Question

use a \\(\chi^2\\)-test to test the claim \\(\sigma < 42\\) at the \\(\alpha = 0.01\\) significance level using sample statistics \\(s = 40.4\\) and \\(n = \\) assume the population is normally distributed.

identify the null and alternative hypotheses.

\\(\bigcirc\\) a. \\(h_0: \sigma > 42\\)
\\(\quad\\) \\(h_a: \sigma \leq 42\\)
\\(\bigcirc\\) b. \\(h_0: \sigma \geq 42\\)
\\(\quad\\) \\(h_a: \sigma < 42\\)
\\(\bigcirc\\) c. \\(h_0: \sigma \leq 42\\)
\\(\quad\\) \\(h_a: \sigma > 42\\)
\\(\bigcirc\\) d. \\(h_0: \sigma < 42\\)
\\(\quad\\) \\(h_a: \sigma \geq 42\\)

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

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

choose the correct conclusion below.

Explanation:

Part 1: Identify the null and alternative hypotheses

In hypothesis testing for a chi - square test about a population standard deviation (or variance), the null hypothesis \(H_0\) is a statement of equality or no change, and the alternative hypothesis \(H_a\) is the claim we are trying to find evidence for. The claim here is \(\sigma< 42\). The null hypothesis will be the complement of the alternative hypothesis in terms of the inequality direction. So the null hypothesis \(H_0:\sigma\geq42\) and the alternative hypothesis \(H_a:\sigma < 42\). So the correct option is B.

Part 2: Identify the standardized test statistic

Step 1: Recall the formula for the chi - square test statistic for a population standard deviation

The formula for the chi - square test statistic \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma_{0}^{2}}\), where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma_{0}\) is the hypothesized population standard deviation from the null hypothesis. We assume \(n\) is missing in the problem statement, but let's assume \(n\) is given (for example, if \(n = 16\), we will proceed with the formula). Wait, the original problem has a typo, but let's assume \(n\) is a positive integer. Let's assume \(n\) is given (for example, if \(n=16\)) and \(\sigma_0 = 42\), \(s = 40.4\)

Step 2: Substitute the values into the formula

\(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma_{0}^{2}}\)
Let's assume \(n = 16\) (we need the correct \(n\) value, but since it's missing in the original problem, we will show the formula application). If \(n = 16\), then \(n-1=15\), \(s^{2}=(40.4)^{2}=1632.16\), \(\sigma_{0}^{2}=42^{2} = 1764\)
\(\chi^{2}=\frac{15\times1632.16}{1764}=\frac{24482.4}{1764}\approx13.879\) (this is with \(n = 16\), if \(n\) is different, the value will change)

Part 3: Identify the critical value(s)

Step 1: Determine the degrees of freedom and the type of test

The degrees of freedom \(df=n - 1\). For a left - tailed test (since \(H_a:\sigma<42\)) with \(\alpha = 0.01\) and \(df=n - 1\), we use the chi - square distribution table to find the critical value \(\chi_{1-\alpha}^{2}\) (because it's a left - tailed test, we look for the value such that \(P(\chi^{2}<\chi_{1 - \alpha}^{2})=\alpha\))

Step 2: Find the critical value from the chi - square table

If \(n = 16\), then \(df=15\). Looking at the chi - square distribution table for \(df = 15\) and \(\alpha=0.01\) (left - tailed), the critical value \(\chi_{0.99}^{2}\) (since \(1-\alpha=0.99\)) for \(df = 15\) is approximately \(5.229\) (using chi - square table: for \(df = 15\), the value corresponding to the lower \(1\%\) tail is \(5.229\))

Part 4: Choose the correct conclusion

Answer:

s:

• Null and alternative hypotheses: B. \(H_{0}:\sigma\geq42\), \(H_{a}:\sigma < 42\)
• Standardized test statistic: Depends on \(n\), formula is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma_{0}^{2}}\) (e.g., if \(n = 16\), \(\approx13.879\))
• Critical value: Depends on \(n\) (e.g., if \(n = 16\), \(\approx5.229\))
• Conclusion: Depends on the comparison of test statistic and critical value (e.g., if test statistic > critical value, fail to reject \(H_0\); if test statistic < critical value, reject \(H_0\))