Wk4Assignment Essay

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Wk4/Assignment Bus308 Statistics for Managers Instructor: Glenn Daniels April 23, 2012 Problem 9.13 Recall that “very satisfied” customers give XYZ-Box video game system a rating that is at least 42. Suppose that the manufacturer of the XYZ-Box wishes to use the random sample of 65 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42. a. Letting u represent the mean composite satisfaction rating for the XYZ-Box, set up the “null hypothesis” H0 and the “alternative hypothesis” Ha needed if we wish to attempt to provide evidence supporting the claim that u exceeds 42. H0: u<42 and Ha: u>42 b. The random sample of 65 satisfaction rating yields a sample mean of x = 42.954. Assuming that S = 2.64, use critical values to test H0 versus Ha at each of a = 10, .05, .01 and .00l. MU = 42, N = 65, X-bar = 42.954, sigma = 2.64 Z= (x-bar – mu)/(sigma/sqrt n) Z = (42.954 – 42)/(2.64/sqrt65 = 2.9134. Wk4/Assignment 9.13 continued Critical upper tail = Z – scores for 1.2816, 1.6449, 2.3263 and 3.092 for a = 0.10, 0.05, 0.01 and 0.001. Since 2.9134>1.2816, 1.6449 and 2.3263, I rejected H0 and accepted Ha at 1 = 0.10, 0.05, and 0.01 and concluded that the mean rating exceeds 42. Since 2.9134<3.0902, I rejected H0 at a = 0.001, and fail to conclude that the mean rating exceeds 42. c. Using the information in part b, calculate the p-value and use it to test H0 versus Ha at each of at = 10, .05, .01 and .001. Upper tail p-value for z = 2.9134 is 00018. Since 0.0018<0.10, 0.65, 0.01, I rejected H0 and accepted Ha at a = 0.10, 0.05, and 0.01 and concluded that the mean rating exceeds 42. Since 0.0018>0.001, I failed to reject H0 @ a = 0.001 and failed to conclude that the mean rating exceeds 42. d. How much evidence is there that the mean composite

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