Question

Suppose you wanted to estimate the difference between two population means correct to within 4.8 at the 92% confidence level. If prior information suggests that both population variances are approximately equal to 12 and you want to select independent random samples of equal size from the populations, how large should the sample sizes be?
Critical Value: 1.75
The sample sizes should be: n1=___n2=_____?

Answer 1

Answer: $n_1=n_2=4$

Step-by-step explanation:

Given : Margin of error : E= 4.8

Confidence level : 92%

Significance level : $1-0.92=0.08$

$\sigma_1^2=\sigma_1^2\approx12$

Two-tailed critical value :-

$z_{\alpha/2}=z_{0.08/2}=z_{0.04}=1.75$

If we want to select independent random samples of equal size from the populations,

Formula for the sample size :

$n_1=n_2=(\dfrac{z_{\alpha/2}}{E})^2(\sigma_1^2+\sigma_2^2)$

Then buy using given values , we have

$n_1=n_2=(\dfrac{1.75}{4.8})^2(12+12)$

Simplify ,

$n_1=n_2=(\dfrac{1.75}{4.8})^2(12+12)=3.190\approx4$  [Round to the next integer.]

Hence, the The sample sizes should be: $n_1=n_2=4$