Example

A random sample of 500 people in a certain county which is about to have a national election were asked whether they preferred “Candidate A” or “Candidate B”.

From this sample, 320 people responded that they preferred Candidate A.

A random sample of 400 people in a second county which is about to have a national election were asked whether they preferred “Candidate A” or “Candidate B”.

From this second county sample, 268 people responded that they preferred Candidate A.

\[\widehat{p}_1 = \frac{320}{500} = 0.64\\ \quad\\ \widehat{p}_2 = \frac{268}{500} = 0.67\\\]

We want to test that

\[H_0:p_1=p_2\qquad H_1:p_1\neq p_2\]

We can change to

\[H_0:p_1-p_2=0\\ H_1:p_1-p_2\neq0\]

For large enough samples,

\[\widehat{p}_1\stackrel{approx}{\sim}N\bigg(p_1,\frac{p_1(1-p_1)}{n_1}\bigg)\]

and

\[\widehat{p}_2\stackrel{approx}{\sim}N\bigg(p_2,\frac{p_2(1-p_2)}{n_2}\bigg)\]

We know that $\widehat{p}_1-\widehat{p}_2$ is normally distributed because it’s a linear combination of normals.

\[\widehat{p}_1-\widehat{p}_2\sim N(?,?)\]

  • Mean:

    \[E[\widehat{p}_1-\widehat{p}_2]=E[\widehat{p}_1]-E[\widehat{p}_2]=p_1-p_2\]

  • Variance:

    \[\begin{align} Var[\widehat{p}_1-\widehat{p}_2]&\stackrel{indep}{=}Var[\widehat{p}_1]-Var[\widehat{p}_2]\\ &=\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} \end{align}\]

We have

\[\frac{\widehat{p}_1-\widehat{p}_2-(p_1-p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}\sim N(0,1)\\\]

Use estimators for $p_1$ and $p_2$ assuming they are the same.

  • Call the common value $p$.
  • Estimate by putting both groups together.

In the example with

\[\widehat{p}_1 = \frac{320}{500} = 0.64\quad\widehat{p}_2 = \frac{268}{500} = 0.67\\\]

We have

\[\widehat{p}=\frac{320+268}{500+400}=\frac{588}{900}=\frac{49}{75}\\ \approx0.6533\]

Use the fact that:

\[\begin{align} Z:&=\frac{\widehat{p}_1-\widehat{p}_2-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}\stackrel{approx}{\sim} N(0,1)\\ &=\frac{\widehat{p}_1-\widehat{p}_2-(p_1-p_2)}{\sqrt{\hat{p}(1-\hat{p})\frac{1}{n_1}+\frac{1}{n_2}}} \end{align}\]

This is two-tailed test with z-critical values.

\[\begin{align} Z&=\frac{0.64-0.67-0}{\sqrt{0.6533(1-0.6533)\bigg(\frac{1}{500}+\frac{1}{500}\bigg)}}\\ &\approx-0.9397 \end{align}\]

Let level of significance $\alpha=0.5$.

png

$Z=-0.9397$ does not fall in the rejection region!