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). 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.
Score on first SAT 470 560 430 590 490 410 460
Score on second SAT 520 620 450 610 530 460 500
Step 2 of 5 : Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.
Solution:
Before | after | Difference |
470 | 520 | -50 |
560 | 620 | -60 |
430 | 450 | -20 |
590 | 610 | -20 |
490 | 530 | -40 |
410 | 460 | -50 |
460 | 500 | -40 |
Data for the paired difference is
-50,-60,-20,-20,-40,-50,-40,
n = 7
Step 2 of 5 : Find the value of the standard deviation of the paired differences.
x | x^2 | |
-50 | 2500 | |
-60 | 3600 | |
-20 | 400 | |
-20 | 400 | |
-40 | 1600 | |
-50 | 2500 | |
-40 | 1600 | |
SUM | -280 | 12600 |
Sample variance s2 =
= [1/(7- 1)][12600 - (-2802/7) ]
= 233.333333333
Now ,
standard deviation s = variance = 15.3
Answer : 15.3