Question

1. two separate samples receive different treatments. after treatment, the first sample has n = 5 with ss = 60, and the second has n = 9 with ss = 84. a. compute the pooled variance for the two samples. b. calculate the estimated standard error for the sample mean difference.

Explanation:

Part a: Compute the pooled variance

Step1: Recall pooled variance formula

The formula for pooled variance \( s_p^2 \) is \( s_p^2=\frac{SS_1 + SS_2}{df_1+df_2} \), where \( df = n - 1 \) for each sample.
For the first sample: \( n_1 = 5 \), so \( df_1=5 - 1 = 4 \); \( SS_1 = 60 \).
For the second sample: \( n_2 = 9 \), so \( df_2=9 - 1 = 8 \); \( SS_2 = 84 \).

Step2: Calculate sum of SS and sum of df

Sum of \( SS \): \( SS_1+SS_2 = 60 + 84=144 \).
Sum of \( df \): \( df_1 + df_2=4 + 8 = 12 \).

Step3: Compute pooled variance

Using the formula: \( s_p^2=\frac{144}{12}=12 \).

Part b: Calculate estimated standard error

Step1: Recall standard error formula

The formula for the estimated standard error (\( s_{M_1 - M_2} \)) for independent samples is \( s_{M_1 - M_2}=\sqrt{\frac{s_p^2}{n_1}+\frac{s_p^2}{n_2}} \), where \( s_p^2 = 12 \) (from part a), \( n_1 = 5 \), \( n_2 = 9 \).

Step2: Substitute values into formula

First, calculate \( \frac{s_p^2}{n_1}=\frac{12}{5}=2.4 \) and \( \frac{s_p^2}{n_2}=\frac{12}{9}=\frac{4}{3}\approx1.333 \).
Then, sum these two values: \( 2.4+\frac{4}{3}=\frac{12}{5}+\frac{4}{3}=\frac{36 + 20}{15}=\frac{56}{15}\approx3.733 \).
Finally, take the square root: \( s_{M_1 - M_2}=\sqrt{\frac{56}{15}}\approx\sqrt{3.733}\approx1.932 \).

Answer:

s:
a. The pooled variance is \(\boldsymbol{12}\).
b. The estimated standard error is approximately \(\boldsymbol{1.93}\) (or more precisely \(\sqrt{\frac{56}{15}}\approx1.932\)).