Question

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

claim: $\sigma^2 \
eq 34.6$; $\alpha = 0.01$
sample statistics: $s^2 = 36.8$, $n = 81$

write the null and alternative hypotheses.
$h_0: \sigma^2 \square$
$h_a: \sigma^2 \square$
(type integers or decimals. do not round.)
calculate the standardized test statistic.
$\chi^2 = \square$ (round to two decimal places as needed.)
determine the p - value.
p - value = $\square$ (round to three decimal places as needed.)
decide whether to reject or fail to reject the null hypothesis and interpret the decision in the context of the claim.
$\square$ $h_0$. there $\square$ enough evidence at the 1% level of significance to $\square$ the claim.

Explanation:

Step1: Define Hypotheses

The claim is $\sigma^2
eq 34.6$, so the null hypothesis $H_0$ is the complement (equality) and alternative $H_a$ is the claim.
$H_0: \sigma^2 = 34.6$
$H_a: \sigma^2
eq 34.6$

Step2: Calculate Test Statistic

The formula for the chi - square test statistic for variance is $\chi^2=\frac{(n - 1)s^2}{\sigma_0^2}$, where $n = 81$, $s^2=36.8$, $\sigma_0^2 = 34.6$.
Substitute values: $(n - 1)=80$, so $\chi^2=\frac{80\times36.8}{34.6}=\frac{2944}{34.6}\approx85.09$

Step3: Find P - value

For a two - tailed test with $\chi^2 = 85.09$ and $df=n - 1 = 80$, use chi - square distribution. Using a calculator or software, the P - value is $2\times(1 - P(\chi^2_{80}\leq85.09))$. From chi - square tables or calculator, $P(\chi^2_{80}\leq85.09)\approx0.765$, so P - value $\approx2\times(1 - 0.765)=0.470$ (rounded to three decimals).

Step4: Decision Rule

Compare P - value ($0.470$) with $\alpha = 0.01$. Since $0.470>0.01$, we fail to reject $H_0$. So there is not enough evidence to support the claim.

Answer:

• Null and Alternative Hypotheses:

$H_0: \sigma^2 = 34.6$
$H_a: \sigma^2
eq 34.6$

• Standardized Test Statistic: $\chi^2\approx85.09$
• P - value: $0.470$
• Decision: Fail to reject $H_0$. There is not enough evidence at the 1% level of significance to support the claim.