From the problem, we know that EBP = 0.03 (3%=0.03) and [latex]{z_{\frac{a}{2}}}[/latex] z_0.05 = 1.645 because the confidence level is 90%.

However, in order to find n, we need to know the estimated (sample) proportion p′. Remember that q′ = 1 – p′. But, we do not know p′ yet. Since we multiply p′ and q′ together, we make them both equal to 0.5 because p′q′ = (0.5)(0.5) = 0.25 results in the largest possible product. (Try other products: (0.6)(0.4) = 0.24; (0.3)(0.7) = 0.21; (0.2)(0.8) = 0.16 and so on). The largest possible product gives us the largest n. This gives us a large enough sample so that we can be 90% confident that we are within three percentage points of the true population proportion. To calculate the sample size n, use the formula and make the substitutions.

n = [latex]{\dfrac{z^2p'q'}{EBP^2}}[/latex] gives n = [latex]{\dfrac{1.645^2(0.5)(0.5)}{0.03^2}}[/latex] = 751.7

Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones.