The minimum sample size required for a test with a confidence interval of [tex]100(1 - \alpha )%[/tex] with a z-score of [tex]z_{ \alpha /2}[/tex] and margin of error of E and a population proportion of p is given by:
[tex]n= \frac{p(1-p)z_{ \alpha /2}^2 \alpha }{E^2} [/tex]
Given p = 58% = 0.58, E = 4% = 0.04, [tex]z_{ \alpha /2}=1.96[/tex]
Therefore,
[tex]n= \frac{0.58(1-0.58)(1.96)^2}{0.04^2} \\ \\ = \frac{0.58(0.42)(3.8416)}{0.0016} = \frac{0.93581376}{0.0016} \\ \\ =584.88=585[/tex]
