One sample has SS = 44 and a second sample has SS = 40. (a) Assuming that n = 7

Question

One sample has SS = 44 and a second sample has SS = 40.

(a) Assuming that n = 7 for both samples, calculate each of the sample variances, then calculate the pooled variance. Because the samples are the same size, you should find that the pooled variance is exactly halfway between the two sample variances. (Use 2 decimal places.)
pooled variance  

(b) Now assume that n = 7 for the first sample and n = 15 for the second. Again, calculate the two sample variances and the pooled variance. You should find that the pooled variance is closer to the variance for the larger sample. (Use 2 decimal places.)
pooled variance

Explanation / Answer

a) here n1 = n2 = 7

Variance of one sample = V(x)
V(x) = ss/(n1-1) = 44/(7-1) = 44/6 = 22/3

Variance of second sample = V(y)
V(y) = ss/(n2-1) = 40/(7-1) = 40/6 = 20/3

Pooled Variance =
= [(n1 - 1)V(x) + (n2 - 1)V(y)] / (n1 + n2 - 2)
=( (7-1)*22/3 +((7-1) *20/3))/(7+7-2) = (22/3 +20/3) /2 = 42/6 = 7

b) here n1 = 7 ,n2 = 15

Variance of one sample = V(x)
V(x) = ss/(n1-1) = = 44/(7-1) = 44/6 = 22/3

Variance of second sample = V(y)
V(y) = ss/(n2-1) = 40/(15-1) = 40/14 = 20/7

Pooled Variance =
= [(n1 - 1)V(x) + (n2 - 1)V(y)] / (n1 + n2 - 2)
=( (7-1)*22/3 +((15-1) *20/7))/(7+15-2)

= (44 + 40)/20

=84/20 =4.2

The pooled variance decreased because the sum of squares in sample 2 remained the same, but n increase, so the variance decreased. Because sample 2 was larger, the weighted average was pulled down.

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