The following numbers are the differences in pulse rate (beats per minute) before and after running for 12 randomly selected people. Positive numbers mean the pulse rate went up. Test the hypothesis that the mean difference in pulse rate was more than 0 , using a significance level of 0.05 . Assume the population distribution is Normal.
\[
24,10,10,11,14,9,0,3,9,39,3 \text {, and } 12 \text { 마 }
\]
Determine the null and alternative hypotheses. Choose the correct answer below.
A.
\[
\begin{array}{l}
H_{0}: \mu=0 \\
H_{a}: \mu> 0
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: \mu< 0 \\
H_{a}: \mu \geq 0
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: \mu=0 \\
H_{a}: \mu< 0
\end{array}
\]
E.
\[
\begin{array}{l}
H_{0}: \mu> 0 \\
H_{a}: \mu \leq 0
\end{array}
\]
c.
\[
\begin{array}{l}
H_{0}: \mu=0 \\
H_{a}: \mu \neq 0
\end{array}
\]
F.
\[
\begin{array}{l}
\mathrm{H}_{0}: \mu \neq 0 \\
\mathrm{H}_{\mathrm{a}}: \mu=0
\end{array}
\]
Find the test statistic.
\[
\mathrm{t}=\square
\]
(Round to two decijhal places as needed.)
Answer
The null and alternative hypotheses are \(H_{0}: \mu=0\) and \(H_{a}: \mu>0\), and the test statistic is \(\boxed{3.96}\).
Step 1 ：Define the null and alternative hypotheses. The null hypothesis \(H_{0}\) is that the mean difference in pulse rate is 0, and the alternative hypothesis \(H_{a}\) is that the mean difference in pulse rate is greater than 0.
Step 2 ：Calculate the sample mean and sample standard deviation. The sample mean is 12.0 and the sample standard deviation is approximately 10.49.
Step 3 ：Calculate the test statistic using the formula: \(t = \frac{{\text{{sample mean}} - \text{{population mean}}}}{{\text{{sample standard deviation}} / \sqrt{n}}}\), where \(n\) is the number of observations. The population mean under the null hypothesis is 0.
Step 4 ：Substitute the values into the formula to get the test statistic: \(t = \frac{{12.0 - 0}}{{10.49 / \sqrt{12}}}\), which simplifies to approximately 3.96.
Step 5 ：The null and alternative hypotheses are \(H_{0}: \mu=0\) and \(H_{a}: \mu>0\), and the test statistic is \(\boxed{3.96}\).
