Question

Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic t.

Test the claim that the mean lifetime of a particular car engine is greater than 220,000 miles. Sample data are summarized as n = 23, x-bar = 226,450 and s=11,500. Use a significance level of 5%. Determine the p-value.


0.993


0.007


0.05


0.018


0.004

Answer 1

Using the t-distribution, it is found that the p-value of the test is 0.007.

At the null hypothesis, it is tested if the mean lifetime is not greater than 220,000 miles, that is:

$H_0: \mu \leq 220000$

At the alternative hypothesis, it is tested if the mean lifetime is greater than 220,000 miles, that is:

$H_1: \mu > 220000$.

We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.
• $\mu$ is the value tested at the null hypothesis.
• s is the standard deviation of the sample.
• n is the sample size.

For this problem:

$\overline{x} = 226450, \mu = 220000, s = 11500, n = 23$

Then, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{226450 - 220000}{\frac{11500}{\sqrt{23}}}$

$t = 2.69$

We have a right-tailed test(test if the mean is greater than a value), with t = 2.69 and 23 - 1 = 22 df.

Using a t-distribution calculator, the p-value of the test is of 0.007.

A similar problem is given at https://brainly.com/question/13873630