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 = 12\\). assume the population is normally distributed.
click the icon to view the chi-square critical values table.
identify the standardized test statistic.
10.776 (round to three decimal places as needed.)
identify the critical value(s).
5.578
(round to three decimal places as needed. use a comma to separate answers as needed.)
choose the correct conclusion below.
\\(\bigcirc\\) a. fail to reject \\(h_0\\). there is enough evidence at the 10% level of significance to reject the claim.
\\(\bigcirc\\) b. fail to reject \\(h_0\\). there is not enough evidence at the 10% level of significance to reject the claim.
\\(\bigcirc\\) c. reject \\(h_0\\). there is enough evidence at the 10% level of significance to reject the claim.
\\(\bigcirc\\) d. reject \\(h_0\\). there is not enough evidence at the 10% level of significance to reject the claim.

Explanation:

Step1: Identify Hypotheses

The claim is $\sigma \geq 39$, so $H_0: \sigma \geq 39$ (null hypothesis) and $H_a: \sigma < 39$ (alternative hypothesis, but wait, earlier options had $H_0: \sigma \geq 39$? Wait, no, the first part: the correct null and alternative for testing $\sigma \geq 39$ at $\alpha = 0.10$ (left - tailed test? Wait, no: when testing $\sigma \geq 39$, the alternative is $\sigma < 39$, so $H_0: \sigma \geq 39$, $H_a: \sigma < 39$. But the test statistic is $\chi^2=\frac{(n - 1)s^2}{\sigma_0^2}$. Here, $n = 12$, $s = 38.6$, $\sigma_0 = 39$. So $\chi^2=\frac{(12 - 1)\times(38.6)^2}{39^2}=\frac{11\times1489.96}{1521}=\frac{16389.56}{1521}\approx10.776$ (matches the given test statistic).

Step2: Critical Value

For a left - tailed test with $\alpha = 0.10$ and $df=n - 1=11$, the critical value from chi - square table is $\chi^2_{1 - \alpha, df}=\chi^2_{0.90,11}=5.578$ (matches the given critical value).

Step3: Compare Test Statistic and Critical Value

The test statistic $\chi^2 = 10.776$ and critical value $\chi^2_{0.90,11}=5.578$. Since $10.776>5.578$ (test statistic is in the non - rejection region for left - tailed test: we reject $H_0$ when test statistic $< $ critical value. Here, test statistic is greater than critical value, so we fail to reject $H_0$). The claim is $H_0: \sigma \geq 39$, so failing to reject $H_0$ means there is not enough evidence to reject the claim. So the correct conclusion is: Fail to reject $H_0$. There is not enough evidence at the 10% level of significance to reject the claim. Which is option B.

Answer:

B. Fail to reject $H_0$. There is not enough evidence at the 10% level of significance to reject the claim.