Recall that \"very satisfied\" customers give the XYZ-Box video game system a ra

Question

Recall that "very satisfied" customers give the 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 69 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42. a Letting mu represent the mean composite satisfaction rating or the XYZ-Box, set up the null hypothesis H_0 and the alternative hypothesis H_a needed if we wish to attempt to provide evidence supporting the claim that mu exceeds 42. H_0: mu 42 versus H_a: mu 42. (b) The random sample of 69 satisfaction ratings yields a sample mean of x = 42.820. Assuming that sigma equals 2.61, use critical values to test H_0 versus H_a at each a = 10, 05, 01, and 001. (Round your answer z_.05 to 3 decimal places and other z-scores to 2 decimal places.) z = 2.609 Reject H_0 with alpha = 10, 05, 01, but not with alpha = 001 c) Using the information in part(b), calculate the-value and use it to test H_0 versus H_a at each of sigma = 10, 05, 01, and 001 (Round your answers to 4 decimal places.) p-value = Since p-value = is less than 10, 05, 01: reject H_0 at those levels of a but not with a = 001

Explanation / Answer

Part a

H0: µ = 42 versus Ha: µ > 42

This is a one tailed test.

Part b

The test statistic formula is given as below:

Z = (Xbar - µ) / [/sqrt(n)]

We are given

Xbar = 42.820, µ = 42, = 2.61, n = 69

Z = (42.82 – 42)/[2.61/sqrt(69)]

Z = 0.82/0.314207

Z = 2.609744

Rejection points

Z.10 = 1.28

Z.05 = 1.645

Z.01 = 2.33

Z.001 = 3.09

(All values are calculated by using standard normal or Z-table)

Reject H0 with = .10, .05, .01, but not with = .001.

Part c

p-value = 0.0045

Since p-value = 0.0045 is less than alpha = .10, .05, .01, reject H0 at those levels of alpha but not with alpha = 0.001.

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