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 66 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 versus ha: µ 42. (b) the random sample of 66 satisfaction ratings yields a sample mean of x⎯⎯=42.880. assuming that σ equals 2.69, use critical values to test h0 versus ha at each of α = .10, .05, .01, and .001. (round your answer z.05 to 3 decimal places and other z-scores to 2 decimal places.) z = rejection points z.10 z.05 z.01 z.001 reject h0 with α = , but not with α = (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. (round your answers to 4 decimal places.) p-value = since p-value = is less than ; reject h0 at those levels of α but not with α = . (d) how much evidence is there that the mean composite satisfaction rating exceeds 42? there is evidence.