Question

Include null and alternate hypothesis, a p-value, and a conclusion PRIVATE about the null hypothesis

Nine students took the SAT. After taking it, they then took a test preparation course and retook the SAT. Can you conclude that the course changes performance on the SAT? (use α = .1)

Before	720	860	850	880	860	710	850	1200	950
After	740	860	840	920	890	720	840	1240	970

Answer 1

Before	After	Difference
720	740	-20
860	860	0
850	840	10
880	920	-40
860	890	-30
710	720	-10
850	840	10
1200	1240	-40
950	970	-20

∑d =    -140  

∑d² =    5200  

n =    9  

Mean , x̅d = Ʃd/n =    -140/9 =    -15.5556

Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] =    √[(5200-(-140)²/9)/(9-1)] =    19.4365

Null and Alternative hypothesis:  

Ho : µd =    0

H1 : µd ≠    0

Test statistic:  

t = (x̅d)/(sd/√n) = (-15.5556)/(19.4365/√9) =    -2.4010

df = n-1 =    8

Critical value :  

Two tailed critical value, t-crit = T.INV.2T(0.1, 8) =    1.860

Reject Ho if t < -1.86 or if t > 1.86  

p-value :  

Two tailed p-value = T.DIST.2T(ABS(-2.401), 8) =    0.0431

Decision:  

p-value < α, Reject the null hypothesis  

Conclusion:  

There is enough evidence to conclude that the course changes performance on the SAT at 0.1 significance level.