#1: The publisher of Golf Illustrated has found through past experience that 60% of the subscribers renew their subscriptions. The publisher selects a random sample of 200 subscribers and finds that 108 plan to renew their subscriptions. a. Compute the 99% confidence interval for the population proportion of subscribers that plan to renew. b. Interpret this confidence interval. c. How many subscribers should the publisher sample in order to estimate to within 5% and with 99% confidence the proportion of all subscribers who plan to renew? #2: An experiment has been conducted to evaluate a new process for producing synthetic diamonds. Six diamonds have been generated by the new process with recorded weights noted. This yields the following results. Sample Size = 6 Sample Mean = .53 carats Sample Standard Deviation = .0559 carats a. Construct the 95% confidence interval for the population mean weight of synthetic diamonds (in carats) using this new synthetic process. b. Interpret this interval. c. How many diamonds should be tested if we wish to generate a 95% confidence interval for the mean weight of synthetic diamonds that is accurate to within .01 carats? #3: The ACTION Paper Company employs a human resources manager who is given responsibility for employee benefits. There is a question about the mean annual dental expense per employee. The manager selects a random sample of 40 employee records for the past year and finds the following. Sample Size = 40 employees Sample Mean = $527 Sample Standard Deviation = $78 Does the sample data provide evidence to conclude that the mean annual dental expense per employee is greater than $500 (using a = .05)? Use the hypothesis testing procedure outlined below. a. Formulate the null and alternative hypotheses. b. State the level of significance. c. Find the critical value (or values), and clearly show the rejection and nonrejection regions. d. Compute the test statistic. e. Decide whether you can reject Ho and accept Ha or not. f. Explain and interpret your conclusion in part e. What does this mean? g. Determine the observed p-value for the hypothesis test and interpret this value. What does this mean? h. Does this sample data provide evidence (with a = 0.05), that the mean annual dental expense per employee is greater than $500? #1: The publisher of Golf Illustrated has found through past experience that 60% of the subscribers renew their subscriptions. The publisher selects a random sample of 200 subscribers and finds that 108 plan to renew their subscriptions. a. Compute the 99% confidence interval for the population proportion of subscribers that plan to renew. b. Interpret this confidence interval. c. How many subscribers should the publisher sample in order to estimate to within 5% and with 99% confidence the proportion of all subscribers who plan to renew? Solution Population proportion, p = 60% = 0.6 Sample size, n = 200 Sample proportion, phat =108/200 = 0.54 Standard error, SE = sqrt {phat *(1- phat)/n} = sqrt{0.54*(1-0.54)/200} = 0.035 a. For 99% confidence interval, z = 2.58 Confidence interval for population proportion = phat z*SE = 0.54 2.58*0.035 = (0.45, 0.63) b. It means that we can say with 99% confidence that the percentage of subscribers renewing their subscriptions lie within 45% and 63% c. z*SE = 5% 2.58*sqrt{0.54*(1-0.54)/n} = 5% sqrt{0.54*(1-0.54)/n} = 0.01938 sqrt(0.2484/n) = 0.01938 n = 661. 3 or 661 approx. #2: An experiment has been conducted to evaluate a new process for producing synthetic diamonds. Six diamonds have been generated by the new process with recorded weights noted. This yields the following results. Sample Size = 6 Sample Mean = .53 carats Sample Standard Deviation = .0559 carats a. Construct the 95% confidence interval for the population mean weight of synthetic diamonds (in carats) using this new synthetic process. b. Interpret this interval. c. How many diamonds should be tested if we wish to generate a 95% confidence interval for the mean weight of synthetic diamonds that is accurate to within .01 carats? Solution Sample size, n = 6 Sample mean, = 0.53 Sample standard deviation, = 0.0559 Standard error, SE = /sqrt(n) = 0.0559/sqrt(6) = 0.023 d. For 95% confidence interval, z = 1.96 Confidence interval for population proportion = z*SE = 0.53 1.96*0.023 = (0.48, 0.58) e. It means that we can say with 95% confidence that the mean weight of synthetic diamonds will lie within 0.48 carats and 0.58 carats f. z*SE = 0.1 1.96* /sqrt(n) = 0.01 1.96* 0.0559/sqrt(n) = 0.01 Sqrt(n) = 10.956 n =120 #3: The ACTION Paper Company employs a human resources manager who is given responsibility for employee benefits. There is a question about the mean annual dental expense per employee. The manager selects a random sample of 40 employee records for the past year and finds the following. Sample Size = 40 employees Sample Mean = $527 Sample Standard Deviation = $78 Does the sample data provide evidence to conclude that the mean annual dental expense per employee is greater than $500 (using a = .05)? Use the hypothesis testing procedure outlined below. a. Formulate the null and alternative hypotheses. b. State the level of significance. c. Find the critical value (or values), and clearly show the rejection and nonrejection regions. d. Compute the test statistic. e. Decide whether you can reject Ho and accept Ha or not. f. Explain and interpret your conclusion in part e. What does this mean? g. Determine the observed p-value for the hypothesis test and interpret this value. What does this mean? h. Does this sample data provide evidence (with a = 0.05), that the mean annual dental expense per employee is greater than $500? Solution Sample size, n = 6 Sample mean