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Next, we take the sample variances and pool them while weighting
each estimate based on the sample’s degrees of freedom, using this
formula:

11 12
df 2
df 1
S 2 Pooled = (S 1 2 ) + (S 2 2 ) = (5.66) + (6.83)
df Total df Total 23 23
+
= 2.71 3.56 = 6.27
Distribution of Means’ Variance: With the estimated population variance in
place, we need to determine the variance of the distribution of means for
each sample. Though we have two samples, we only have one estimate of
the population variance (S 2 Pooled ), so we will need to use that to calculate the
variance of both distributions of means. In formula form, the estimated
variance for the distribution of means for the brain-training sample is:
S 2
S 2 = Pooled
M 1
N 1
6.27
S 2 = = 0.52
12
M 1
Now we need to do the same for the no brain-training sample,
S 2
S 2 = Pooled
M 2
N 2
6.27
S 2 = = 0.48
M 2
13
Note that here we are dividing by the sample size and not by degrees
of freedom. That’s because a distribution of means’ variance depends
on the sample size.

S 2
S 2 = Pooled
M 1
N 1
S 2
S 2 = Pooled
M 2
N 2

Want to review the ↗ Distribution of Differences Between Means’ Variance and Standard Deviation:
As we’ve done with other hypothesis testing examples, we need the stan-
logic behind the
distribution of means’ dard error to build a comparison distribution. To get the standard error, we
variance? SEE CHAPTER 7. first calculate the comparison distribution’s variance. With two distribu-
tions of means, we need to account for both samples’ variation. Greater
variation in either sample will create greater variation in the differences.
Here’s what we mean: If we subtract two numbers (e.g., 5 − 2), the
difference is 3. If either the 5 or the 2 change (i.e., vary), the outcome

354 S TATIS TI c S F OR L IFE

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