Question

If the population size is 24,000 and the sample size is 400, and \(p = 0.7\), what is the sampling distribution of the sample proportion \(\hat{p}\)?

• A. Exactly Normal with \(\hat{p} = 0.7\) and \(\sigma_{\hat{p}} = 0.023\)
• B. Approximately Normal with \(\hat{p} = 0.7\) and \(\sigma_{\hat{p}} = 0.0935\)
• C. Approximately Normal with \(\hat{p} = 0.7\) and \(\sigma_{\hat{p}} = 0.023\)
• D. Exactly binomial with \(\mu_{\hat{p}} = 289\) and \(\sigma_{\hat{p}} = 9.17\)

Hint:

Check \(np\) and \(nq\) for normal approximation validity, then use \(\sigma_{\hat p} = \sqrt{pq/n}\) since the sampling fraction is small.

Answer 1

The Correct Option is C

Step 1: Start from the fact that the number of successes \(X\) in the sample follows a Binomial(\(n=400, p=0.7\)) distribution (approximately, treating sampling as with replacement since \(n/N\) is small), and the sample proportion is \(\hat{p} = X/n\).

Step 2: Since \(n\) is large and both \(np = 280\) and \(nq = 120\) exceed the usual rule-of-thumb value of 5, the Binomial distribution of \(X\), and hence of \(\hat{p}\), can be approximated closely by a Normal distribution (De Moivre-Laplace / CLT). It is an approximation, so we never call it "exactly Normal".

Step 3: Variance of \(\hat{p}\) is $Var(\hat{p}) = \dfrac{pq}{n} = \dfrac{(0.7)(0.3)}{400} = 0.000525$, giving standard deviation $\sigma_{\hat p} = \sqrt{0.000525} = 0.0229$, rounding to $0.023$. The finite population correction factor $\sqrt{\dfrac{N-n}{N-1}} = \sqrt{\dfrac{23600}{23999}} \approx 0.9917$ barely changes this value, since $n/N$ is only about $1.7\%$.

Step 4: Mean stays $E(\hat p) = p = 0.7$.

So $\hat p$ is approximately $N(0.7, 0.023^2)$.

\[\boxed{\hat{p}\ \text{is approximately Normal with}\ \hat p = 0.7,\ \sigma_{\hat p} = 0.023}\]