Answer:
Since the calculated value of t = -0.215 falls in the critical region so we accept Ha that SAT prep course improves the students' verbal SAT scores and reject the null hypothesis at significance level 0.05.
e. These results support the claim that the SAT prep course improves the students' verbal SAT scores.
Step-by-step explanation:
Student 1 2 3 4 5 6 7
Score on first SAT 500 380 560 430 450 360 560
Score on second SAT 540 470 580 450 480 400 600
Difference d -40 -90 -20 -20 -30 -40 -40 ∑ -280
d² 1600 8100 400 400 900 1600 1600 ∑14600
a. Let the hypotheses be
H0: ud= 0 against the claim Ha: ud ≠0
The degrees of freedom = n-1= 7-1= 6
The significance level is 0.05
The test statistic is
t= d`/sd/√n
The critical region is ║t║≤ t (0.025,6) = ±2.447
d`= ∑di/n= -280/7= -4
Sd²= ∑(di-d`)²/n-1 = 1/n-1 [∑di²- (∑di)²n]
= 1/6[14600-(-4)²/7] = [14600-2.2857/6]= 2432.952
b. Sd= 49.3249= 49.325
Therefore
c. t= d`/ sd/√n
t= -4/ 49.325/√7
t= -4/18.6435 = -0.2145= -0.215
d. Since the calculated value of t = -0.215 falls in the critical region so we accept Ha that SAT prep course improves the students' verbal SAT scores and reject the null hypothesis at significance level 0.05
e. These results support the claim that the SAT prep course improves the students' verbal SAT scores.
