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Sampling distribution of proportion

Short answer

The sampling distribution of proportion is the distribution of sample proportions p̂ from all possible samples. Its mean equals population proportion P and standard error is sqrt(PQ/n), with finite population correction sqrt((N−n)/(N−1)). Example: P=0.05, n=100 gives SE=0.0218.

Define sampling distribution of proportion with example.

Sampling distribution of proportion: if all possible samples of size n are drawn from a population in which the proportion having a certain attribute is P, and the sample proportion p̂ = x/n (x = number of successes in the sample) is calculated for each, the probability distribution of p̂ is the sampling distribution of the proportion.

Properties:

  • Mean: E(p̂) = P
  • Standard error: SE(p̂) = √(PQ/n), where Q = 1 − P
  • For finite population (without replacement): SE = √(PQ/n) × √((N − n)/(N − 1))
  • For large n (nP ≥ 5 and nQ ≥ 5) it is approximately normal, so z = (p̂ − P)/√(PQ/n).

Example

A factory knows that 5% of the items it makes are defective (P = 0.05). Samples of n = 100 items are taken.

  • Mean of p̂ = 0.05
  • SE = √(0.05 × 0.95/100) = √0.000475 = 0.0218

Probability that the sample proportion defective exceeds 8%:

z = (0.08 − 0.05)/0.0218 = 1.38

P(p̂ > 0.08) = P(Z > 1.38) = 0.0838

So only about 8.4% of such samples will have more than 8% defectives.

✓ Answer checked by Claude Sonnet 5.5

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