1. If he uses a previous estimate of 22%
The problem states within 2 percentage points so the error is 0.02. The z score for this is 2.575.
The solution would be: 0.22(1-0.22) (2.575/0.02)^2 = 0.1716 (16576.5625) = 2844.538125; therefore we need to round up and the sample size would be 2845.
2. If he does not use any prior estimates.
The sample size required to obtain 1 – α * 100% confidence interval for p with a margin of error E, without a previous estimate of p, is given by: n = 0.25 (z α/2 divided by E) ^2
Therefore, n = 0.25 (2.575/0.02)^2
= 0.25 (16576.5625)
=4144.14; but we need to round up. So the sample size is 4145.
