Question

test the claim about the population variance $\sigma^2$ at the level of significance $\alpha$. assume the population is normally distributed.

claim: $\sigma^2 \geq 8.8$; $\alpha = 0.05$
sample statistics: $s^2 = 7.09$, $n = 22$

write the null and alternative hypotheses.
$h_0: \sigma^2 \geq 8.8$
$h_a: \sigma^2 < 8.8$
(type integers or decimals. do not round.)
calculate the standardized test statistic.
$\chi^2 = \square$ (round to two 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 when testing a claim about the population variance \(\sigma^{2}\) is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma_{0}^{2}}\), where \(n\) is the sample size, \(s^{2}\) is the sample variance, and \(\sigma_{0}^{2}\) is the hypothesized population variance under the null hypothesis.

Step2: Identify the values

We are given that \(n = 22\), \(s^{2}=7.09\), and from the null hypothesis \(H_{0}:\sigma^{2}\geq8.8\), we take \(\sigma_{0}^{2}=8.8\). First, calculate \(n - 1\): \(n-1=22 - 1=21\).

Step3: Substitute the values into the formula

Substitute \(n - 1 = 21\), \(s^{2}=7.09\), and \(\sigma_{0}^{2}=8.8\) into the formula \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma_{0}^{2}}\).

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Answer:

\(16.92\)