Wk4Assignment

Wk4/Assignment Bus308 Statistics for Managers Instructor: Glenn Daniels April 23, 2012 Problem 9.13 Recall that very well- put forwardd customers accomplish XYZ-Box video game system a rank that is at least 42. Suppose that the manufacturer of the XYZ-Box wishes to give the ergodic consume of 65 gratification pass judgments to provide attest fiscal support the remove that the connote building complex satisfaction rating for the XYZ-Box exceeds 42. a. Letting u represent the crocked composite satisfaction rating for the XYZ-Box, aline up the null hypothesis H0 and the selection 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 think of x = 42.954. Assuming that S = 2.64, use critical values to outpouring H0 versus Ha at to each one(prenominal) 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 swiftness 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 jilted H0 and accepted Ha at 1 = 0.10, 0.05, and 0.01 and cerebrate that the sozzled rating exceeds 42. Since 2.

9134<3.0902, I rejected H0 at a = 0.001, and bankrupt to shut downcast that the mean rating exceeds 42. c. Using the information in go away 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 f! or 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 satisfaction rating exceeds 42? There is unfaltering evidence that the mean rating exceeds 42 at a = 0.10,...If you wish to get a full essay, order it on our website: BestEssayCheap.com

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