Question

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

claim: $\sigma^2 \
eq 34.9$; $\alpha = 0.05$
sample statistics: $s^2 = 37.7$, $n = 91$

...

write the null and alternative hypotheses.
$h_0: \sigma^2 = 34.9$
$h_a: \sigma^2 \
eq 34.9$
(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 chi - square test statistic for variance

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

Step2: Identify the values

We are given that \(n = 91\), \(s^{2}=37.7\), and \(\sigma^{2}=34.9\).

Step3: Substitute the values into the formula

First, calculate \(n - 1\): \(n-1=91 - 1=90\).
Then, substitute into the formula: \(\chi^{2}=\frac{90\times37.7}{34.9}\)
Calculate the numerator: \(90\times37.7 = 3393\)
Then, divide by the denominator: \(\chi^{2}=\frac{3393}{34.9}\approx97.22\)

Answer:

\(97.22\)