One-Sample Z-Test for Proportion

191. Two-Sample Z-Test for Proportion

Example

Your friend is convinced studying doesn’t help. You believe it does, so you take 2 simple random samples of your Stat course: one for students who studied > 4 hrs and one for those who studied < 4 hrs. Of the 120 students who studied > 4 hrs, 103 passed. Of the 80 who studied less, 57 passed. At a = 0.05 test your claim.

\begin{aligned} 𝐻_{0} & : 𝑝_{> 4} - 𝑝_{< 4} \leq 0 \\ 𝐻_{𝑎} & : 𝑝_{> 4} - 𝑝_{< 4} > 0 \end{aligned}

𝑧 = \frac{{\hat{𝑝}}_{> 4} - {\hat{𝑝}}_{< 4}}{\sqrt{{\hat{𝑝}}_{𝑐} (1 - {\hat{𝑝}}_{𝑐}) (\frac{1}{𝑛_{> 4}} + \frac{1}{𝑛_{< 4}})}}

Where:

${\hat{𝑝}}_{> 4} = \frac{103}{120} = 0.8583$
${\hat{𝑝}}_{< 4} = \frac{57}{80} = 0.7125$
$𝑛_{> 4} = 120$
$𝑛_{< 4} = 80$
${\hat{𝑝}}_{𝑐} = \frac{𝑥_{> 4} + 𝑥_{< 4}}{𝑛_{> 4} + 𝑛_{< 4}} = \frac{103 + 57}{120 + 80} = 0.8$

𝑧 = \frac{0.8583 - 0.7125}{\sqrt{0.8 (1 - 0.8) (\frac{1}{120} + \frac{1}{80})}} = 2.5253

{\hat{𝑝}}_{> 4} - {\hat{𝑝}}_{< 4} = 0.8583 - 0.7125 = 0.1458

𝑃 (𝑧 \geq 2.5253) = 0.0058

\begin{aligned} 𝑧 -value & > critical-value \\ 2.5253 & > 1.6449 \end{aligned}

\begin{aligned} 𝑝 -value & < 𝛼 -value \\ 0.0058 & < 0.05 \end{aligned}

Reject $𝐻_{0}$ . At $𝛼 = 0.05$ there is sufficient evidence that students who studied more than 4 hours passed at a higher rate.