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The tables I’ll use throughout this are: 1. (TCO B) The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 28.00 inches and a standard deviation of 1.4 inches. What is the probability that the average length of a steel sheet from a sample of 36 units is more than 27.65 inches long? Z = (x-mu)/(sd/sqrt(N))z = (27.65-28)/(1.4/sqrt(36))z = -1.5prob(z > -1.5).9332 2. (TCO B) Quality Progress, February 2005, reports on improvements in customer satisfaction and loyalty made by Bank of America. A key measure of customer satisfaction is the response (on a scale from 1 to 10) to the question: "Considering all the business you do with Bank of America, what is your overall satisfaction with Bank of America?" Here, a response of 9 or 10 represents "customer delight." Historically, the percentage of Bank of America customers expressing customer delight has been 48%. Suppose that we wish to use the results of a survey of 350 Bank of America customers to justify the claim that more than 48% of all current Bank of America customers would express customer delight. The survey finds that 193 of 350 randomly selected Bank of America customers express customer delight. I, for the sake of argument, we assume that the proportion of customer delight is p = .48, calculate the probability of observing asample proportion greater than or equal to 193/350 = .55. That is, calculate P( .55). Get the Z score:z = (phat-p)/sqrt(p*(1-p)/N)z = (0.55-0.48)/sqrt(0.48*0.52/350)z = 2.621258prob(z > 2.621258)0.0044 3. (TCO C) Until this year the mean braking distance of a Nikton automobile moving at 60 miles per hour was 175 feet. Nikton engineers have developed what they consider a better braking system. They test the new brake system on a random sample of 36 cars and determine the sample mean braking distance is x bar equals 167 ft. (assume the population standard deviation is known to be 30 feet). Compute the 98% confidence interval for the mean braking distance. (Points: 1)The z value is 2.326.Mean – z*sd/sqrt(N) to Mean + z*sd/sqrt(N)167-2.326*30/sqrt(36) to 167+2.326*30/sqrt(36)[155.4,178.6] 4. (TCO C) Until this year the mean braking distance of a Nikton automobile moving at 60 miles per hour was 175 feet. Nikton engineers have developed what they consider a better braking system. They test the new brake system on a random sample of 36 cars and determine the sample mean braking distance is x bar 167 ft (assume the population standard deviationis known to be 30 feet). How many cars should be tested if Nikton wants to be 98% confident of being within 4 feet of the population mean braking distance? (Points: 1)N = (z*sd/e)^2N = (2.326*30/4)^2306 5. (TCO C) You are the shipping manager at the Postham Company. You are interested in the percentage of packages delivered damaged. A simple random sample of 500 packages yields 18 delivered damaged and 482 delivered undamaged. Compute the 90% confidence interval for the percentage of packages that are delivered damaged in the population. (Points: 1)The z value is 1.6449.The interval goes from:p-z*sqrt(p*(1-p)/N) to p+z*sqrt(p*(1-p)/N)18/500-1.6449*sqrt(18/500*(1-18/500)/500) to 18/500+1.6449*sqrt(18/500*(1-18/500)/500)[0.022, 0.050] 6. (TCO D) CD Camera produces a new camera that it claims can take an average of more than 7 photographs per second. A test of 16 randomly selected cameras yielded the following sample values: x bar equals 7.15 photos/second, s=.52 photos/sec. Use = alpha = 0.10. Determine the critical value rule. Note: You will be doing several problems using the CD Camera problem situation. Be sure to check the given data each time, from problem to problem, because the given numbers will likely be different. (Points: 1)Df = n-1 = 15The one tailed value, from the table, is:t > 1.341 7. (TCO C) You are the shipping manager at the Postham Company. You are interested in the percentage of packages delivered damaged. A simple random sample of 500 packages yields 16 delivered damaged and 484 delivered undamaged. How many packages should be sampled in order to be 90% confident of being within 0.9% of the actual population percentage of packages delivered damaged. (Points: 1)N = p*(1-p)*(z/e)^2N = 16/500*(1-16/500)*(1.6449/0.009)^2N = 10351,035 8. (TCO D) CD Camera produces a new camera that it claims can take an average of more than 7 photographs per second. A test of 16 randomly selected cameras yielded the following sample values: x bar =7.15 seconds, s=.52 photos/sec. Use alpha= 0.10. Formulate the null and alternative hypotheses.Note: You will be doing several problems using the CD Camera problem situation. Be sure to check the given data each time, from problem to problem, because the given numbers will likely be different. (Points: 1)H0: u ≤ 7, Ha: u > 79. (TCO D) CD Camera produces a new camera that it claims can take an average of more than 7 photographs per second. A test of 16 randomly selected cameras yielded the following sample values: x bar =7.15 seconds, s=.52 photos/sec. Use alpha= 0.10.. Compute the value of the test statistic. Note: You will be doing several problems using the CD Camera problem situation. Be sure to check the given data each time, from problem to problem, because the given numbers will likely be different. (Points: 1)t = (x-mu)/(s/sqrt(N))t = (7.15-7)/(0.52/sqrt(16))t=1.154 10. (TCO D) CD Camera produces a new camera that it claims can take an average of more than 7 photographs per second. A test of 16 randomly selected cameras yielded the following sample values: x bar =7.15 seconds, s=.52 photos/sec. Use alpha= 0.10. Calculate the p-value.Note: You will be doing several problems using the CD Camera problem situation. Be sure to check the given data each time, from problem to problem, because the given numbers will likely be different. (Points: 1)t = (x-mu)/(s/sqrt(N))t = (7.15-7)/(0.52/sqrt(16))t=1.154Df = n-1 = 15Using the table:p-value=.1333 11. (TCO D) Historically, the publisher of Golf Illustrated has found that 64% of its subscribers renew their subscriptions. Because of the advent of the Internet and cable television, the publisher is concerned that the proportion of subscribers who renew their subscriptions may be declining. The publisher selects a random sample of 200 subscribers and finds that 116 plan to renew their subscriptions. Use (a=0.01). p = the population of subscribers to Golf Illustrated who renew. Determine the critical value rule.Note: You will be doing several problems using the Golf Illustrated problem situation. Be sure to check the given data each time, from problem to problem, because the given numbers will likely be different. (Points: 1)One tail, alpha = 0.01, using the table:z < -2.33 12. (TCO D) Historically, the publisher of Golf Illustrated has found that 64% of its subscribers renew their subscriptions. Because of the advent of the Internet and cable television, the publisher is concerned that the proportion of subscribers who renew their subscriptions may be declining. The publisher selects a random sample of 200 subscribers and finds that 116 plan to renew their subscriptions. Use (a=0.01). p = the population of subscribers to Golf Illustrated who renew. Compute the value of the test statistic.Note: You will be doing several problems using the Golf Illustrated problem situation. Be sure to check the given data each time, from problem to problem, because the given numbers will likely be different. (Points: 1)Z = (phat-p)/sqrt(p*(1-p)/N)z = (116/200-0.64)/sqrt(0.64*(1-0.64)/200)z = -1.77 13. (TCO D) Historically, the publisher of Golf Illustrated has found that 64% of its subscribers renew their subscriptions. Because of the advent of the Internet and cable television, the publisher is concerned that the proportion of subscribers who renew their subscriptions may be declining. The publisher selects a random sample of 200 subscribers and finds that 116 plan to renew their subscriptions. Use (a=0.01). p = the population of subscribers to Golf Illustrated who renew. Calculate the p-value.Note: You will be doing several problems using the Golf Illustrated problem situation. Be sure to check the given data each time, from problem to problem, because the given numbers will likely be different. (Points: 1)Z = (phat-p)/sqrt(p*(1-p)/N)z = (116/200-0.64)/sqrt(0.64*(1-0.64)/200)z = -1.77 Using the table:p-value = .0385 14. (TCO D) Historically, the publisher of Golf Illustrated has found that 64% of its subscribers renew their subscriptions. Because of the advent of the Internet and cable television, the publisher is concerned that the proportion of subscribers who renew their subscriptions may be declining. The publisher selects a random sample of 200 subscribers and finds that 116 plan to renew their subscriptions. Use (a=0.01). p = the population of subscribers to Golf Illustrated who renew. Formulate the null and alternative hypotheses.Note: You will be doing several problems using the Golf Illustrated problem situation. Be sure to check the given data each time, from problem to problem, because the given numbers will likely be different. (Points: 1)H0: p ≤ 0.64, Ha: p < 0.64 15. (TCO D) Part a: For the CD Camera problem situation: interpret the p-value you calculated, in the terms of the problem. The p value of 0.1333 is very high (much higher than alpha). We don’t reject the null hypothesis, so we cannot say that the mean is greater than 7.Part b: For the CD Camera problem situation: compare the test statistic you calculated to the corresponding critical value (NOTE: be sure to use the value of alpha given in the test statistic question when calculating the critical value you will compare to the test statistic) and interpret the decision in the terms of the problem. (Points: 1)The t value of 1.154 is lower than the critical value of 1.341. Therefore, we don’t reject the null hypothesis, so we cannot say that the mean is greater than 7.16. (TCO D) Part a: For the Golf Illustrated problem situation: interpret the p-value you calculated, in the terms of the problem. The p value of 0.0385 is higher than the alpha level for this problem, too. Therefore, we don’t reject the null hypothesis. We cannot say that the percentage has declined.Part b: For the Golf Illustrated problem situation: compare the test statistic you calculated to the corresponding critical value rule (NOTE: be sure to use the value of alpha given in the test statistic question when calculating the critical value you will compare to the test statistic) and interpret the decision in the terms of the problem. (Points: 1) The test statistic of -1.77 is not lower than the critical value of -2.33. Therefore, we don’t reject the null hypothesis. We cannot say that the percentage has declined. ................
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