posted by HN on .
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 are the null and alternative hypotheses for this study? what is the critical value for this this?
Ho: sample mean = warranty mean
Ha: sample mean > warranty mean
Z = (sample mean - warranty mean)/standard error (SE) of difference between means
SEdiff = √(SEmeanS^2 + SEmeanW^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.