Prob and StatsSampling DistributionsS-IC.A.1

Sampling Distribution of a Proportion

Sample proportions center on p with spread √(p(1 − p)/n).

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Poll 1,000 voters and 52% support a measure; poll another 1,000 and you might get 49%. The sample proportion $\hat{p}$ is itself a random variable. Its distribution over all possible samples, the sampling distribution, follows from the binomial distribution and is close to a normal curve. You saw this kind of variation in simulations.

Key idea

For random samples of size $n$ from a population with proportion $p$, the sample proportion $\hat{p}$ has $$\mu_{\hat{p}} = p, \qquad \sigma_{\hat{p}} = \sqrt{\dfrac{p(1 - p)}{n}}.$$ When $np \ge 10$ and $n(1 - p) \ge 10$, the sampling distribution of $\hat{p}$ is approximately normal.

Why it works

The number of successes $X$ in the sample is binomial (when the population is large, or at least 10 times the sample), with mean $np$ and SD $\sqrt{np(1 - p)}$. Since $\hat{p} = X/n$, divide both by $n$: the mean becomes $p$ and the SD becomes $\sqrt{np(1 - p)}/n = \sqrt{p(1 - p)/n}$. With enough expected successes and failures, the binomial bars fill in a bell shape, so a normal curve fits.
Each sample gives one proportion of heads. Take 1,000 samples, then move n: the proportions always center on 0.5, and their spread shrinks like 1/√n.

Worked example: Center and spread

40% of a city's adults have a library card. For random samples of 150 adults, describe the sampling distribution of $\hat{p}$.
  1. Center: $\mu_{\hat{p}} = 0.4$.
  2. Spread: $$\sigma_{\hat{p}} = \sqrt{\dfrac{0.4 \cdot 0.6}{150}} = \sqrt{0.0016} = 0.04$$
  3. Shape: $np = 60$ and $n(1 - p) = 90$ are both at least 10, so approximately normal.

Answer: About normal, mean 0.4, SD 0.04.

Worked example: A probability

In that setting, how likely is a sample with $\hat{p} \ge 0.48$? ($P(Z < 2.00) = 0.9772$.)
  1. $z = \dfrac{0.48 - 0.4}{0.04} = 2$.
  2. $$P(\hat{p} \ge 0.48) = P(Z \ge 2) = 1 - 0.9772 = 0.0228$$

Answer: About $0.023$: a sample this high would be unusual.

Worked example: A bigger sample

What happens to $\sigma_{\hat{p}}$ with samples of 600 instead of 150?
  1. n is 4 times as large, so $\sigma_{\hat{p}}$ is divided by $\sqrt{4} = 2$.
  2. $\sqrt{0.24/600} = \sqrt{0.0004} = 0.02$.

Answer: It halves, to 0.02.

Common mistake

Confusing the distribution of one sample's data with the distribution of the statistic. One sample's data are just yeses and nos (0s and 1s); they are not bell-shaped at all. The bell shape describes $\hat{p}$ across many different samples. Check: is the question about individuals, or about a statistic computed from a whole sample?

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.

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Sources and further reading

  • OpenStax, Introductory Statistics 2e, section 4.3 Binomial Distribution
  • OpenStax, Introductory Statistics 2e, section 8.3 A Population Proportion
  • Common Core State Standards for Mathematics, S-IC.A.1 (Understand statistics as a process for making inferences about population parameters based on a random sample)
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