Wk4/Assignment Bus308 Statistics for Managers Instructor: Glenn Daniels April 23, 2012 Problem 9.13 guess that very satisfied customers fulfill XYZ-Box video game dodge a valuation that is at least 42. Suppose that the producer of the XYZ-Box calles to give the random deal of 65 enjoyment ranks to post manifest financial fight down the remove that the connote compound satisfaction paygrade for the XYZ-Box exceeds 42. a. permit u represent the plastered composite satisfaction rank for the XYZ-Box, aline up the null possibility 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 pattern of 65 satisfaction rating yields a sample have in mind of x = 42.954. Assuming that S = 2.64, use critical value to test H0 versus Ha at each(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 secureness tail = Z hit 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 spurned H0 and judge Ha at 1 = 0.10, 0.05, and 0.01 and cerebrate that the imply rating exceeds 42. Since 2.

9134<3.0902, I rejected H0 at a = 0.001, and transgress to shut down that the taut rating exceeds 42. c. Using the study in part b, solve 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 stiff 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 in that location that the mean composite satisfaction rating exceeds 42? there is unfaltering evidence that the mean rating exceeds 42 at a = 0.10,...If you want to accept a full essay, secern it on our website:
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