e-Statistics

Testing a Proportion

Here we want to compare the population proportion $ p$ with a particular value $ p_0$. Then our hypothesis testing problem becomes

$ H_A:\hspace{0.05in}p$ $ p_0 =$

For example, in order for vaccine to be approved for widespread use, it must be established that the probability $ p$ of serious adverse reaction must be less than $ p_0$. In this case we should set `` $ H_0:\: p = p_0$ versus $ H_A:\: p < p_0$'' to see whether we can reject $ H_0$ in favor of $ H_A$.

Let $ X =$ be the frequency of the specified type in categorical data of size $ n =$ . In the above vaccine example $ X$ will be the number of participants who suffered adverse reaction among $ n$ participants. Then the test statistic is calculated as

$ Z = \displaystyle\frac{X - np_0}{\sqrt{np_0(1-p_0)}}
\left(= \displaystyle\frac{(X/n) - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\right) =$ .

The appropriateness of the test is established by adequately large sample size $ n$, ensured by $ np_0 > 5$ and $ n(1-p_0) > 5$. Provided that the null hypothesis $ H_0$ is true, the test statistic $ Z$ has approximately the standard normal distribution. Then we can proceed to construct the p-value = , and reject $ H_0$ when the p-value is less than $ \alpha$ of your choice.

When $ H_0$ is rejected, we want to further investigate the confidence interval for the population proportion $ p$. We have the point estimate of $ p$ as $ \hat{p} = X/n$, and obtain the $ (1-\alpha)$-level confidence interval

$ \displaystyle
\hat{p} \pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} =$ ( , )

can be calculated for the population proportion.