In: Statistics and Probability
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)
ANSWER::
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.
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