An insurance agent says the standard deviation of the total hospital charges for patients involved in a crash in which the vehicle struck a construction barricade is less than $\$ 3900$. A random sample of 22 total hospital charges for patients involved in this type of crash has a standard deviation of $\$ 4500$. At $\alpha=0.10$ can you support the agent's claim? Use the P-value method to test the claim.
Identify the null and alternative hypotheses. Let $\sigma$ be the population standard deviation.
A.
\[
\begin{array}{l}
H_{0} \sigma> \$ 3900 \\
H_{a} \sigma \leq \$ 3900
\end{array}
\]
$\checkmark \mathrm{C}$
\[
\begin{array}{l}
H_{0}: \sigma \geq \$ 3900 \\
H_{2}: \sigma< \$ 3900
\end{array}
\]
Identify the standardized test statistic
(Round to two decimat places as needed)
\[
\text { B. } \begin{array}{l}
H_{0}: \sigma< \$ 3900 \\
H_{2}: \sigma \geq \$ 3900
\end{array}
\]
\[
\begin{array}{l}
H_{0} \sigma \leq \$ 3900 \\
H_{a} \sigma> \$ 3900
\end{array}
\]
Answer
The final answer is: The null and alternative hypotheses are: \[H_{0}: \sigma \geq \$ 3900\] \[H_{a}: \sigma<\$ 3900\] The standardized test statistic is approximately \(\boxed{27.96}\). The p-value is approximately \(\boxed{0.1413}\). Since the p-value is greater than the significance level of 0.10, we do not reject the null hypothesis. The data does not provide enough evidence to support the insurance agent's claim that the standard deviation of the total hospital charges is less than $3900.
Step 1 ：Identify the null and alternative hypotheses. Let \(\sigma\) be the population standard deviation. The null hypothesis is typically a statement of no effect or no difference. The alternative hypothesis is what you might believe to be true or hope to prove true. In this case, the insurance agent's claim that the standard deviation of the total hospital charges is less than $3900 is our alternative hypothesis. The null hypothesis is the opposite of the alternative hypothesis. The null and alternative hypotheses are: \[H_{0}: \sigma \geq \$ 3900\] \[H_{a}: \sigma<\$ 3900\]
Step 2 ：Identify the standardized test statistic. The test statistic is a quantity derived from the sample data and used in the test procedure. In this case, we are testing a claim about a standard deviation or variance, so we use the chi-square distribution. The formula for the test statistic in a chi-square test for a standard deviation or variance is \((n - 1)s^2 / \sigma^2\), where n is the sample size, s is the sample standard deviation, and \(\sigma\) is the population standard deviation.
Step 3 ：Calculate the test statistic. Given that n = 22, s = 4500, and \(\sigma\) = 3900, the test statistic is approximately 27.96 (rounded to two decimal places as required by the question).
Step 4 ：Calculate the p-value. The p-value is approximately 0.1413.
Step 5 ：Compare the p-value with the significance level. The p-value is greater than the significance level of 0.10, so we do not reject the null hypothesis. This means that the data does not provide enough evidence to support the insurance agent's claim that the standard deviation of the total hospital charges is less than $3900.
Step 6 ：The final answer is: The null and alternative hypotheses are: \[H_{0}: \sigma \geq \$ 3900\] \[H_{a}: \sigma<\$ 3900\] The standardized test statistic is approximately \(\boxed{27.96}\). The p-value is approximately \(\boxed{0.1413}\). Since the p-value is greater than the significance level of 0.10, we do not reject the null hypothesis. The data does not provide enough evidence to support the insurance agent's claim that the standard deviation of the total hospital charges is less than $3900.
