Mathfolis

Sampling Distribution of a Sample Proportion

Unit 5 · Sampling Distributions

What AP Stats asks here

p^\hat{p} is the sample proportion drawn from a population with true proportion pp. Its sampling distribution centers at pp and has SE p(1−p)/n\sqrt{p(1-p)/n}. Before using the normal approximation, check the large-counts condition np≥10np \ge 10 and n(1−p)≥10n(1-p) \ge 10, plus the 10% rule n<0.10Nn < 0.10 N.

Sampling distribution of p^\hat{p}

Center
μp^=p\mu_{\hat{p}} = p
Standard error
σp^=p(1−p)n\sigma_{\hat{p}} = \sqrt{\dfrac{p(1-p)}{n}}

Conditions for the normal approximation

Large counts
np≥10 AND n(1−p)≥10np \ge 10 \text{ AND } n(1-p) \ge 10
10% condition
n<0.10⋅Nn < 0.10 \cdot N
AP Tip: When p is unknown (planning stage), use p = 0.5 to maximize p(1−p) and get the worst-case (largest) sample-size estimate.
Caution: When p is small, the large-counts condition fails for moderate n. Use the exact binomial distribution instead of the normal approximation.
Type 1

Describe the sampling distribution

Verify the conditions, then write p^∼N(p,p(1−p)/n)\hat{p} \sim N(p, \sqrt{p(1-p)/n}) with both numerical values plugged in.

Example 1
A poll surveys an SRS of n=400n = 400 voters from a city of 100,000 where true support is p=0.55p = 0.55. Which best describes the sampling distribution of p^\hat{p}? (A) p^∼N(0.45,0.025)\hat{p} \sim N(0.45, 0.025) (B) p^∼N(0.55,0.0249)\hat{p} \sim N(0.55, 0.0249) — large counts and 10% condition both satisfied. (C) p^∼N(0.55,0.25)\hat{p} \sim N(0.55, 0.25) (D) Cannot apply normal approximation because n is too small.

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Type 2

Compute P(p^>c)P(\hat{p} > c)

Standardize the threshold with σp^\sigma_{\hat{p}} and use the z-table.

Example 2
With p=0.55p = 0.55 and n=400n = 400 (so σp^≈0.0249\sigma_{\hat{p}} \approx 0.0249), what is P(p^>0.60)P(\hat{p} > 0.60)? (A) ≈ 0.022 (B) ≈ 0.05 (C) ≈ 0.50 (D) ≈ 0.95

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Type 3

Large counts and 10% conditions

Both npnp and n(1−p)n(1-p) must be at least 10. When either fails, the sampling distribution is skewed and the normal approximation gives wrong probabilities.

Example 3
A pollster surveys n=50n = 50 people about a rare event with p=0.05p = 0.05. Should p^\hat{p} be treated as approximately normal? (A) Yes — n is large enough. (B) No — np=2.5<10np = 2.5 < 10 fails the large-counts condition; the sampling distribution is right-skewed. Use the exact binomial instead. (C) Yes — the 10% condition is satisfied. (D) No — n(1−p)=47.5<50n(1-p) = 47.5 < 50.

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