We are given
Margin of error = 5%
at a 95% confidence interval
We are to find the sample size
The sample size(n) is given as
[tex]n=\frac{z^2p(1-p)}{E^2}[/tex]
Where z = z score for 95% confidence interval
Using
Hence, z = 1.96
E = Margin of error = 5%
To maximize the sample size,
Then, p must be 0.5
Hence,
P = 0.5
Substituting the values we get
[tex]n=\frac{1.96^2\times0.5(1-0.5)}{(0.05)^2}[/tex]
Simplifying we have
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