Question.3017 - Complete exercises 9.3, 9.13, 9.14, 9.25, 9.48, 9.55 For problems requiring computations, please ensure that your Excel file includes the associated cell computations and/or statistics output; this information is needed in order to receive full credit on these problems. 9.3 If you use a 0.10 level of significance in a two-tail hypothesis test, what is your decision rule for rejecting a null hypothesis that the population mean is 500 if you use the Z test? 9.13 Do students at your school study more than, less than, or about the same as students at other business schools? BusinessWeek reported that at the top 50 business schools, students studied an average of 14.6 hours per week. (Data extracted from “Cracking the Books,” Special Report/Online Extra, www.businessweek.com, March 19, 2007.) Set up a hypothesis test to try to prove that the mean number of hours studied at your school is different from the 14.6-hour- per-week benchmark reported by BusinessWeek. a. State the null and alternative hypotheses. b. What is a Type I error for your test? c. What is a Type II error for your test? 9.14 The quality-control manager at a light bulb factory needs to determine whether the mean life of a large shipment of light bulbs is equal to 375 hours. The population standard deviation is 100 hours. A random sample of 64 light bulbs indicates a sample mean life of 350 hours. a. At the 0.05 level of significance, is there evidence that the mean life is different from 375 hours? b. Compute the p-value and interpret its meaning. c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs. d. Compare the results of (a) and (c). What conclusions do you reach? 9.25 A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages are labeled as 8 ounces, the company wants the packages to contain a mean of 8.17 ounces so that virtually none of the packages contain less than 8 ounces. A sample of 50 packages is selected periodically, and the packaging process is stopped if there is evidence that the mean amount packaged is different from 8.17 ounces. Suppose that in a particular sample of 50 packages, the mean amount dispensed is 8.159 ounces, with a sample standard deviation of 0.051 ounce. a. Is there evidence that the population mean amount is different from 8.17 ounces? (Use a 0.05 level of significance.) b. Determine the p-value and interpret its meaning. 9.48 Southside Hospital in Bay Shore, New York, commonly conducts stress tests to study the heart muscle after a person has a heart attack. Members of the diagnostic imaging department conducted a quality improvement project with the objective of reducing the turnaround time for stress tests. Turnaround time is defined as the time from when a test is ordered to when the radiologist signs off on the test results. Initially, the mean turnaround time for a stress test was 68 hours. After incorporating changes into the stress-test process, the quality improvement team collected a sample of 50 turnaround times. In this sample, the mean turnaround time was 32 hours, with a standard deviation of 9 hours. (Data extracted from E. Godin, D. Raven, C. Sweetapple, and F. R. Del Guidice, “Faster Test Results,” Quality Progress, January 2004, 37(1), pp. 33–39.) a. If you test the null hypothesis at the 0.01 level of significance, is there evidence that the new process has reduced turnaround time? b. Interpret the meaning of the p-value in this problem. 9.55 The U.S. Department of Education reports that 46% of full-time college students are employed while attending college. (Data extracted from “The Condition of Education 2009,” National Center for Education Statistics, nces.ed. gov.) A recent survey of 60 full-time students at Miami University found that 29 were employed. a. Use the five-step p-value approach to hypothesis testing and a 0.05 level of significance to determine whether the proportion of full-time students at Miami University is different from the national norm of 0.46. b. Assume that the study found that 36 of the 60 full-time students were employed and repeat (a). Are the conclusions the same?
Answer Below:
1) Complete exercises 9.3, 9.13, 9.14, 9.25, 9.48, 9.55. 2) For problems requiring computations, please ensure that your Excel file includes the associated cell computations and/or statistics output; this information is needed in order to receive full credit on these problems. 3) Submit output to the instructor in one Excel file by the end of Module 6. -------------------------------------------------------------------------------------------- 9.3 If you use a 0.10 level of significance in a (two-tail) hypothesis test, what is your decision rule for rejecting a null hypothesis that the population mean is 500 if you use the Z test? I used the normsinv function in Excel. I used p = 0.05 because we want the cutoff for the bottom and top 5%. -1.644853625 =NORMSINV(0.05) Therefore we reject H0 if z < -1.645 or z > 1.645 9.13 Do students at your school study more, less, or about the same as at other business schools? Business Week reported that at the top 50 business schools, students studied an average of 14.6 hours (data extracted from “Cracking the Books,” SPECIAL REPORT/Online Extra, www.businessweek.com, March 19, 2007). Set up a hypothesis test to try to prove that the mean number of hours studied at your school is different from the 14.6 hour benchmark reported by Business Week. a. State the null and alternative hypotheses. H0: mu = 14.6 HA: mu ≠ 14.6 b. What is a Type I error for your test? Saying that the mean is statistically different from 14.6 when it's actually the same. c. What is a Type II error for your test? Saying that the mean is no different from 14.6 when it's actually different. 9.14 The quality control manager at a lightbulb factory needs to determine whether the mean life of a large shipment of lightbulbs is equal to the specified value of 375 hours. State the null and alternative hypotheses. n 64 μ 375 s 100 x-bar 350 (1) Formulate the hypotheses: Ho: μ = 375 Ha: μ ≠ 375 (2) Decide the test statistic and the level of significance: z (Two-tailed), α = 0.05 Lower Critical z- score = -1.959963986 Upper Critical z- score = 1.959963986 (3) State the decision Rule: Reject Ho if |z| > 1.9600 (4) Calculate the value of test statistic: SE = s/√n = 12.5 z = (x-bar - μ)/SE = -2 (5) Compare with the critical value and make a decision: Since 2.0000 > 1.9600 we reject Ho and accept Ha Decision: It appears that the average life of the bulbs is different from 375 hours (b) p- value = 0.0455. Since p-value<alpha, we reject the true null hypothesis. (c) n 64 x-bar 350 s 100 c 95% Standard Error, SE = σ/√n = 12.5 z- score = -1.959963986 Width of the confidence interval = z * SE = -24.49954983 Lower Limit of the confidence interval = x-bar - |width| = 325.5004502 Upper Limit of the confidence interval = x-bar + |width| = 374.4995498 The confidence interval is [325.50 hours, 374.50 hours] (d) 375 hours lies outside the confidence interval in (c). This means we reject Ho. The same conclusion was reached in (a) also. Both (a) and (c) mean the same thing. 9.25 A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages are labeled as 8 ounces, the company wants the packages to contain a mean of 8.17 ounces so that virtually none of the packages contain less than 8 ounces. A sample of 50 packages is selected periodically, and the packaging process is stopped if there is evidence that the mean amount packaged is different from 8.17 ounces. Suppose that in a particular sample of 50 packages, the mean amount dispensed is 8.159 ounces, with a sample standard deviation of 0.051 ounce. a. Is there evidence that the population mean amount is different from 8.17 ounces? (Use a 0.05 level of significance.) H0: mu = 8.17 HA: mu ≠ 8.17 First I calculated the critical value of z at 0.05. Since it's a two-tailed test, I divided alpha by 2 NORMSINV(0.05/2) -1.959963986 Then I calculated z z = (xbar - mu)/(sigma/√n) z = (8.17 - 8.159)/(0.051/√50) X-bar 8.159 mu 8.17 sigma 0.051 n 50 z = -1.525132273 The direction doesn't matter in a 2-tailed test. Since |z| < |-1.96|, there is insufficient evidence to reject the null hypothesis. The mean is not statistically different from 8.17 b. Determine the p-value and interpret its meaning. p = NORMSDIST(1.53) p = 0.9369916355 To convert this into a one-tailed p value, I subtracted from 1. That shows the "tail" of the distribution. 1 - 0.9364 0.06300836446 And to convert it into a two-tailed p value, I multiplied by 2 0.636 * 2 0.1260167289 Interpretation: There's a 12.60% chance that we could see a difference this far from 8.17 (in either direction) due to random sampling error, if the true population mean was 8.17 9.48 Southside Hospital in Bay Shore, New York, commonly conducts stress tests to study the heart muscle after a person has a heart attack. Members of the diagnostic imaging department conducted a quality improvement project with the objective of reducing the turnaround time for stress tests. Turnaround time is defined as the time from when a test is ordered to when the radiologist signs off on the test results. Initially, the mean turnaround time for a stress test was 68 hours. After incorporating changes into the stress-test process, the quality improvement team collected a sample of 50 turnaround times. In this sample, the mean turnaround time was 32 hours, with a standard deviation of 9 hours. (Data extracted from E. Godin, D. Raven, C. Sweetapple, and F. R. Del Guidice, “Faster Test Results,”Quality Progress, January 2004, 37(1), pp. 33–39.) a. If you test the null hypothesis at the 0.01 level of significance, is there evidence that the new process has reduced turnaround time? b. Interpret the meaning of the p-value in this problem. H0: the new process has higher or equal turnaround time than the previous process. Ha: the new process has reduced turnaround time Ha is the claim X-bar 32 mu 68 sigma 9 n 50 z = -28.28427125 p 0 alpha 0.01 Critical Z -2.326347874 a. Rejection Region, Z<Z critical Since Z lies in the crirical region, we reject the null hypothesis that the new process has higher or equal turnaround time than the previous process. (b) p- value = 0.0000000. Since p-value<alpha, we reject the null hypothesis. At 0.01 level of significane, there is enough evidence to support the claim that the new process has lower turnaround time 9.55) The U.S. Department of Education reports that 46% of full-time college students are employed while attending college. (Data extracted from “The Condition of Education 2009,”National Center for Education Statistics, nces.ed. gov.) A recent survey of 60 full-time students at Miami University found that 29 were employed. a. Use the five-step p-value approach to hypothesis testing and a 0.05 level of significance to determine whether the proportion of full-time students at Miami University is different from the national norm of 0.46. b. Assume that the study found that 36 of the 60 full-time students were employed and repeat (a). Are the conclusions the same? a. H0: p=0.46 Ha: p ≠ 0.46 p 0.46 q 0.54 alpha 0.05 Number of items of interest 29 n 60 p' 0.4833333333 Standard Error 0.06434283177 Z Test Statistic 0.3626407588 p-value 0.6415633706 Right Tailed p-value 0.3584366294 Actual p-value 0.7168732588 It is a two tailed test Lower Critical value -1.959963986 Upper Critical Value 1.959963986 Decision rule: p-value < 0.05, reject H0 Decision: Since p-value = 0.7169 > 0.05, fail to reject H0. There is not enough evidence that the proportion of full-time students at Miami University is different from the national norm of 0.46. b. H0: p=0.46 Ha: p ≠ 0.46 p 0.46 q 0.54 alpha 0.05 Number of items of interest 36 n 60 p' 0.6 Standard Error 0.06434283177 Z Test Statistic 2.175844553 p-value 0.985216557 Right Tailed p-value 0.01478344301 Actual p-value 0.02956688602 It is a two tailed test Lower Critical value -1.959963986 Upper Critical Value 1.959963986 Decision rule: p-value < 0.05, reject H0 Decision: Since p-value = 0.0296 < 0.05, we reject H0. There is enough evidence that the proportion of full-time students at Miami University is different from the national norm of 0.46.More Articles From Statistics
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