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.
\\(\square\\) (round to three decimal places as needed.)

Explanation:

Step1: Recall Chi - Square Test for Variance

The formula for the chi - square test statistic for a hypothesis test about 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 population standard deviation under the null hypothesis.

Step2: Identify Values

We are given that \(n = 19\), \(s=41.8\), and under the null hypothesis \(H_{0}:\sigma\geq41\), so we use \(\sigma = 41\) (since we use the value from the null hypothesis for the calculation of the test statistic).

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 \(\sigma^{2}\)

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

Step6: Calculate the Test Statistic

Now, substitute the values into the formula \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\):

$$
LATEXBLOCK0
$$

Answer:

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

For the null and alternative hypotheses, the correct option is:
D. \(H_{0}:\sigma\geq41\)
\(H_{a}:\sigma < 41\)