Question

Two separate samples receive different treatments. After treatment, the first sample has n = 6 with SS = 236, and the second has n 5=12 with SS = 340.
a. Find the pooled variance for the two samples.
b. Compute the estimated standard error for the sample mean difference.

Answer 1

Answer: a) 33.88 and b) 2.91.

Step-by-step explanation:

Since we have given that

Sample n₁ = 6

SS = 236

Sample n₂ = 12

SS = 340

So, we need to find the following:

a. Find the pooled variance for the two samples.

$S_{p^2}=\dfrac{SS_1+SS_2}{n_1+n_2-1}$

So, it becomes,

$S_p^2=\dfrac{340+236}{12+6-1}=\dfrac{576}{17}=33.88$

b. Compute the estimated standard error for the sample mean difference.

$SE=S_p\times \sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}\\\\SE=\sqrt{33.88}\times \sqrt{\dfrac{1}{6}+\dfrac{1}{12}}\\\\SE=5.82\times \sqrt{\dfrac{2+1}{12}}\\\\SE=5.82\times \sqrt{\dfrac{3}{12}}\\\\SE=5.82\times \sqrt{\dfrac{1}{4}}\\\\SE=5.82\times \dfrac{1}{2}\\\\SE=2.91$

Hence, a) 33.88 and b) 2.91.