BUS 308 Statistics for Managers
January 28, 2013
9.13) Recall that “very satisfied” customers give XYZ-Box video game system a ratting at least 42. Suppose that the manufacturer of 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 u 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: mu 42
b. The random sample of 65 satisfaction rating yields a sample mean of x = 42.954. Assuming that s equals 2.64, use critical values to test H0 versus Ha at each of a = .10, .05, .01, and .001.
z = (xbar - µ)/(σ/√n)
z = (42.954 - 42 )/(2.64/√65)
z = 0.954 / (2.64/8.0623)
z = 2.9134
alpha z-crit result
0.10 1.282 significant
0.05 1.645 significant
0.01 2.326 significant
0.001 3.09 not significant
c. Using the information in part b, calculate the p-value and use it to test H0 versus Ha at each of a = .10, .05, .01, and .001.
Upper 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 0.0018 > 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 not at a = 0.001 We can be at least 99% certain, but not as much as 99.9% certain.
How do we decide whether to use a z test or a t test when testing a hypothesis about a population mean?
Assuming the population is approximately normally distributed, we can use the z-test when the sample size is large...