Question

A prior study determined the point estimate of the population proportion as 58% ( = 0.58). The analysts decide to conduct a second study on the same topic and would like its margin of error, E, to be 4% when its confidence level is 95% (z*-score of 1.96). What is the minimum sample size that should be used so the estimate of will be within the required margin of error of the population proportion? n = (1 – ) • 12 23 585 808

Answer 1

Answer:

585

Step-by-step explanation:

just took test

Answer 2

The minimum sample size required for a test with a confidence interval of $100(1 - \alpha )%$ with a z-score of $z_{ \alpha /2}$ and margin of error of E and a population proportion of p is given by:

$n= \frac{p(1-p)z_{ \alpha /2}^2 \alpha }{E^2}$

Given p = 58% = 0.58, E = 4% = 0.04, $z_{ \alpha /2}=1.96$

Therefore, 

$n= \frac{0.58(1-0.58)(1.96)^2}{0.04^2} \\ \\ = \frac{0.58(0.42)(3.8416)}{0.0016} = \frac{0.93581376}{0.0016} \\ \\ =584.88=585$