Answer: 3
Step-by-step explanation:
Given : Pooled variance : [tex]\sigma^2=18[/tex]
Sample sizes of each sample = [tex]n_1=n_2=4[/tex]
We know that the standard error for the sample mean difference is given by :-
[tex]S.E.=\sqrt{\sigma^2(\dfrac{1}{n_1}+\dfrac{1}{n_2})}\\\\=\sqrt{(18)(\dfrac{1}{4}+\dfrac{1}{4})}\\\\=\sqrt{(18)(\dfrac{1}{2})}=\sqrt{9}=3[/tex]
Hence, the estimated standard error for the sample mean difference =3
