Systolic blood pressure levels above $120 \mathrm{~mm} \mathrm{Hg}$ are considered to be high. For the 100 systolic blood pressure levels listed in the accompanying data set, the mean is $123.90000 \mathrm{~mm} \mathrm{Hg}$ and the standard deviation is 14.63737 $\mathrm{mm} \mathrm{Hg}$. Assume that a simple random sample has been selected. Use a 0.10 significance level to test the claim that the sample is from a population with a mean greater than $120 \mathrm{~mm} \mathrm{Hg}$.
Click the icon to view the data set of systolic blood pressure levels.
(Type integers or decimals. Do not round.)
Identify the test statistic.
2.66
(Round to two decimal places as needed.)
Identify the P-value.
(Round to three decimal places as needed.)

Step 1 ：Given that the systolic blood pressure levels above 120 mm Hg are considered to be high. For the 100 systolic blood pressure levels listed in the accompanying data set, the mean is 123.9 mm Hg and the standard deviation is 14.63737 mm Hg. We are to use a 0.10 significance level to test the claim that the sample is from a population with a mean greater than 120 mm Hg.

Step 2 ：The null hypothesis is that the population mean is 120 mm Hg, and the alternative hypothesis is that the population mean is greater than 120 mm Hg.

Step 3 ：The test statistic for a one-sample t-test is calculated as \((\text{sample mean} - \text{population mean}) / (\text{sample standard deviation} / \sqrt{\text{sample size}})\).

Step 4 ：Substituting the given values into the formula, we get a test statistic of approximately 2.66.

Step 5 ：The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. Since this is a one-tailed test, we can use the t-distribution to find the P-value.

Step 6 ：Using the t-distribution, we find that the P-value is approximately 0.005.

Step 7 ：Final Answer: The test statistic is approximately \(\boxed{2.66}\) and the P-value is approximately \(\boxed{0.005}\).