 95% level of confidence. s 3.8 8.73 (a) X t 15.3 2.0227 14.085 16.515 n 40 p (1 – ps ) 0.675(0.325) (b) ps Z s 0.675 1.96 0.530 p 0.820 n 40 Z 2 2 1.96 5 2 2 (c) n 2 2 = 24.01 Use n = 25 e 2 Z 2 p(1– p) 1.96 2 (0.5)(0.5) (d) n 2 2 = 784 Use n = 784 e (0.035) (e) If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 784) should be used. s 380 8.74 (a) X t 1759 2.6490 1,638.69 1,879.31 n 70 p (1– ps ) 0.60(0.40) (b) p s Z s 0.60 1.96 0.485 p 0.715 n 70 Solutions to End-of-Section and Chapter Review Problems 163 s 4 8.75 (a) X t 9.7 2.0639 8.049 11.351 n 25 ps (1 – ps ) 0.48(0.52) (b) ps Z 0.48 1.96 0.284 p 0.676 n 25 Z 2 2 1.96 2 4.5 2 (c) n 2 2 = 34.57 Use n = 35 e 1.5 Z 2 p(1– p) 1.6452 (0.5)(0.5) (d) n 2 2 = 120.268 Use n = 121 e (0.075) (e) If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 121) should be used. Z 2 2 2.582 182 8.76 (a) n 2 2 = 86.26 Use n = 87 e 5 Note: If the t-value used is carried out to 2.5758, the value of n is 85.986 and only 86 women would be sampled. Z 2 p(1– p) 1.6452 (0.5)(0.5) (b) n 2 2 = 334.07 Use n = 335 e (0.045) (c) If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 335) should be used. s $11.39 8.77 (a) X t $28.52 1.9949 $25.80 $31.24 n 70 p (1 – ps ) 0.40(0.60) (b) ps Z s 0.40 1.645 0.3037 p 0.4963 n 70 Z 2 2 1.96 10 2 2 (c) n 2 2 = 96.04 Use n = 97 e 2 Z 2 p(1– p) 1.6452 (0.5)(0.5) (d) n 2 2 = 422.82 Use n = 423 e (0.04) (e) If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 423) should be used. s $9.22 8.78 (a) X t $21.34 1.9949 $19.14 $23.54 n 70 ps (1 – ps ) 0.3714(0.6286) (b) ps Z 0.3714 1.645 n 70 0.2764 p 0.4664 Z 2 2 1.96 102 2 (c) n 2 2 = 170.74 Use n = 171 e 1.5 Z 2 p(1– p) 1.6452 (0.5)(0.5) (d) n 2 2 = 334.08 Use n = 335 e (0.045) (e) If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 335) should be used. 164 Chapter 8: Confidence Interval Estimation s $7.26 8.79 (a) X t $38.54 2.0010 $36.66 $40.42 n 60 p (1 – ps ) 0.30(0.70) (b) ps Z s 0.30 1.645 0.2027 p 0.3973 n 60 Z 2 2 1.96 2 82 (c) n 2 2 = 109.27 Use n = 110 e 1.5 Z 2 p(1– p) 1.6452 (0.5)(0.5) (d) n 2 2 = 422.82 Use n = 423 e (0.04) (e) If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 423) should be used. Z 2 p (1 – p ) 1.96 2 (0.5) (0.5) 8.80 (a) n = 384.16 Use n = 385 e2 (0.05) 2 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. ps (1 – ps ) 0.84(0.16) (b) ps Z 0.84 1.96 n 50 0.7384 p 0.94161 (c) 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 according perform a cost-benefit analysis to decide if he/she want to sell the Ice Melt product. 7 8.81 (a) ps 0.14 50 ps (1 – ps ) N –n 0.14(0.86) 1000 – 50 ps Z 0.14 1.2816 n N –1 50 1000 – 1 p 0.2013 (b) Since the upper bound is higher than the tolerable exception rate of 0.15, the auditor should request a larger sample. Z 2 2 1.962 302 8.82 (a) n0 = 138.292 e2 52 n0 N 138.292 25056 n = 137.53 Use n = 138 n0 ( N –1) 138.292 (25056 –1) 12 (b) ps 0.087 138 ps (1 – ps ) N – n ps Z n N –1 0.087(0.913) 25, 056 –138 0.087 1.645 138 25, 056 –1 0.0476 p 0.1263 Solutions to End-of-Section and Chapter Review Problems 165 8.82 (c) Using Excel, we find the t-value for 95% confidence and 137 degrees cont. of freedom is t = 1.9774. s N –n $34.55 25,056 –138 X t $93.70 1.9774 n N –1 138 25,056 – 1 $87.90 X $99.50 s N –n (d) N X N t n N –1 $34.55 25, 056 –138 25, 056 $93.70 25, 056 1.9774 138 25, 056 –1 $2,202,427.61 Population Total $2,493,066.79 D 241 (D – D) 2 6432.12 (e) D 1.7463768 sD 6.85199 n 138 n–1 137 s N –n ND Nt D n N –1 $6.85199 25,056 –138 25,056 $1.7463768 25,0561.9774 138 25,056 –1 $14,937.30 Total Difference in the Population $72,577.14 Z 2 2 2.582 4002 8.83 (a) n = 26.6256 Use n = 27 e2 2002 s N –n $384.62 258 – 27 (b) N X N t 258 $3,054 258 2.77872 n N –1 27 258 –1 $737,655.50 Population Total $838,275.58 166 Chapter 8: Confidence Interval Estimation The Springville Herald Case SH8.1 (a) The advantages of random digit dialing is that this method allows the survey to reach households that have unlisted telephone numbers. One of the disadvantages is that additional telephone calls would be made due to calls to commercial and other non- residential telephone numbers as well as calls to households outside the Springville Herald distribution area. The other disadvantage would be the inability to reach individuals who might be interested in purchasing the Springville Herald , but do not have a telephone in their household. (b) An alternative approach would involve the use of a mail survey of households with a telephone follow-up to contact non-responding households. The advantage of this approach is that the mail survey would be less expensive to conduct and would be able to reach some households without phones. However, the rate of response would undoubtedly be lower than for a telephone survey. SH8.2 Note: 95% confidence intervals used Question 1: Yes: ps = .8670 .834 to .900 Question 2: Yes: ps = .3864 .3355 to.4373 Question 3: Monday - Saturday: ps = .1323 .0754 to .1894 Question 3: Sunday only: ps = .1838 .1187 to .2489 Question 3: Every day: ps = .6838 .6056 to .7620 Question 4: Every day: ps = .3611 .2970 to .4252 Question 4: Most days: ps = .4398 .3736 to .5060 Question 4: Occasionally: ps = .1991 .1458 to .2524 Question 5: Every Sunday: ps = .6389 .5757 to .7039 Question 5: 2-3 Sundays a month: ps = .25 .1923 to .3077 Question 5: Once a month: ps = .1111 .0692 to .1530 Question 6: Convenience store/Delicatessen: ps = .3426 .2793 to .4059 Question 6: Stationery/Candy store: ps = .4398 .3736 to .5060 Question 6: Vending machine: ps = .0972 .0577 to .1367 Question 6: Supermarket: ps = .0602 .0285 to .0919 Question 6: Other: ps = .0602 .0285 to .0919 Question 7: Yes: ps = .2130 .1584 to .2676 Question 8: X = 3.7348 s = .3602 3.6278 to 3.8418 Question 9: Yes: ps = .3920 .3410 to .4430 Question 10: Yes : ps = .1875 .1467 to .2283 A large proportion of households read the paper (estimated to be between 83.4 and 90%) but it is estimated that only between 33.55% and 43.73% receive a home subscription. The bulk of those who have a subscription receive the newspaper every day. Those who purchase the newspaper are most likely to do so either every day or most days and most Sundays. They are most likely to purchase the newspaper in a convenience store/delicatessen or stationery/ candy store. However, we estimate that only between 15.84% and 26.76% would consider a discounted trial subscription. We should in the future investigate the reasons why readers do not wish to receive a trial subscription. Of those who would consider a discounted trial subscription, we estimate the average price that they would be willing to pay to be between $3.63 and $3.84. We also estimate that only between 34.1% and 44.3% of readers read another daily newspaper. Finally, between 14.67 and 22.83% of the readers would consider prepaying for six months in order to receive a restaurant discount card.

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