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).
$\begin{array}{cccccccccc}9.4 & 13.8 & 10.2 & 20.6 & 18.2 & 21.1 & 16.7 & 21.8 \\ 7.1 & 20.1 & 22.1 & 9.1 & 13.9 & 21.6 & 15.6 & 13.4 & \\ 15.8 & 20.1 & 12.8 & 9.3 & 9.6 & 8.4 & 12.4 & 9.1 & \\ 20.9 & 20.9 & 19.5 & 7.9 & 7.6 & 21.5 & 8.3 & 20.6 & \end{array}$
(b) Identify the standardized test statistic. Use technology.
$z=1.72$ (Round to two decimal places as needed.)
(c) Find the P-value. Use technology.
\[
P=
\]
(Round to three decimal places as needed.)
Answer
Final Answer: The P-value is \(\boxed{0.043}\).
Step 1 ：The question is asking for the P-value of a one-sample z-test. The z-score has already been provided as 1.72. The P-value is the probability that we would observe such an extreme test statistic assuming the null hypothesis is true. In this case, the null hypothesis is that the mean time it takes smokers to quit smoking permanently is 13 years.
Step 2 ：To find the P-value, we need to find the area to the right of the z-score in the standard normal distribution. This is because we are conducting a one-tailed test where we are testing if the mean time to quit smoking is greater than 13 years.
Step 3 ：Using the z-score of 1.72, we find the P-value to be approximately 0.043.
Step 4 ：Final Answer: The P-value is \(\boxed{0.043}\).
