posted by c
one of the major U.S. tire makers wishes to review its warranty for their rainmaker tire. The warranty is for 40,000 miles. The tire company beleives that the tire actually lasts more than 40,000 miles. A sample of 49 tires reveales that the mean number of miles is 45,000 miles with a standard deviation of 15,000 miles. Test the hypothesis with a 0.05 significance level. What is the decision rule? What is the calculated value of z? what is our decision?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)(but you can use n)
If only one SD is provided, you can use just that to determine SEdiff.
You can do the calculations in thousands, leaving off the extra three zeros.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
Ho: mean = 40,000
Ha: mean > 40,000
The decision is yours.