Question

use a \\(\chi^2\\)-test to test the claim \\(\sigma \geq 39\\) at the \\(\alpha = 0.10\\) significance level using sample statistics \\(s = 38.6\\) and \\(n = 1\\) (note: likely a typo, probably \\(n\\) has a proper value). assume the population is normally distributed. click the icon to view the chi-square critical values table. identify the null and alternative hypotheses. \\(\bigcirc\\) a. \\(h_0: \sigma \geq 39\\), \\(h_a: \sigma < 39\\) \\(\bigcirc\\) b. \\(h_0: \sigma \leq 39\\), \\(h_a: \sigma > 39\\) \\(\bigcirc\\) c. \\(h_0: \sigma > 39\\), \\(h_a: \sigma \leq 39\\) \\(\bigcirc\\) d. \\(h_0: \sigma < 39\\), \\(h_a: \sigma \geq 39\\) identify the standardized test statistic. \\(\square\\) (round to three decimal places as needed.)

Explanation:

Step1: Recall Chi - Square Test Formula

The formula for the chi - square test statistic for a test of a population standard deviation (or variance) is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\), where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma\) is the hypothesized population standard deviation.

Step2: Identify Values

We are given that \(s = 38.6\), \(\sigma=39\), but we need to know the sample size \(n\). Wait, the original problem has a typo in \(n = 1\)? Maybe it's a typo, but assuming that the correct \(n\) (let's assume \(n\) is a valid sample size, maybe \(n = 15\) or other, but since the user's problem has \(n = 1\) which is incorrect for a chi - square test (as \(n-1\) would be 0), but maybe it's a typo. Let's assume that the correct \(n\) is, for example, if we take \(n = 15\) (maybe a common sample size), but wait, the user's problem: let's re - check. Wait, the original problem's \(n\) is written as \(n = 1\)? That can't be. Maybe it's a typo, like \(n=15\). Let's proceed with the formula.

Wait, the claim is \(\sigma\geq39\), so \(\sigma^{2}\geq39^{2}=1521\). The sample standard deviation \(s = 38.6\), so \(s^{2}=38.6^{2}=1489.96\).

If we assume that the correct \(n\) (let's say \(n = 15\) for example, since \(n = 1\) is invalid), then \(n-1=14\).

Then \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}=\frac{14\times1489.96}{1521}=\frac{20859.44}{1521}\approx13.714\). But since the user's \(n\) is written as \(n = 1\), which is wrong, but maybe it's a typo. Wait, maybe the original problem has \(n = 15\). Let's check the correct approach.

Wait, the key formula is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\). Let's assume that the \(n\) in the problem is a typo, and let's say \(n = 15\) (a common sample size). Then:

\(n-1 = 14\), \(s = 38.6\), so \(s^{2}=38.6\times38.6 = 1489.96\), \(\sigma = 39\), so \(\sigma^{2}=39\times39 = 1521\).

\(\chi^{2}=\frac{(14)\times1489.96}{1521}=\frac{20859.44}{1521}\approx13.714\)

But since the user's \(n\) is \(n = 1\), which is incorrect (as \(n\geq2\) for a chi - square test of variance), there must be a typo. However, if we proceed with the formula, assuming that the correct \(n\) is, for example, \(n = 15\) (maybe the original problem had \(n = 15\) and a typo in writing \(n = 1\)), the calculation is as above.

But let's check the null hypothesis first. The claim is \(\sigma\geq39\), so the null hypothesis \(H_{0}:\sigma\geq39\) and the alternative hypothesis \(H_{a}:\sigma\lt39\) (option A is correct for the null and alternative hypotheses).

For the test statistic, using the formula \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\). Let's assume that the correct \(n\) is \(n = 15\) (since \(n = 1\) is invalid). Then:

\(n-1=14\), \(s = 38.6\), \(\sigma = 39\)

\(s^{2}=38.6^{2}=1489.96\)

\(\chi^{2}=\frac{14\times1489.96}{39^{2}}=\frac{14\times1489.96}{1521}=\frac{20859.44}{1521}\approx13.714\)

If the correct \(n\) is different, say \(n = 20\), then \(n - 1=19\), \(\chi^{2}=\frac{19\times1489.96}{1521}=\frac{28309.24}{1521}\approx18.612\). But since the user's \(n\) is written as \(n = 1\), which is an error, we can only provide the formula and the way to calculate.

But let's assume that the \(n\) is a typo and the correct \(n\) is, for example, \(n = 15\) (a common sample size). Then the test statistic is approximately \(13.714\).

Answer:

For the null and alternative hypotheses: A. \(H_{0}:\sigma\geq39\), \(H_{a}:\sigma\lt39\)

For the test statistic (assuming \(n = 15\)): \(\approx13.714\) (If \(n\) is different, recalculate with the correct \(n\))