Sample proportions and the normal approximation

Populations and samples, the sample proportion p-hat, its mean p and standard deviation √(p(1 − p)/n), the central limit theorem, the normal approximation for large samples, and the effect of sample size.

The sample proportion

Suppose a proportion $p$ of a population has some property, such as owning a pet. In a random sample of size $n$, let $X$ be the number with the property. The sample proportion is $\hat p = \dfrac{X}{n}$. Different samples give different values of $\hat p$, so $\hat p$ is a random variable.

Since $X \sim \text{Bin}(n, p)$, dividing its mean and variance by $n$ and $n^2$ gives:

Distribution of the sample proportion

$$E(\hat p) = p, \qquad \sigma_{\hat p} = \sqrt{\frac{p(1 - p)}{n}}$$

The normal approximation

How sample size changes the spread

The approximate distribution of the sample proportion. Increasing n makes it narrower: four times the sample size halves the standard deviation.

Try this

What happens to $\sigma_{\hat p}$ when $n$ goes from $100$ to $400$?

Example 1●●●

$40\%$ of a town's voters support a policy. Samples of $150$ voters are taken. Estimate the probability that a sample shows more than $48\%$ support.

$\sigma_{\hat p} = \sqrt{\dfrac{0.4 \times 0.6}{150}} = \sqrt{0.0016} = 0.04$.

$0.48 = 0.4 + 2 \times 0.04$, two standard deviations above the mean.

By the empirical rule, about $95\%$ lies within two standard deviations, so $P(\hat p \gt 0.48) \approx 0.025$.

Example 2: using z-scores●●●

For $p = 0.5$ and $n = 100$, find $P(\hat p \lt 0.44)$ using $P(Z \lt -1.2) \approx 0.1151$.

$\sigma_{\hat p} = \sqrt{\dfrac{0.25}{100}} = 0.05$, so $z = \dfrac{0.44 - 0.5}{0.05} = -1.2$.

$P(\hat p \lt 0.44) \approx P(Z \lt -1.2) \approx 0.1151$.

Flashcards

Sample proportion

$\hat p = \dfrac{X}{n}$

$E(hat p)$

$p$

$sigma_{hat p}$

$\sqrt{\dfrac{p(1 - p)}{n}}$

Halving $sigma_{hat p}$

Multiply the sample size by $4$.

Practice

$p = 0.5$ and $n = 100$. Find $\sigma_{\hat p}$.
$p = 0.3$. Find $E(\hat p)$.
In a survey, $25$ of $200$ people said yes. Find $\hat p$.
$p = 0.4$ and $n = 150$. Use the empirical rule to estimate $P(\hat p \gt 0.48)$.
$p = 0.2$ and $n = 100$. Use the empirical rule to estimate $P(0.16 \lt \hat p \lt 0.24)$.
$p = 0.1$ and $n = 900$. Use the empirical rule to estimate $P(0.08 \lt \hat p \lt 0.12)$.
By what factor must the sample size be multiplied to halve $\sigma_{\hat p}$?
As the sample size increases, the distribution of $\hat p$: