11.1: z-Test for Comparing Proportions

In Section 9.4 we covered a one-sample test for proportions using the \(z\) approximation to the binomial distribution. Here we want to compare a proportion \(p_{1}\) in one population with \(p_{2}\) in another population, a two-sample test for proportions, also using the \(z\) approximation to the binomial distribution. Define

\[\hat{p}_{1} = \frac{x_{1}}{n_{1}} \mbox{\ \ \ \ and\ \ \ \ } \hat{p}_{2} = \frac{x_{2}}{n_{2}}\]

where \(x_{1}\) and \(x_{2}\) are the number of items of interest in the samples from the two populations and \(n_{1}\) and \(n_{2}\) are their sample sizes. Also define the corresponding \(q_{1} = 1 - p_{1}\), \(q_{2} = 1 - p_{2}\), \(\hat{q}_{1} = 1 - \hat{p}_{1}\) and \(\hat{q}_{2} = 1 -\hat{ }p_{2}\). The hypotheses we want to test is

\[ H_{0}: p_{1} = p_{2} \hspace*{3em} H_{1}: p_{1} \neq p_{2} \]

which is equivalent to

\[ H_{0}: p_{1} - p_{2} = 0 \hspace*{3em} H_{1}: p_{1} - p_{2} \neq 0 \]

If \(n_{1}p_{1}, n_{1}q_{1}, n_{2}p_{2}\), and \(n_{2}q_{2}\) are all \(> 5\) then the appropriate normal distribution will provide a good approximation to the relevant binomial distribution and we can use the following test statistic to test the hypotheses

\[ z_{\rm test} = \frac{\hat{p}_{1} - \hat{p}_{2}}{\sqrt{\bar{p}\bar{q} \left( \frac{1}{n_{1}} + \frac{1}{n_{2}} \right)}} \]

where

\[ \bar{p} = \frac{x_{1}+x_{2}}{n_{1}+n_{2}} \hspace*{3em} \bar{q} = 1 -\bar{p} \]

are the proportions of items of interest and not of interest in the two samples combined.

Example 11.1 : In a nursing home study we are interested in the proportions of nursing homes that have vaccination rates of less than 80\(\%\). The two populations we want to compare are small nursing homes and large nursing homes. In a sample of 34 small nursing homes, 12 were found to have a vaccination rate of less than 80\(\%\). In a sample of 24 large nursing homes, 17 were found to have a vaccination rate of less than 80\(\%\). At \(\alpha = 0.05\) is there a difference in the proportions of small and large nursing homes with vaccination rates of less than 80\(\%\)?

Solution :

0. Data reduction.

First define: population 1 = small nursing homes and population 2 = large nursing homes. Then compute the proportions:

\[ \hat{p}_{1} = \frac{x_{1}}{n_{1}} = \frac{12}{34} = 0.35 \hspace{4em} \hat{p}_{2} = \frac{x_{2}}{n_{2}} = \frac{17}{24} = 0.71 \] \[ \bar{p} = \frac{x_{1} + x_{2}}{n_{1} + n_{2}} = \frac{12+17}{34+24} = \frac{29}{58} = 0.5 \hspace{4em} \bar{q} = 1 - \bar{p} = 1 - 0.5 = 0.5 \]

1. Hypotheses.

\[H_{0}: p_{1} = p_{2} \hspace*{3em} H_{1}: p_{1} \neq p_{2}\]

2. Critical statistic.

Use Table F, the last (\(z\)) line in the column for a two-tailed test at \(\alpha = 0.05\): \(z_{\rm crit} = \pm 1.96\)

3. Test statistic.

\[\begin{eqnarray*} z_{\rm test} & = & \frac{\hat{p}_{1} - \hat{p}_{2}}{\sqrt{\bar{p}\bar{q} \left( \frac{1}{n_{1}} + \frac{1}{n_{2}} \right)}} \\ z_{\rm test} & = & \frac{0.35 - 0.71}{\sqrt{(0.5)(0.5) \left( \frac{1}{34} + \frac{1}{24} \right)}} \\ z_{\rm test} & = & \frac{-0.36}{0.1333} \\ z_{\rm test} & = & -2.7 \end{eqnarray*}\]

4. Decision.

Reject \(H_{0}\).

5. Interpretation.

There is enough evidence, from a \(z\) proportions test at \(\alpha = 0.05\) to support the observation that large nursing homes have worse vaccination rates than small nursing homes. Make sure your parents end up in a small nursing home. (Note that rejection of \(H_{0}\) in a one-tail test allows us to believe the direction of difference given by the sample data.)

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