Step 1 :The teacher's null hypothesis (H0) is that the variation in her students' test scores is equal to or greater than 2.31, the variation in previous terms. Her alternative hypothesis (H1) is that the variation in her students' test scores is less than 2.31.
Step 2 :She conducts a hypothesis test and calculates her test statistic to be \( \chi^{2}=16.062 \).
Step 3 :She looks up the critical value for this test and finds it to be \( \chi^{2}=11.292 \).
Step 4 :The test statistic of \( \chi^{2}=16.062 \) is greater than the critical value of \( \chi^{2}=11.292 \). Therefore, the test statistic falls in the critical (rejection) region for this test.
Step 5 :Since the test statistic falls in the critical (rejection) region, there is sufficient evidence to reject the null hypothesis and support the alternative hypothesis.
Step 6 :Final Answer: \(\boxed{\text{D. The test statistic falls in the critical (rejection) region for this test. Therefore, there is sufficient evidence to support her claim.}}\)