You can calculate the P-value for a chi-square test using technology. After calculating the standardized test statistic, use the cumulative distribution function (CDF) to calculate the area under the curve Use the P-value method to test the claim.
A hospital spokesperson claims that the standard deviation of the waiting times experienced by patients in its minor emergency department is no more than 0.5 minutes. A random sample of 25 waiting times has a standard deviation of 0.6 minutes. At $\alpha=0.10$, is there enough evidence to reject the spokesperson's claim?
A.
\[
\begin{array}{l}
H_{0}: \sigma \leq 0.5 \\
H_{a}: \sigma> 0.5
\end{array}
\]
c.
\[
\begin{array}{l}
H_{0} \sigma> 0.5 \\
H_{a} . \sigma \leq 0.5
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0} \quad \sigma< 0.5 \\
H_{2}: \sigma \geq 0.5
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: \sigma \geq 0.5 \\
H_{a} \quad \sigma< 0.5
\end{array}
\]
Identify the standardized test statistic.
$x^{2}=34.56$ (Round to two decimal places as needed)
Identify the P-value.
$P=\square$ (Round to four decimal places as needed)
Answer
The final answer is: The P-value is approximately \(\boxed{0.0752}\).
Step 1 ：State the null hypothesis \(H_{0}: \sigma \leq 0.5\) and the alternative hypothesis \(H_{a}: \sigma>0.5\).
Step 2 ：The standardized test statistic is given as \(x^{2}=34.56\).
Step 3 ：Calculate the P-value using the cumulative distribution function (CDF) of the chi-square distribution with degrees of freedom equal to the sample size minus 1. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.
Step 4 ：The calculated P-value is approximately 0.0752. This is the probability of observing a test statistic as extreme as 34.56, assuming the null hypothesis is true.
Step 5 ：Since the P-value is greater than the significance level of 0.10, we do not have enough evidence to reject the null hypothesis. Therefore, we cannot reject the spokesperson's claim that the standard deviation of the waiting times is no more than 0.5 minutes.
Step 6 ：The final answer is: The P-value is approximately \(\boxed{0.0752}\).
