9.13 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 65 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42. a: Letting μ represent the mean composite satisfaction rating for the XYZ-Box, set up the null hypothesis H0 and the alternative hypothesis Ha needed if we wish to attempt to provide evidence supporting the claim that μ exceeds 42.
H0: μ ≤ 42
Ha: μ> 42 b: The random sample of 65 satisfaction ratings yields a sample mean of = 42.954. Assuming that s equals 2.64, use critical values to test H0 versus Ha at each of α = .10, .05, .01, and .001. if and only if the test statistic z is greater than the critical value za α = .10= 1.28 z=(42.954-42)/(2.64/√65)
=.05 =1.64 z=2.913
=.01 =2.32
=.001=3.09

Because z = 2.913 is greater than the critical value z10 = 1.28, we can reject H0: μ = 42 in favor of Ha : μ > 42 by setting α equal to .10.
Because z = 2.913 is greater than the critical value z05 = 1.64, we can reject H0: μ = 42 in favor of Ha : μ > 42 by setting α equal to .05.
Because z = 2.913 is greater than the critical value z01 = 2.32, we can reject H0: μ = 42 in favor of Ha : μ > 42 by setting α equal to .01.
Because z = 2.913 is less than the critical value z001 = 3.09, we can accept H0: μ = 42 in favor of Ha : μ > 42 by setting α equal to .001.

using minitab
One-Sample T

Test of mu = 42 vs > 42

95% Lower N Mean StDev SE Mean Bound T P
65 42.954 2.640 0.327 42.407 2.91 0.002 c: Using the information in part b, calculate the p-value and use it to test H0 versus Ha at each of α = .10, .05, .01, and .001. using minitab
P=.002
If the p-value for testing