Question

use a \\(\chi^2\\)-test to test the claim \\(\sigma < 41\\) at the \\(\alpha = 0.01\\) significance level using sample statistics \\(s = 41.8\\) and \\(n = 19\\). assume the population is normally distributed.

identify the null and alternative hypotheses.

a. \\(h_0: \sigma > 41\\)
\\(h_a: \sigma \leq 41\\)

b. \\(h_0: \sigma \leq 41\\)
\\(h_a: \sigma > 41\\)

c. \\(h_0: \sigma < 41\\)
\\(h_a: \sigma \geq 41\\)

d. \\(h_0: \sigma \geq 41\\)
\\(h_a: \sigma < 41\\)

identify the standardized test statistic.
18.709 (round to three decimal places as needed.)

identify the critical value(s).
(round to three decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Recall Chi - Square Test for Variance

For a hypothesis test about the population standard deviation \(\sigma\) (or variance \(\sigma^{2}\)), the null hypothesis \(H_{0}\) and alternative hypothesis \(H_{a}\) are set such that the null hypothesis contains the equality. The claim here is \(\sigma < 41\), so the alternative hypothesis \(H_{a}:\sigma < 41\), and the null hypothesis \(H_{0}\) is the complement of the alternative hypothesis in terms of the inequality direction that includes equality, so \(H_{0}:\sigma\geq41\).

Step2: Calculate the Critical Value

The formula for the chi - square test statistic for testing a claim about the standard deviation is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma_{0}^{2}}\), but for finding the critical value, we use the chi - square distribution with degrees of freedom \(df=n - 1\). Here, \(n = 19\), so \(df=19 - 1=18\). The significance level \(\alpha = 0.01\), and since the alternative hypothesis is left - tailed (\(H_{a}:\sigma < 41\)), we need to find the critical value \(\chi_{1-\alpha}^{2}\) with \(df = 18\).

Looking up in the chi - square distribution table or using a calculator, for \(df = 18\) and \(\alpha=0.01\) (left - tailed), the critical value \(\chi_{0.99}^{2}\) (because \(1-\alpha=0.99\)) with \(df = 18\) is found as follows:

Using the chi - square distribution formula or a calculator, the critical value \(\chi_{0.99}^{2}\) for \(df = 18\) is approximately \(7.015\) (we can use a chi - square calculator: for a left - tailed test with \(\alpha = 0.01\) and \(df = 18\), the critical value is the value such that \(P(\chi^{2}<\text{critical value})=\alpha = 0.01\), or \(P(\chi^{2}\geq\text{critical value})=0.99\)).

Answer:

The critical value is \(7.015\)