One of the primary advantages of a repeated-measures design compared to independ
ID: 3152065 • Letter: O
Question
One of the primary advantages of a repeated-measures design compared to independent-measures, is that it reduces the overall variability by removing variance caused by individual differences. The following data are from a research study comparing two treatment condition
(a) Assume that the data are from an independent-measures study using two separate samples, each with n = 6 participants. Compute the pooled variance and the estimated standard error for the mean difference. (Use 3 decimal places.)
sp2 =
sMD =
(b) Now assume that the data are from a repeated-measures study using the same sample of n = 6 participants in both treatment conditions. Compute the variance for the sample of difference scores and the estimated standard error for the mean difference. (You should find that the repeated-measures design substantially reduces the variance and the standard error.) (Use 3 decimal places.)
s2 =
sMD =
Treatment 1 Treatment 2 Difference 12 14 8 8 3 5 M = 8.333 13 15 3 8 0 4 2 M = 9.667 M = 1.333 SS= 85.33333 SS= 83.33333 SS= 5.33333Explanation / Answer
a)
Calculating the standard deviations of each group,
s1 = 4.131182236
s2 = 4.082482905
Thus, the pooled standard deviation is given by
Sp = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]
As n1 = 6 , n2 = 6
Then
Sp = 4.106904755
Sp^2 = 16.86666667 [ANSWER, POOLED VARIANCE]
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Thus, the standard error of the difference is
SMD = Sp sqrt (1/n1 + 1/n2) = 2.371122566 [ANSWER]
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b)
Calculating the standard deviation of the differences (third column):
s = 0.984731928
Hence,
s^2 = 0.96969697 [ANSWER]
Here, n = 6.
Thus, the standard error of the difference is sD = s/sqrt(n):
sMD = 0.402015126 [ANSWER]
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