Question

1. two separate samples, each with n = 9 individuals, receive different treatments. after treatment, the first sample has ss = 546 and the second has ss = 606. a. find the pooled variance for the two samples. b. compute the estimated standard error for the sample mean difference.

Explanation:

Part a: Pooled Variance

Step1: Recall the formula for pooled variance

The formula for pooled variance ($s_p^2$) when two samples have sizes $n_1$ and $n_2$ (here $n_1 = n_2 = 9$) and sum of squares $SS_1$ and $SS_2$ is:
$$s_p^2=\frac{SS_1 + SS_2}{df_1 + df_2}$$
where $df = n - 1$ for each sample.

Step2: Calculate degrees of freedom

For each sample, $df = 9 - 1 = 8$. So $df_1 + df_2 = 8 + 8 = 16$.

Step3: Substitute values into the formula

$SS_1 = 546$, $SS_2 = 606$. So:
$$s_p^2=\frac{546 + 606}{16}=\frac{1152}{16}=72$$

Part b: Estimated Standard Error

Step1: Recall the formula for standard error of mean difference

The formula for the estimated standard error ($s_{M_1 - M_2}$) for two independent samples is:
$$s_{M_1 - M_2}=\sqrt{\frac{s_p^2}{n_1}+\frac{s_p^2}{n_2}}$$

Step2: Substitute values

We know $s_p^2 = 72$, $n_1 = n_2 = 9$. So:
$$s_{M_1 - M_2}=\sqrt{\frac{72}{9}+\frac{72}{9}}=\sqrt{8 + 8}=\sqrt{16}=4$$

Answer:

a. The pooled variance is $\boldsymbol{72}$.
b. The estimated standard error is $\boldsymbol{4}$.