Question

use a \\(chi^2\\)-test to test the claim \\(sigma^2 = 0.55\\) at the \\(alpha = 0.10\\) significance level using sample statistics \\(s^2 = 0.526\\) and \\(n = 18\\). assume the population is normally distributed.

identify the null and alternative hypotheses.

a \\(h_0: sigma^2 = 0.55\\) \\(h_a: sigma^2 \
eq 0.55\\)

b \\(h_0: sigma^2 \geq 0.55\\) \\(h_a: sigma^2 < 0.55\\)

c \\(h_0: sigma^2 \leq 0.55\\) \\(h_a: sigma^2 > 0.55\\)

d \\(h_0: sigma^2 \
eq 0.55\\) \\(h_a: sigma^2 = 0.55\\)

identify the standardized test statistic

\\(\square\\) (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the chi - square test statistic for variance

The formula for the chi - square test statistic when testing a claim about a population variance \(\sigma^{2}\) 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 of \(n\), \(s^{2}\), and \(\sigma^{2}\)

We are given that \(n = 18\), \(s^{2}=0.526\), and \(\sigma^{2}=0.55\).

Step3: Substitute the values into the formula

First, calculate \(n - 1\): \(n-1=18 - 1=17\).
Then, substitute into the formula: \(\chi^{2}=\frac{(17)\times(0.526)}{0.55}\)
Calculate the numerator: \(17\times0.526 = 8.942\)
Then, divide by \(0.55\): \(\chi^{2}=\frac{8.942}{0.55}\approx16.26\)

Answer:

The standardized test statistic is approximately \(16.26\)