In: Statistics and Probability
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)d=(verbal SAT scores prior to taking the prep course)−(verbal SAT scores after taking the prep course). Use a significance level of α=0.01α=0.01 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 | 370 | 380 | 450 | 500 | 360 | 400 | 360 |
Score on second SAT | 420 | 480 | 500 | 580 | 400 | 460 | 400
Step 2 of 5 : Find the value of the standard deviation of the paired differences. Round your answer to one decimal place. |
the necessary calculation table :-
Score on first SAT (xi) | Score on second SAT (yi) | ||
370 | 420 | -50 | 100 |
380 | 480 | -100 | 1600 |
450 | 500 | -50 | 100 |
500 | 580 | -80 | 400 |
360 | 400 | -40 | 400 |
400 | 460 | -60 | 0 |
360 | 400 | -40 | 400 |
sum= - 420 | sum=3000 |
sample size (n) = 7
the standard deviation of the paired differences be:-
hypothesis:-
[ claim ]
test statistic be:-
df = (n-1) = (7-1) = 6
p value = P(t<-7.09) = 0.0002
[ in any blank cell of excle type =T.DIST(-7.09,6,TRUE) ]
decision:-
p value = 0.0002 < 0.01 (alpha)
we reject the null hypothesis.There is enough evidence to claim that the SAT prep course improves the students' verbal SAT scores.
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