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 test of 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 under the null hypothesis $H_{0}:\sigma\geq39$, we will use $\sigma = 39$ (since we use the value from the null hypothesis for the calculation). First, calculate $s^{2}$: $s^{2}=(38.6)^{2}=38.6\times38.6 = 1489.96$. Then, calculate $(n - 1)$: $n-1=12 - 1=11$.

Step3: Substitute into Formula

Substitute the values into the formula: $\chi^{2}=\frac{(12 - 1)\times(38.6)^{2}}{39^{2}}=\frac{11\times1489.96}{1521}$. First, calculate the numerator: $11\times1489.96 = 16389.56$. Then, divide by the denominator: $\frac{16389.56}{1521}\approx10.775$.

Answer:

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