3.2 Sampling Distributions for Sample Proportions — Answers and Rubrics

$\mu_{\hat p}=p=0.40$. $\sigma_{\hat p}\approx\sqrt{0.40(0.60)/200}=\sqrt{0.0012}=0.034641$, about 3.46 percentage points. The approximation uses the small sampling fraction. The actual statistic is $79/200=0.395$; it varies from sample to sample, whereas the complete records establish the fixed population p=0.40.

The SRS is random. Since $200\leq 0.10(20{,}000)=2{,}000$, the sampling fraction is small enough for the usual independence approximation; without-replacement observations are not literally independent. Expected counts are $np=200(0.40)=80$ members who renewed online and $n(1-p)=200(0.60)=120$ members who did not renew online, both at least 10. Thus an approximately normal sampling distribution is appropriate under the stated model.

(i) $3{,}000>2{,}000$, so the usual no-finite-population-correction SD approximation is unsupported. (ii) Expected counts are 4 and 196; 4<10 fails the success-count normality condition. (iii) A voluntary poll has no random selection, so the SRS sampling model is unsupported even if its counts are large. Independent with-replacement draws do not create the without-replacement dependence, so that 10% check is unnecessary; the expected-count check is still needed for a normal sample-proportion model.

$z=(0.45-0.40)/0.034641=1.44338$. The upper-tail normal area is about $1-\Phi(1.44338)=0.07446$. Across repeated SRSs of 200 members, the average sample online-renewal proportion is 0.40 and the typical deviation from 0.40 is about 0.03464, or 3.46 percentage points. About 7.45% of those size-200 samples would have an online-renewal proportion at least 0.45 under this approximate model. It is a probability for whole-sample proportions, not for one member.