The sampling distribution of p̂

A proportion is a mean

Code each sampled invoice as 1 if it is paid late and 0 if not. The sample proportion late,

$$\hat{p}=\frac{\text{number late}}{n}=\frac{X}{n}$$

is the mean of those 0/1 values — so everything Module 8 has said about $\bar{X}$ applies to $\hat{p}$, once the population mean and variance of a 0/1 variable are known. A single 0/1 value with $P(1)=p$ has mean $p$ and variance $p(1-p)$ (Module 6's table method on two rows). Substituting into L8.2's results:

$$E(\hat{p})=p,\mathrm{Var}(\hat{p})=\frac{p(1-p)}{n},\mathrm{SD}(\hat{p})=\sqrt{\frac{p(1-p)}{n}}$$

$\hat{p}$ is unbiased for $p$, and its standard error shrinks with $\sqrt{n}$ like every other mean. Equivalently, $X$ is binomial, $\hat{p}=X/n$ is the binomial divided by $n$, and L6.3's scaling rule divides $np$ by $n$ and $np(1-p)$ by $n^2$ — the same two formulas.

When it is normal

A 0/1 population is as far from normal as a population can be, so L8.3 never applies and the shape comes from the CLT — with the binomial's own condition standing in for "$n>30$": $\hat{p}$ is approximately normal when

$$np\ge 5\text{and}n(1-p)\ge 5$$

— the conditions of L7.5, because this is the normal approximation to the binomial, rescaled. No continuity correction is applied to $\hat{p}$ in this course: it is treated as a mean, not a count.

Worked example — late payers at Northfield

Across Northfield Credit's customer base, 20 % of invoices are paid late: $p=0.20$. A quality reviewer samples $n=400$ invoices. Then $np=80$ and $n(1-p)=320$, both far above 5, so

$$\hat{p}\approx N\left(0.20,\frac{0.20\times 0.80}{400}\right),\mathrm{SE}(\hat{p})=\sqrt{\frac{0.16}{400}}=\sqrt{0.0004}=0.02$$

Probability the sample shows more than 23 % late.

$$z=\frac{0.23-0.20}{0.02}=1.50,P(\hat{p}>0.23)=1-\Phi (1.50)=1-0.9332=0.0668$$

About one sample in fifteen would show 23 % or more late even though the true rate is 20 %. By symmetry, $P(\hat{p}<0.17)=0.0668$ as well; the middle 95 % of sample proportions lie within $1.96\times 0.02=\pm 0.039$ of 0.20 — Module 12's confidence interval, seen from the population's side.

Which $p$ goes in the SE

The SE formula uses the population proportion $p$. In this module $p$ is given, and the calculation is exact. In Module 12, $p$ is unknown — the whole point of the exercise — and the sample's $\hat{p}$ stands in for it in a confidence interval, while the hypothesised $p_0$ is used in a test. Which one goes under the root is a mark on every proportion question; here, with $p$ known, it is 0.20.

Notation

IBS1 writes the sample proportion $p^{\prime }$; the course, the handouts and the paper write $\hat{p}$ ("p-hat"). They are the same quantity. Read $p^{\prime }$ in the textbook as $\hat{p}$ and write $\hat{p}$ on the paper.