Sample Proportion

sample proportion ( $\hat{p}$ )
- estimates the population proportion $p$
  - proportion of the population that would do x
- single point estimate of $p$
  - don't know how close to $p$ $\hat{p}$ is, as p is unknown
  - we use mathematical arguments based on the Sampling Distributions of $\hat{p}$ to make statements
    - like margin of error, range
    - 0.43 accurate within 0.03 19 times out of 20
      - (0.40, 0.46) = 95% confidence interval (CI) for $p$
for large sample sizes, $\hat{p}$ distribution is approximately normal
- normal approximation is best when n is large and $p = 0.5$
- reasonable with
  - $n \hat{p} \geq 15 and n (1 - \hat{p}) \geq 15$
  - n large, p cant be close to 0 or 1
$\hat{p}$ estimates $p$
- binomial random variable $\hat{p} = \frac{X}{n}$
  - $E (\hat{p}) = p, V a r (\hat{p}) = \frac{p (1 - p)}{n}$
  - $S E (\hat{p}) = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}$ as we can't use statistic $p$
- number of individual in the sample with the characteristic of interest / total number of individuals in the sample
- for large sample sizes $\hat{p}$ is approximately normal

Suppose $X \sim B i n (n, p)$ . Let $Y = \frac{X}{n}$ represent the sample proportion of successes

Mean$$\begin{align}
E(Y)= E\left( \frac{X}{n} \right) & =\frac{1}{n}E(X)=\frac{1}{n}np=p
\end

V a r i a n c e $ $ \begin{aligned} V a r (Y) = V a r (\frac{X}{n}) & = \frac{1}{n^{2}} V a r (X) \\ = \frac{1}{n^{2}} n p (1 - p) \\ = \frac{p (1 - p)}{n} = \frac{p q}{n} \end{aligned}