11. Comparing Proportions

11.2 Confidence Interval for the Difference between Two Proportions

The form of the confidence interval is

    \[ (\hat{p}_{1} - \hat{p}_{2}) - E < (p_{1} - p_{2}) < (\hat{p}_{1} - \hat{p}_{2}) + E \]

with

    \[ E = z_{\cal{C}} \sqrt{\frac{\hat{p}_{1}\hat{q}_{1}}{n_{1}} + \frac{\hat{p}_{2}\hat{q}_{2}}{n_{2}} } \]

where, as usual you can get z_{\cal{C}} from the last line of the t Distribution Table.

Example 11.2 : Using the data from Example 11.1, find the 95\% confidence interval for p_{1} - p_{2}.

Solution : The relevant numbers from Example 11.1 are: n_{1} = 34, \hat{p}_{1} = 0.35, \hat{q}_{1} = 1 - 0.35 = 0.65 and n_{2} = 24, \hat{p}_{2} = 0.71, \hat{q}_{1} = 1 - 0.71 = 0.29.

Compute (after finding z_{95\%} = 1.96 from the t Distribution Table)

    \begin{eqnarray*} E & = & z_{95\%} \sqrt{\frac{\hat{p}_{1}\hat{q}_{1}}{n_{1}} + \frac{\hat{p}_{2}\hat{q}_{2}}{n_{2}} }\\ E & = & 1.96 \sqrt{\frac{(0.35)(0.65)}{34} + \frac{(0.71)(0.29)}{24} }\\ E & = & 0.242 \end{eqnarray*}

and

    \[ \hat{p}_{1} - \hat{p}_{2} = 0.35 - 0.71 = -0.36 \]

So

    \begin{eqnarray*} (\hat{p}_{1} - \hat{p}_{2}) - E & < (p_{1} - p_{2}) < & (\hat{p}_{1} - \hat{p}_{2}) + E \\ -0.36 - 0.242 & < (p_{1} - p_{2}) < & -0.36 + 0.242 \\ -0.602 & < (p_{1} - p_{2}) < & -0.118 \end{eqnarray*}

with 95\% confidence. (Note that this corresponds with the rejection of H_{0} in Example 11.1 since 0 is not in the confidence interval.)

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Introduction to Applied Statistics for Psychology Students Copyright © 2022 by Gordon E. Sarty is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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