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: Check the conditions for the Central Limit Theorem to apply to the sample proportion. Here \(n = 400\), \(p = 0.7\), \(q = 1-p = 0.3\). Compute \(np = 400 \times 0.7 = 280\) and \(nq = 400 \times 0.3 = 120\). Both are much greater than 5, so the normal approximation to the sampling distribution of \(\hat{p}\) is valid.

Step 2: Since \(\hat{p}\) is a discrete random variable built from a finite sample, its exact distribution is binomial (scaled), not truly normal. The normal distribution is only an approximation, however good, so the distribution is "approximately Normal", never "exactly Normal".

Step 3: Compute the standard error. Since the sampling fraction \(n/N = 400/24000 \approx 0.0167\) is very small (well below 5%), the finite population correction can be ignored, so:

\[\sigma_{\hat{p}} = \sqrt{\dfrac{pq}{n}} = \sqrt{\dfrac{0.7 \times 0.3}{400}} = \sqrt{\dfrac{0.21}{400}} = \sqrt{0.000525} \approx 0.0229 \approx 0.023\]

Step 4: Mean of \(\hat{p}\) is \(E(\hat{p}) = p = 0.7\). So the sampling distribution of \(\hat{p}\) is approximately Normal with mean 0.7 and standard error 0.023.

Final Answer: Approximately Normal with \(\hat{p} = 0.7\) and \(\sigma_{\hat{p}} = 0.023\).