7.3 A population proportion -- rrc math 1020  (Page 3/24)

Solution b

Solution b

From the problem, we know that EBP = 0.03 (3%=0.03) and $z_{\frac{\alpha }{2}}$ 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.

$n=\frac{z^2{p}^{\prime }{q}^{\prime }}{EB{P}^2}$ gives $n=\frac{{1.645}^2(0.5)(0.5)}{{0.03}^2}=751.7$

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.