Posted by **HN** on Monday, December 10, 2012 at 11:38am.

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?

- statistic -
**PsyDAG**, Monday, December 10, 2012 at 12:12pm
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.

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