For a large-sample proportion, which is the correct expression for the variance of p-hat?

For a large-sample proportion, which is the correct expression for the variance of p-hat?

• A. p(1-p)/n
• B. p(1-p)
• C. np(1-p)
• D. n/(p(1-p))

Answer: (A)

When you take a large-sample proportion, the number of successes X among n independent trials with success probability p follows a binomial distribution: X ~ Binomial(n, p). The proportion is p-hat = X/n. For this proportion, the mean is p and the variance is Var(p-hat) = Var(X)/n^2. Using Var(X) = np(1-p), we get Var(p-hat) = [np(1-p)] / n^2 = p(1-p) / n. This is the quantity that tells you how spread out the sample proportion is around the true p, and its square root, the standard error, is sqrt(p(1-p)/n). In large samples, p-hat is well approximated by a normal distribution with mean p and variance p(1-p)/n. The other expressions don’t match this spread: Var(X) = np(1-p) is the variance of the count, not the proportion; p(1-p) would not account for the shrinking effect of increasing n; and n/(p(1-p)) isn’t a variance form for this setup.

When you take a large-sample proportion, the number of successes X among n independent trials with success probability p follows a binomial distribution: X ~ Binomial(n, p). The proportion is p-hat = X/n. For this proportion, the mean is p and the variance is Var(p-hat) = Var(X)/n^2.

Using Var(X) = np(1-p), we get Var(p-hat) = [np(1-p)] / n^2 = p(1-p) / n. This is the quantity that tells you how spread out the sample proportion is around the true p, and its square root, the standard error, is sqrt(p(1-p)/n). In large samples, p-hat is well approximated by a normal distribution with mean p and variance p(1-p)/n.

The other expressions don’t match this spread: Var(X) = np(1-p) is the variance of the count, not the proportion; p(1-p) would not account for the shrinking effect of increasing n; and n/(p(1-p)) isn’t a variance form for this setup.