Recall that "very satisfied" customers give the XYZ-Box videogame system a rating that is at least 42. Suppose that themanufacturer of the XYZ-Box wishes to use the random sample of 64satisfaction ratings to provide evidence supporting the claim thatthe mean composite satisfaction rating for the XYZ-Box exceeds42.
(a) Letting µ represent the meancomposite satisfaction rating for the XYZ-Box, set up the nullhypothesis H0 and the alternative hypothesisHa needed if we wish to attempt to provideevidence supporting the claim that µ exceeds 42.
H0: µ (Click to select)=??>Ha: µ (Click toselect)?=>?
(b) The random sample of 64 satisfactionratings yields a sample mean of x¯=42.970x¯=42.970. Assuming that ?equals 2.67, use critical values to test H0versus Ha at each of ? = .10, .05,.01, and .001. (Round your answer z.05to 3 decimal places and other z-scores to 2 decimalplaces.)
z =
| Rejection points |
| z.10 | |
| z.05 | |
| z.01 | |
| z.001 | |
|
Reject H0 with ? = (Click toselect).001.10.10, .05, .01.01, .001 , but not with ?=(Click to select).10.10, .05, .01.01, .001.001
(c) Using the information in part (b),calculate the p-value and use it to testH0 versus Ha at each of? = .10, .05, .01, and .001. (Round your answersto 4 decimal places.)
| p-value = |
| Since p-value = is less than (Click toselect).10, .05, .01.01, .001.001.10 ; rejectH0 at those levels of ? but not with? = (Click to select).10.001.01, .001.10, .05, .01. |
(d) How much evidence is there that the meancomposite satisfaction rating exceeds 42?
There is (Click to select)very strongextremely strongnoweakstrongevidence.
