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. \\(\bigcirc\\) a. $h_0: \sigma > 41$ $h_a: \sigma \leq 41$ \\(\bigcirc\\) b. $h_0: \sigma \leq 41$ $h_a: \sigma > 41$ \\(\bigcirc\\) c. $h_0: \sigma < 41$ $h_a: \sigma \geq 41$ \\(\bigcirc\\) d. $h_0: \sigma \geq 41$ $h_a: \sigma < 41$ identify the standardized test statistic. (round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for the chi - square test statistic for variance

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.

Step2: Identify the values of \(n\), \(s\), and \(\sigma_{0}\)

We are given that \(n = 19\), \(s=41.8\), and from the null hypothesis \(H_{0}:\sigma\geq41\), we take \(\sigma_{0} = 41\).

Step3: Calculate \((n - 1)\)

First, calculate \(n-1\): \(n-1=19 - 1=18\).

Step4: Calculate \(s^{2}\)

Calculate \(s^{2}\): \(s^{2}=(41.8)^{2}=41.8\times41.8 = 1747.24\).

Step5: Calculate \(\sigma_{0}^{2}\)

Calculate \(\sigma_{0}^{2}\): \(\sigma_{0}^{2}=41^{2}=1681\).

Step6: Substitute the values into the formula

Substitute \(n - 1 = 18\), \(s^{2}=1747.24\), and \(\sigma_{0}^{2}=1681\) into the formula \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma_{0}^{2}}\):

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Answer:

The standardized test statistic is approximately \(18.710\) (rounded to three decimal places).