The previous problem demonstrates that removing individual differences can subst

Question

The previous problem demonstrates that removing individual differences can substantially reduce variance and lower the standard error. However, this benefit only occurs if the individual differences are consistent across treatment conditions. In problem 21, for example, the participants with the highest scores in the more-sleep condition also had the highest scores in the less-sleep condition. Similarly, participants with the lowest scores in the first condition also had the lowest scores in the second condition. To construct the following data, we started with the scores in problem 21 and scrambled the scores in treatment 1 to eliminate the consistency of the individual differences.

Number of Academic Problems

Student

Following Nights with Above Average Sleep

Following Nights with Below Average Sleep

A

10

13

B

8

14

C

5

13

D

5

5

E

4

9

F

10

6

G

11

6

H

3

6

[Enlarge Table]

Treat the data as if the scores are from an independent-measures study using two separate samples, each with  participants. Compute the pooled variance, the estimated standard error for the mean difference, and the independent-measures t statistic. Using a two-tailed test with , is there a significant difference between the two sets of scores? Note: The scores in each treatment are the same as in Problem 21. Nothing has changed.

Now assume that the data are from a repeated-measures study using the same sample of  participants in both treatment conditions. Compute the variance for the sample of difference scores, the estimated standard error for the mean difference and the repeated-measures t statistic. Using a two-tailed test with , is there a significant difference between the two sets of scores? (You should find that removing the individual differences with a repeated-measures t no longer reduces the variance because there are no consistent individual differences.)

Explanation / Answer

through indpependent sample

here

degree of freedom =n1+n2-2 =14

from above pooled std deviation =Sp=((n1-1)s12+(n2-1)s22/(n1+n2-2))1/2 =3.4641

std error of difference =(s12/n1+s22/n2)1/2 =1.7321

hence test stat t=(x1-x2)/std error =-1.1547

for 0.05 level and 14 degree of freedom critival value of t =2.1448

as test stat is not in rejection zone, we can not accept that there is a significant difference between the two sets of scores

2) paired test

for 7degree of freedom and 0.05 level ; critical value of t =2.3646

as test stat is not in rejection zone, we can not accept that there is a significant difference between the two sets of scores

from both test it is found that removing the individual differences with a repeated-measures t no longer reduces the variance because there are no consistent individual differences

S. no 1st 2nd 1 10 13 2 8 14 3 5 13 4 5 5 5 4 9 6 10 6 7 11 6 8 3 6 total 56.000 72.000 mean 7.000 9.000 std deviation(S) 3.117 3.780

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