Question 9 of 10
- Use the tadistribution and the sample results to complete the test of the hypotheses. Use a $5 \%$ significance level. Assume the results come from a random sample, and if the sample size is small, assume the underlying distribution is relatively normal.

Test $H_{0}: \mu=100$ vs $H_{a}: \mu< 100$ using the sample results $\bar{x}=91.7, s=12.5$, with $n=30$.
(a) Give the test statistic and the p-value.
Round your answer for the test statistic to two decimal places and your answer for the $p$-value to three decimal places.
test statistic $=$
$p-$ value $=$
(b) What is the conclusion?
Reject $H_{0}$.
Donot reject $H_{0}$.

Step 1 ：Given values are: population mean (\( \mu \)) = 100, sample mean (\( \bar{x} \)) = 91.7, sample standard deviation (s) = 12.5, and sample size (n) = 30.

Step 2 ：Calculate the test statistic using the formula: \( t_{stat} = \frac{\bar{x} - \mu}{s / \sqrt{n}} \).

Step 3 ：Substitute the given values into the formula to get: \( t_{stat} = \frac{91.7 - 100}{12.5 / \sqrt{30}} \approx -3.64 \).

Step 4 ：Calculate the p-value. Since it's a left-tailed test, we use the cumulative distribution function (cdf) instead of the survival function (sf).

Step 5 ：The p-value is approximately 0.00053.

Step 6 ：Since the p-value is less than the significance level of 0.05, we reject the null hypothesis.

Step 7 ：This means that we have enough evidence to support the claim that the population mean is less than 100.

Step 8 ：The final answer is: The test statistic is approximately \(\boxed{-3.64}\) and the p-value is approximately \(\boxed{0.00053}\). We \(\boxed{reject}\) the null hypothesis.