In: Statistics and Probability
Student |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
Pre-chess memory score |
510 |
610 |
640 |
675 |
600 |
550 |
610 |
625 |
450 |
720 |
575 |
675 |
Post-chess memory score |
850 |
790 |
850 |
775 |
700 |
775 |
700 |
850 |
690 |
775 |
540 |
680 |
Test the claim at the α = 0.01 level of significance that students who participated in the chess program achieve higher memory scores after completion of the program.
Here we are comparing two data sets where the sample subjects are same. So this is a dependent samples t-test. We will use t-dist to compare the mean of the difference of scores
d = Pre -post
We want to test whether the post memoery is higher or not. So in terms of 'd' we are testing whether 'd <0' or not.
Pre-chess memory score | Post-chess memory score | d | d^2 | |
1 | 510 | 850 | -340 | 115600 |
2 | 610 | 790 | -180 | 32400 |
3 | 640 | 850 | -210 | 44100 |
4 | 675 | 775 | -100 | 10000 |
5 | 600 | 700 | -100 | 10000 |
6 | 550 | 775 | -225 | 50625 |
7 | 610 | 700 | -90 | 8100 |
8 | 625 | 850 | -225 | 50625 |
9 | 450 | 690 | -240 | 57600 |
10 | 720 | 775 | -55 | 3025 |
11 | 575 | 540 | 35 | 1225 |
12 | 675 | 680 | -5 | 25 |
Total | -1735 | 383325 | ||
Mean | -144.583 | |||
SD | 109.7406 |
Mean =
SD =
State the null and alternate hypotheses [3]
: d = 0 .(There is no change in the scores)
: d < 0 .(The scores are higher after the completion of the program.)
Check the assumptions [4]
Give the test statistic [2]
Test Stat =
null difference = 0
n = 12
Test Stat = -4.564
Give the P-value and the conclusion reached about the null hypothesis based on the P-value. [2]
p- value =P( > |test stat| ) ..........this is one tailed so we check only on one side
= P( > 4.56)
p-value = 0.00041 .............using t-dist tables
Summarize the final conclusion in the context of the claim. [2]
Since p-value < 0.01
We reject the null hypothesis at 0.01 level.
We conclude that the chess program significantly improved the memory of the students.