We can solve this problem by referring to the standard probability distribution tables for z.
We are required to find for the number of samples given the proportion (P = 5% = 0.05) and confidence level of 95%. This would give a value of z equivalent to:
z = 1.96
Since the problem states that it should be within the true proportion then p = 0.5
Now we can find for the sample size using the formula:
n = (z^2) p q /E^2
where,
p = 0.5
q = 1 – p = 0.5
E = estimate of 5% = 0.05
Substituting:
n = (1.96^2) 0.5 * 0.5 / 0.05^2
n = 384.16
Around 385students are required.
