# Tests Concerning a Population Proportion

Posted by Beetle B. on Tue 18 July 2017


Let $$X$$ be the number of successes.

If $$n<<N$$, then $$X$$ is approximately binomial (i.e. approximating as if we are sampling with replacement). If $$n$$ is large, $$X$$ and $$\hat{p}=X/n$$ are approximately normal.

## Large Sample Tests

If $$np_{0}\ge10$$ and $$n(1-p_{0})\ge10$$:

$$H_{0}:p=p_{0}$$

Let the test statistic be:

\begin{equation*} z=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}} \end{equation*}
• $$H_{a}:p>p_{0}\implies z\ge z_{\alpha}$$
• $$H_{a}:p<p_{0}\implies z\le-z_{\alpha}$$
• $$H_{a}:p\ne p_{0}\implies z\ge z_{\alpha/2}$$ or $$z\le-z_{\alpha/2}$$

### $$\beta$$ and Sample Size Determination

When $$H_{0}$$ is true, $$Z$$ above is approximately standard normal. But what if $$p=p'\ne p_{0}$$? $$Z$$ is still approximately normal (linear in $$\hat{p}$$). But the mean is not 0 and $$\sigma\ne 1$$:

\begin{equation*} E(Z)=\frac{p'-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}} \end{equation*}
\begin{equation*} V(Z)=\frac{p'(1-p')/n}{p_{0}(1-p_{0})/n} \end{equation*}

Let:

\begin{equation*} A=\frac{p_{0}-p'}{\sqrt{\frac{p'(1-p')}{n}}},B_{\alpha}=\frac{z_{\alpha}\sqrt{p_{0}(1-p_{0})/n}}{\sqrt{p'(1-p')/n}} \end{equation*}

Then:

• $$H_{a}:p>p_{0},\beta(p')=\Phi(A+B_{\alpha})$$
• $$H_{a}:p<p_{0},\beta(p')=1-\Phi(A-B_{\alpha})$$
• $$H_{a}:p\ne p_{0},\beta(p')=\Phi(A+B_{\alpha/2})-\Phi(A-B_{\alpha/2})$$

The sample size $$n$$ for $$\beta(p')=\beta$$ (one-tailed):

\begin{equation*} n=\left[\frac{1}{p'-p_{0}}\left(z_{\alpha}\sqrt{p_{0}(1-p_{0})}+z_{\beta}\sqrt{p'(1-p')}\right)\right]^{2} \end{equation*}

For two tailed, replace $$\alpha$$ with $$\frac{1}{2}\alpha$$

## Small Sample Tests

When $$n$$ is small, use the Binomial Distribution directly. If $$H_{0}$$ is true, then $$p=p_{0}$$ and the pmf is $$\textrm{Bin}(n,p_{0})$$ The Type I error can be calculated directly.

If $$p=p'>p_{0}$$, then the pmf=$$\textrm{Bin}(n,p')$$ You can calculate the Type II error directly.