Using the t-distribution, it is found that the p-value of the test is 0.007.
At the null hypothesis, it is <u>tested if the mean lifetime is not greater than 220,000 miles</u>, that is:

At the alternative hypothesis, it is <u>tested if the mean lifetime is greater than 220,000 miles</u>, that is:
.
We have the <u>standard deviation for the sample</u>, thus, the t-distribution is used. The test statistic is given by:
The parameters are:
is the sample mean.
is the value tested at the null hypothesis.
- s is the standard deviation of the sample.
- n is the sample size.
For this problem:

Then, the value of the test statistic is:



We have a right-tailed test(test if the mean is greater than a value), with <u>t = 2.69</u> and 23 - 1 = <u>22 df.</u>
Using a t-distribution calculator, the p-value of the test is of 0.007.
A similar problem is given at brainly.com/question/13873630