Z-test for Population Proportion

Population Proportion

Population Proportion (p): Describes a percentage value of a group in a population.

\boxed{ \text{$\hat p$: } \begin{cases} \hat p &= \frac{X}{n} \\ \hat p &\sim N( p, \sqrt{ \frac{p(1-p)}{n} } ) \end{cases} } \\ \small\textit{where $X$ is \# of outcomes w/ a characteristic} \\ \small\textit{dist. only true if $np$ and $n(1-p) \ge 5$} \\

Hypothesis Testing for p

  1. Define population parameter (p), hypotheses, and \alpha
  2. Check if np_0 \ge 5 and n(1-p_0) \ge 5
  3. Compute \hat p
  4. Calculate test statistic

\boxed{ Z = \frac{ \hat p - p_0 }{ \sqrt{ \frac{ p_0 (1-p_0) }{ n } } } \sim N(0,1) } \\ \small\textit{$p = p_0$ under $H_0$}

\boxed{ \text{Reject $H_0$ if: } \begin{cases} H_1 : p \ne p_0 &\quad |Z| > z_{\alpha / 2 } &\quad 2P(z>|Z|) < \alpha \\ H_1 : p > p_0 &\quad Z > Z_{\alpha} &\quad P(z > Z) < \alpha \\ H_1 : p < p_0 &\quad Z < -Z_{\alpha} &\quad P(z < Z) < \alpha \end{cases} } \\ \small\textit{(you can use Z or P-value to decide)}

Example

Q: In a survey of 800 adults, 500 said that their favorite sport is football. Does this sample provide sufficient evidence to conclude that the population proportion of adults who favorite sport is football is greater than 0.6? (\alpha = 0.1)

A:

Let p: pop. proportion of adults whose favorite sport is football.

H_0: p \le 0.6 and H_1: p > 0.6.

Check distribution:

Estimating p:

\hat p = 500 / 800 = 0.625

Calculate test statistic and critical value:

Z = \frac{ \hat p - p_0 }{ \sqrt{ \frac{ p_0 (1-p_0) }{ n } } } = \frac{ 0.625 - 0.6 }{ \sqrt{ \frac{ 0.6 (1-0.6) }{ 800 } } } = 1.44

Z_{\alpha} = Z_{0.1} = 1.28

  1. Z > Z_\alpha, so reject H_0 at \alpha = 0.1, or
  2. \text{P-value} = P(Z > 1.44) = 0.0749 \to \text{P-value} < \alpha, so reject H_0

\boxed{ \text{C.I. for $p$: } \hat p \pm Z_{\alpha / 2} \sqrt{ \frac{ \hat p (1 - \hat p) }{ n } } } \\ \small\textit{doesn't guarantee (0,1)}