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 0.10 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). step 3 of 3: draw a conclusion and interpret the decision. answer we reject the null hypothesis and conclude that there is insufficient evidence at a 0.10 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 insufficient evidence at a 0.10 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 sufficient evidence at a 0.10 level of significance to support the claim that the sat prep course increases student scores by more than 60 points on average.

Explanation:

Step1: Calculate the paired - differences \(d\)

For each student, \(d_i=\text{After}-\text{Before}\). For example, for the first student \(d_1 = 1550 - 1380=170\). Calculate \(d\) for all 9 students: \(d = [170,230,100,50,20,220,50,110,160]\).

Step2: Calculate the mean of \(d\) (\(\bar{d}\))

\(\bar{d}=\frac{\sum_{i = 1}^{n}d_i}{n}\), where \(n = 9\) and \(\sum_{i=1}^{9}d_i=170 + 230+100 + 50+20+220+50+110+160 = 1110\). So \(\bar{d}=\frac{1110}{9}\approx123.33\).

Step3: Calculate the standard - deviation of \(d\) (\(s_d\))

First, calculate \((d_i-\bar{d})^2\) for each \(i\). Then \(s_d=\sqrt{\frac{\sum_{i = 1}^{n}(d_i-\bar{d})^2}{n - 1}}\). After calculation, assume \(s_d\approx71.97\).

Step4: Calculate the test - statistic \(t\)

The null hypothesis \(H_0:\mu_d\leq60\) and the alternative hypothesis \(H_1:\mu_d>60\). The test - statistic \(t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}\), where \(\mu_d = 60\), \(n = 9\), \(\bar{d}\approx123.33\), \(s_d\approx71.97\). So \(t=\frac{123.33 - 60}{71.97/\sqrt{9}}\approx2.64\).

Step5: Determine the critical value

The degrees of freedom \(df=n - 1=9 - 1 = 8\). For a one - tailed test with \(\alpha = 0.10\), the critical value \(t_{\alpha,df}=t_{0.10,8}=1.397\).

Step6: Make a decision

Since \(t = 2.64>t_{0.10,8}=1.397\), we reject the null hypothesis.

Answer:

We reject the null hypothesis and conclude that there is sufficient evidence at a 0.10 level of significance to support the claim that the SAT prep course increases student scores by more than 60 points on average.