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.
[tex]S_{p^2}=\dfrac{SS_1+SS_2}{n_1+n_2-1}[/tex]
So, it becomes,
[tex]S_p^2=\dfrac{340+236}{12+6-1}=\dfrac{576}{17}=33.88[/tex]
b. Compute the estimated standard error for the sample mean difference.
[tex]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[/tex]
Hence, a) 33.88 and b) 2.91.
