Question

An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first two times they took the verbal SAT. Before taking the SAT for the second time, each student took a course to try to improve his or her verbal SAT scores. Do these results support the claim that the SAT prep course improves the students' verbal SAT scores? Let d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.05 for the test. Assume that the verbal SAT scores are normally distributed for the population of students both before and after taking the SAT prep course.

Student 1   2 3 4 5 6   7

Score on first SAT 360 440 520 490 510 490 480

Score on second SAT 400 520 590 550 550 520 520

Step 1 of 5: State the null and alternative hypotheses for the test.

Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5: Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5: Make the decision for the hypothesis test.

Answer 1

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
360	400	-40.000	130.612
440	520	-80.000	816.327
520	590	-70.000	344.898
490	550	-60.000	73.469
510	550	-40.000	130.612
490	520	-30.000	459.184
480	520	-40.000	130.612

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	3290	3650	-360.000	2085.714

1)

Ho :   µd=   0
Ha :   µd <   0
  

2)

mean of difference ,    D̅ =ΣDi / n =   -51.4286
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    18.6

3)

std error , SE = Sd / √n =    18.6445   / √   7   =   7.0470      
                          
t-statistic = (D̅ - µd)/SE = (   -51.42857143   -   0   ) /    7.0470   =   -7.298

4)

Degree of freedom, DF=   n - 1 =    6  
t-critical value , t* =        -1.943   [excel function: =t.inv(α,df) ]

reject Ho if t <-1.943

5)

Reject the null hypothesis