Score every person in your sample 1 for yes and 0 for no, and the sample proportion p̂ is the sample mean of those 0s and 1s. So every result from this chapter carries straight over: p̂ is centered on the true p, its spread is √(p(1−p)/n), and its shape drifts toward normal as n grows. Two things are genuinely new. The spread is locked to the center — there is no separate σ to look up — and p̂ can only land on the n+1 rungs 0/n, 1/n, …, n/n. Move p, move n, and draw real samples.
Watch the p̂'s pile up
exact distribution of p̂your simulated p̂'snormal modeltrue p±1 SE
One sample, up close
This is where a p̂ comes from: n independent yes/no answers, counted. Every dot is one person.
Press Draw 1 sample to take a sample.
Which p is hardest to pin down?
Fair coinp = 0.50 · n = 40 · press “Draw 200 samples” to begin
A situation to sample from
Set it yourself
0.50
40
True p0.50
Sample size n40
Expected np20.0
Expected n(1−p)20.0
Where the p̂'s actually landed
Top number: measured from the p̂'s you piled up. Bottom line: the exact answer from the binomial, and what the normal model would have promised.
Mean of the p̂'s—p = 0.500
SD of the p̂'s—SE = 0.0791
Within ±1 SE——
Within ±2 SE——
How many people would you need?
The normal model says p̂ lands within 1.96 SE of p in about 95% of samples. Name how close you want to get and solve 1.96√(p(1−p)/n) = m for n, giving n = 1.96² p(1−p) / m².
0.030
A proportion is a mean of 0s and 1s. Write 1 for every yes and 0 for every no. The mean of that column of 0s and 1s is the fraction of yeses — it is p̂. So p̂ inherits everything: it is centered on p, it wobbles by σ/√n, and it goes bell-shaped as n grows. You never have to look σ up, because for a 0/1 variable the population SD is forced to be √(p(1−p)) — the center dictates the spread. Divide by √n and you have the standard error of a proportion, all of it derived from a single number p.
Four ways this goes wrong.(1) A big n is not enough by itself. The normal model needs np ≥ 10 and n(1−p) ≥ 10 — roughly ten expected yeses and ten expected noes. With p = 0.02 and n = 200 you have n = 200 but only 4 expected yeses, and the distribution of p̂ is visibly right-skewed. Set that scenario and look. (2) p̂ is discrete. It can only be one of n+1 values, so with n = 5 the “distribution” is six spikes and coverage percentages jump around; no smooth curve is even possible. (3) Population size is irrelevant. There is no N in √(p(1−p)/n) — 1,000 people pins down a town and a nation equally well, provided the sample is random. (The one caveat: this assumes independent draws, which holds if you sample with replacement or take no more than about 10% of a finite population.) (4) None of this survives bias. If only volunteers answer, p̂ is centered on the wrong p, and raising n just tightens the pile around that wrong number. Everything on this page is a statement about p̂ wandering around a fixed p — not about p wandering.