Sampling Distribution of p̂
Derive the mean (p) and standard deviation (√(p(1−p)/n)) of the sampling distribution of a sample proportion. Verify the Large Counts condition (np ≥ 10, n(1−p) ≥ 10) for normality and apply.
A lesson of AP Statistics, in Sampling Distributions, about 10 minutes.
Sampling Distribution of a Sample Proportion
By the end of this lesson, you will be able to describe the sampling distribution of p̂, apply the Central Limit Theorem for proportions, and use the sampling distribution to find probabilities.
What Is p̂?
The sample proportion (p̂, read "p-hat") is the fraction of a sample with a specific characteristic:
$$\hat{p} = \frac{\text{number with characteristic}}{n}$$
Example: In a sample of 200 voters, 120 support Candidate A. p̂ = 120/200 = 0.60.
The population proportion is p (the true proportion we're trying to estimate or test).
The Sampling Distribution of p̂
If you took many, many samples of size n from the same population:
- Each sample would give a (slightly) different p̂
- The distribution of all those p̂ values is the sampling distribution of p̂
Key properties (when conditions are met):
Center: The sampling distribution is centered at p (the true population proportion). $$\mu_{\hat{p}} = p$$
p̂ is an unbiased estimator of p.
Spread (standard deviation): $$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$$
Larger n → smaller standard deviation → less variation in p̂ values → more precise estimates.
Shape: When the sample size is large enough, the sampling distribution of p̂ is approximately normal.
Conditions for Normal Approximation
For the sampling distribution of p̂ to be approximately normal:
1. Random: The sample must be random.
2. Independent: Either sampling with replacement, or n ≤ 10% of the population.
3. Large counts (Success/Failure condition): $$np \geq 10 \quad \text{AND} \quad n(1-p) \geq 10$$
When these conditions hold: p̂ is approximately N(p, √(p(1-p)/n))
The Central Limit Theorem for Proportions
CLT (proportions): For a large enough sample, the sampling distribution of p̂ is approximately normal with:
- Mean = p
- Standard deviation = √(p(1-p)/n)