Question

an sat prep course claims to increase student scores by more than 60 points, on average. to test this claim, 9 students who have previously taken the sat are randomly chosen to take the prep course. their sat scores before and after completing the prep course are listed in the following table. test the claim at the 0.01 level of significance assuming that the population distribution of the paired differences is approximately normal. let ( d = ) (scores after completing the prep course) ( - ) (scores before completing the prep course).

sat scores
after prep course	1600	1520	1550	1220	1310	1220	1230	1450	1250

step 3 of 3: draw a conclusion and interpret the decision.

answer

• we reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to support the claim that the sat prep course increases student scores by more than 60 points on average.
• we reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance to support the claim that the sat prep course increases student scores by more than 60 points on average.
• we fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to support the claim that the sat prep course increases student scores by more than 60 points on average.

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Explanation:

To solve this, we first need to recall the steps of a hypothesis test for paired differences. Let's assume we already calculated the test statistic and critical value (or p - value) in the previous steps.

Step 1: Recall the Hypotheses

The null hypothesis \(H_0:\mu_d\leq60\) and the alternative hypothesis \(H_1:\mu_d > 60\) (since the claim is that the prep course increases scores by more than 60 points, so the mean of the differences \(d\) is more than 60).

Step 2: Decision Rule

For a one - tailed test at \(\alpha = 0.01\) with \(n - 1=8\) degrees of freedom (since \(n = 9\) paired observations), the critical value for a t - test (since the population of paired differences is approximately normal) can be found from the t - distribution table. The critical value \(t_{\alpha,df}=t_{0.01,8}\approx2.896\).

If we calculated the test statistic \(t\) (from the paired differences) and found that \(t>t_{\alpha,df}\) (or the p - value \(<\alpha\)), we reject the null hypothesis.

Step 3: Conclusion

From the options, when we reject the null hypothesis, we conclude that there is sufficient evidence at the 0.01 level of significance to support the claim that the SAT prep course increases student scores by more than 60 points on average. The second option is incorrect because if we reject the null hypothesis, we have sufficient evidence (not insufficient). The third option is incorrect because failing to reject the null hypothesis would not lead to the conclusion of sufficient evidence to support the alternative claim.

Answer:

We reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to support the claim that the SAT prep course increases student scores by more than 60 points on average.