Given:
Margin of error = 0.04
Confidence level = 95%
Let's answer the following questions:
• (A). Assume that nothing is known about the percentage to be estimated.
Since we are to assume that nothing about the percentage is know, we have:
p = q = 0.5
E = 0.04
Given a 95% confidence interval, we have:
Significance level = 1 - 0.95 = 0.05
Using the z-table, for a two tailed test, we have:
[tex]z_{\frac{0.05}{2}}=1.96[/tex]Hence, the sample size will be:
[tex]\begin{gathered} n=pq(\frac{z_{\frac{\alpha}{2}}}{E})^2 \\ \\ n=(0.5)(0.5)(\frac{1.96}{0.04})^2 \\ \\ n=0.25*2401 \\ \\ n=600.25\approx600 \end{gathered}[/tex]Therefore, the required sample size is 600
• (b,). Here, we have the following:
p = 60% = 0.60
q = 1 - p = 1 - 0.60 = 0.40
95% confidence interval, z = 1.96
E = 0.04
To solve for n, we have:
[tex]\begin{gathered} n=(0.6)(0.4)(\frac{1.96}{0.04})^2 \\ \\ n=576.24\approx576 \end{gathered}[/tex]Here, the sample size is 576
(C).
No, the added knowledge in part B does not have much of an effect on the sample size.
It only slightly reduces the sample size.
ANSWER:
• (A). 600
,• (B). 576
,• (C). No, it only slightly reduces the sample size.