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.
$\chi^2 = 16.92$ (round to two decimal places as needed.)
determine the p - value.
p - value = $\square$ (round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for P - value in chi - square test for variance

The test statistic for testing a claim about the population variance \(\sigma^{2}\) is given by \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma_{0}^{2}}\), 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. For a left - tailed test (since \(H_{a}:\sigma^{2}<8.8\)), the P - value is the probability that a chi - square random variable with \(df=n - 1\) degrees of freedom is less than the calculated test statistic \(\chi^{2}\).

First, we calculate the degrees of freedom: \(df=n - 1=22-1 = 21\). The calculated test statistic \(\chi^{2}=16.92\) and the null hypothesis is \(H_{0}:\sigma^{2}\geq8.8\), alternative hypothesis \(H_{a}:\sigma^{2}<8.8\) (left - tailed test).

Step2: Use chi - square distribution to find P - value

We need to find \(P(\chi_{21}^{2}<16.92)\). We can use a chi - square distribution table or a statistical software (such as a TI - 84 Plus calculator or R).

Using a calculator (for example, in R, we can use the command pchisq(16.92,21)):

The function pchisq(x,df) gives the cumulative probability for a chi - square distribution with df degrees of freedom up to \(x\).

Calculating \(P(\chi_{21}^{2}<16.92)\):

Using a chi - square table or calculator, we know that for \(df = 21\) and \(\chi^{2}=16.92\), the cumulative probability (P - value) is approximately \(0.212\) (after rounding to three decimal places).

Answer:

\(0.212\)