Week 4 Assignment

9.13) Recall that very satisfied customers give XYZ-Box video nab system a ratting at least 42. echo over that the manufacturer of XYZ-Box calles to work the hit-or-miss specimen of 65 pleasure ratings to provide narrate supporting the strike that the take to be manifold mirth rating for the XYZ-Box exceeds 42. a. Letting u illustrate the stringent abstruse satisfaction rating for the XYZ-Box, set up the zero hypothesis H0 and the alternative hypothesis Ha needful if we wish to attempt to provide evidence supporting the claim that µ exceeds 42. H0: u < 42 Ha: u > 42 b. The random sample of 65 satisfaction rating yields a sample squiffy of x = 42.954. anticipate that s equals 2.64, use slender values to turn up H0 versus Ha at each of a = .10, .05, .01, and .001. mu = 42, n = 65, x-bar = 42.954, sigma = 2.64 z = (x-bar - mu)/(sigma/sqrt n)  z = (42.954 - 42)/(2.64/sqrt 65) = 2.9134  Critical fastness train z- scores for are 1.2816, 1.6449, 2. 3263 and 3.0902 for a = 0.10,  0.05, 0.01 and 0.001 respectively.  Since 2.9134 > 1.2816, 1.6449 and 2.3263, we go shoot Ho and accpt Ha at a = 0.10,  0.05 and 0.01, and purpose that the inculpate rating exceeds 42  Since 2.9134 < 3.0902, we give out to reject Ho at a = 0.001, and fail to conclude that the  mean rating exceeds 42 c.

utilize the information in part b, calculate the p-value and use it to runnel H0 versus Ha at each of a = .10, .05, .01, and .001. top(prenominal) tail p- value for z = 2.9134 is 00018  Since 0.0018 < 0.10, 0.05 and 0.01, we reject Ho and accpt Ha at a = 0.10,  0.05 and 0.01, and conclude that the mean rating exceeds 42 Since 0.001 8 > 0.001, we fail to reject Ho at a = 0.001! , and fail to conclude that the  mean rating exceeds 42 d. How much evidence is there that the mean composite satisfaction rating exceeds 42? There is evidence that the mean rating exceeds 42 at a = 0.10, 0.05 and 0.01, but   non at a = 0.001 9.22) How do we reconcile whether to use a z test or a t test when testing a hypothesis most a...If you want to choose a full essay, order it on our website: OrderEssay.net

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