Question

use a $chi^2$-test to test the claim $sigma geq 39$ at the $alpha = 0.10$ significance level using sample statistics $s = 38.6$ and $n = 12$. assume the population is normally distributed. click the icon to view the chi-square critical values table. identify the null and alternative hypotheses. a $h_0: sigma geq 39$ $h_a: sigma < 39$ b $h_0: sigma leq 39$ $h_a: sigma > 39$ c $h_0: sigma > 39$ $h_a: sigma leq 39$ d $h_0: sigma < 39$ $h_a: sigma geq 39$ identify the standardized test statistic. (round to three decimal places as needed.)

Explanation:

Step1: Recall Chi - Square Test Statistic Formula

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 $n = 12$, $s=38.6$, and from the null hypothesis $H_{0}:\sigma\geq39$, we use $\sigma = 39$ (since we assume the null hypothesis is true when calculating the test statistic).

Step3: Calculate $(n - 1)s^{2}$ and $\sigma^{2}$

First, calculate $n - 1=12 - 1 = 11$.
Then, $s^{2}=(38.6)^{2}=38.6\times38.6 = 1489.96$.
So, $(n - 1)s^{2}=11\times1489.96 = 16389.56$.
And $\sigma^{2}=(39)^{2}=1521$.

Step4: Calculate the Test Statistic

Now, substitute these values into the formula: $\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}=\frac{16389.56}{1521}\approx10.775$.

Answer:

The standardized test statistic is approximately $\boldsymbol{10.775}$.