The lengths of time (in years) it took a random sample of 32 former smokers to quit smoking permanently are listed. Assume the population standard deviation is 6.5 years. At $\alpha=0.05$, is there enough evidence to reject the claim that the mean time it takes smokers to quit smoking permanently is 13 years? Complete parts (a) through (e).
$\begin{array}{ccccccccc}9.4 & 13.8 & 10.2 & 20.6 & 18.2 & 21.1 & 16.7 & 21.8 \\ 7.1 & 20.1 & 22.1 & 9.1 & 13.9 & 21.6 & 15.6 & 13.4 \\ 15.8 & 20.1 & 12.8 & 9.3 & 9.6 & 8.4 & 12.4 & 9.1 \\ 20.9 & 20.9 & 19.5 & 7.9 & 7.6 & 21.5 & 8.3 & 20.6\end{array}$
D.
\[
\begin{array}{l}
H_{0}: \mu \leq 13 \text { (claim) } \\
H_{a} \cdot \mu> 13
\end{array}
\]
E.
\[
\begin{array}{l}
H_{0}: \mu> 13 \text { (claim) } \\
H_{a} \cdot \mu \leq 13
\end{array}
\]
F.
\[
\begin{array}{l}
H_{0}: \mu \neq 13 \text { (claim) } \\
H_{a}: \mu=13
\end{array}
\]
(b) Identify the standardized test statistic. Use technology.
\[
z=
\]
(Round to two decimal places as needed.)
Answer
Round the standardized test statistic to two decimal places. The final answer is \(\boxed{1.72}\).
Step 1 ：Given that the population mean (\(\mu\)) is 13, the population standard deviation (\(\sigma\)) is 6.5, and the sample size (\(n\)) is 32. The sample data is [9.4, 13.8, 10.2, 20.6, 18.2, 21.1, 16.7, 21.8, 7.1, 20.1, 22.1, 9.1, 13.9, 21.6, 15.6, 13.4, 15.8, 20.1, 12.8, 9.3, 9.6, 8.4, 12.4, 9.1, 20.9, 20.9, 19.5, 7.9, 7.6, 21.5, 8.3, 20.6].
Step 2 ：Calculate the sample mean (\(\bar{x}\)) from the given data. The sample mean is 14.98125.
Step 3 ：Substitute all the values into the z-score formula to calculate the standardized test statistic. The formula for the z-score is \(z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\).
Step 4 ：Substitute \(\bar{x} = 14.98125\), \(\mu = 13\), \(\sigma = 6.5\), and \(n = 32\) into the formula to get \(z = 1.7242526895087347\).
Step 5 ：Round the standardized test statistic to two decimal places. The final answer is \(\boxed{1.72}\).
