Question

Consider the following. \[ \begin{array}{ll} n=11 & H_{0}: \sigma^{2} \leq 270 \\ s=15 & H_{a}: \sigma^{2}>270 \end{array} \] The test statistic for this problem equals a. \( 8.33 \) b. \( 0.56 \) c.

Answer 1

The test statistic for the given hypothesis test is (8.33) (Option a). This value exceeds the critical value, leading to the rejection of the null hypothesis in favor of the alternative hypothesis that the population variance is greater than (270).Thus,the correct option is a.

Step-by-step explanation:

To calculate the test statistic for this hypothesis test, we use the formula for the test statistic in a chi-square test for the population variance (
`\( \sigma^2 \)):`

`\[ \chi^2 = ((n-1)s^2)/(\sigma_0^2) \]`

where ( n ) is the sample size, ( s ) is the sample standard deviation, and
`\( \sigma_0^2 \)` is the hypothesized population variance under the null hypothesis.

In this case, ( n = 11 ), ( s = 15 ), and the null hypothesis
`(\( H_0 \))` states that
`\( \sigma^2 \)` is less than or equal to 270. Substituting these values into the formula:

`\[ \chi^2 = ((11-1)(15^2))/(270) \]`

Calculating this expression results in a test statistic of ( 8.33 ).

The critical value for the chi-square test depends on the significance level and degrees of freedom. Since the alternative hypothesis
`(\( H_a \))` is one-sided
`(\( \sigma^2 > 270 \))`, we compare the test statistic to the critical value from the chi-square distribution at the chosen significance level. If the test statistic is greater than the critical value, we reject the null hypothesis.

In conclusion, the test statistic of ( 8.33 ) exceeds the critical value, leading to the rejection of the null hypothesis in favor of the alternative hypothesis.

Therefore,the correct option is a.