We are asked to find the required sample size for a proportion.
Given information:
Margin of error = 4% = 0.04
Confidence level = 90% = 0.90
Significance level = 1 - 0.90 = 0.10
The two-sided z-score corresponding to α = 0.10 is found to be 1.645
The required sample size is given by
[tex]n=p(1-p)\cdot(\frac{z}{E})^2[/tex]
Where p is the estimated proportion and can be assumed to be 0.50 when it is not given.
z is the z-score and E is the margin of error
[tex]\begin{gathered} n=0.50(1-0.50)\cdot(\frac{1.645}{0.04})^2 \\ n=422.8 \end{gathered}[/tex]
Round up the answer to a whole number.
[tex]n=423[/tex]
Therefore, the smallest sample size required to obtain the desired margin of error is 423
