Question

one sample has ss=36 and a second sample has ss=18

If n=4 for both samples, find each of the sample variances and compute 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.

The first sample has ________ (choose one of the following 12.00, 9.00, 6.00, 3.00), and the second has s^2=______((choose one of the folloeing 12.00, 6.00, 3.00, 4.50). The pooled variance is s^2p=________(9.00, 6.00,12.00)

Now assume that n=4 for the first sample and n=7 for the second. Agsin
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.

The first sample has s^2=______(6.00,12.00, 9.00,3.00),and the second has s^2=________(3.00,6.00,4.50, 12.00). The pooled variance is s^2p=_______(9.00,3.00,12.00,6.00)

Answer 1

The ss of the two samples are: $ss_{1}=36$ and $ss_{2}=18$ with $n_{1}=n_{2}=4$.

Their variance can be calculated as -

$s_{1}^{2}=\frac{ss_{1}}{n_{1}-1}=\frac{36}{3}=12.00$ and

$s_{2}^{2}=\frac{ss_{2}}{n_{2}-1}=\frac{18}{3}=6.00$

The pooled variance is

$s_{p}^{2}=\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}= \frac{3\times 12.00+3\times 6.00}{8-2}=\frac{54}{6}=9.00$

which is equal to $\frac{1}{2}(s_{1}^{2}+s_{2}^{2})=\frac{1}{2}(12.00+6.00)=9.00$ , i.e. the pooled variance is halfway between the two sample variances.

Now, for the given ss of the two samples : $ss_{1}=36$ and $ss_{2}=18$ with $n_{1}=4 \hspace{0.06cm};\hspace{0.06cm}n_{2}=7$ .

Their variance can be calculated as -

$s_{1}^{2}=\frac{ss_{1}}{n_{1}-1}=\frac{36}{3}=12.00$ and

$s_{2}^{2}=\frac{ss_{2}}{n_{2}-1}=\frac{18}{6}=3.00$

The pooled variance is

$s_{p}^{2}=\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}= \frac{3\times 12.00+6\times 3.00}{11-2}=\frac{54}{9}=6.00$

which is closer to $s_{1}^{2}$, i.e. the pooled variance is closer to variance of the first (larger) sample.