In a study of pulse rates of men, a simple random sample of 149 men results in a standard deviation of 11.4 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.05 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute; see the accompanying StatCrunch display for this test. What do the results indicate about the effectiveness of using the range rule of thumb with the "normal range" from 60 to 100 beats per minute for estimating $\sigma$ in this case? Assume that the simple random sample is selected from a normally distributed population. (i) Click the icon to view the StatCrunch display. and altemative hypotheses. \[ \begin{array}{l} H_{0}: \sigma=10 \\ H_{1}: \sigma \neq 10 \end{array} \] (Type integers or decimals. Do not round.) Question Viewer Identify the test statistic. \[ 192.34 \] (Round to two decimal places as needed.) Identify the P-value. (Round to three decimal places as needed.)
Solution
Step 1 ：State the null and alternative hypotheses. The null hypothesis \(H_{0}: \sigma=10\) and the alternative hypothesis \(H_{1}: \sigma \neq 10\).
Step 2 ：Identify the test statistic. The test statistic is \(192.34\).
Step 3 ：Identify the degrees of freedom. The degrees of freedom for this test would be the sample size minus 1, which is \(149 - 1 = 148\).
Step 4 ：Since our alternative hypothesis is two-sided (\(\sigma \neq 10\)), we need to perform a two-tailed test. This means we will find the area to the right of our test statistic (\(192.34\)) and multiply it by 2 to get the P-value.
Step 5 ：Calculate the P-value. The P-value for the given test statistic and degrees of freedom is \(0.017\).
Step 6 ：Final Answer: The P-value for the given test statistic and degrees of freedom is \(\boxed{0.017}\).
