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.

To test the hypothesis for this study, we need to state the null hypothesis (H0) and the alternative hypothesis (Ha).

In this case, the null hypothesis (H0) will be that the mean number of miles the rainmaker tire lasts is 40,000 miles or less. The alternative hypothesis (Ha) will be that the mean number of miles the rainmaker tire lasts is more than 40,000 miles.

H0: µ ≤ 40,000 (Mean number of miles ≤ 40,000)
Ha: µ > 40,000 (Mean number of miles > 40,000)

Next, we need to determine the critical value for this hypothesis test. The critical value is based on the chosen significance level, which is given as 0.05 in this case (equivalent to a significance level of 5%).

Since the alternative hypothesis is one-tailed (indicating that the mean number of miles is greater than 40,000), we will use the Z-distribution table to find the critical value at the 0.05 level of significance.

The critical value for a one-tailed test at a 0.05 significance level is 1.645 (rounded to three decimal places).

So, the critical value for this hypothesis test is 1.645.