Wk4Assignment

Wk4/Assignment
Bus308 Statistics for Managers
Instructor: Glenn Daniels
April 23, 2012

Problem 9.13
Recall that very satisfied customers accomplish XYZ-Box video game system a rating that is at least 42. Suppose that the manufacturer of the XYZ-Box wishes to give the random consume of 65 satisfaction ratings to provide evidence financial support the remove that the connote composite 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 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(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 rejected H0 and accepted Ha at 1 = 0.10, 0.05, and 0.01 and cerebrate that the mean rating exceeds 42. Since 2.



9134<3.0902, I rejected H0 at a = 0.001, and bankrupt to shut down 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 satisfaction rating exceeds 42?
There is unfaltering evidence that the mean rating exceeds 42 at a = 0.10,...If you want to get a full essay, order it on our website: Ordercustompaper.com



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