Confidence Interval for the Difference Between Two Proportions

Has retention rate at WMU been changing? Suppose that a random sample of 200 entering students in 1989 showed 74% were still enrolled 3 years later. Another random sample of 200 entering students in 1999 showed that 66% were still enrolled 3 years later. This constitutes an 8% change in 3-year retention rate. However, the 8% difference is based on random sampling, and is only an estimate of the true difference. What is the likely size of the error of estimation?

The calculation of the standard error for the difference in proportions parallels the calculation for a difference in means .

$$\mbox{SE of} \;\; (\hat{p_1}-\hat{p_2}) = \sqrt{ \mbox{SE}_1^2 + \mbox{SE}_2^2}$$ (7.5)

where $\mbox{SE}_1$ and $\mbox{SE}_2$ are the SE's of $\hat{p_1}$ and $\hat{p_2}$, respectively. For the retention rates, let $\hat{p_1}=.74$ with standard error $\mbox{SE}_1=\sqrt{(.74)(1-.74)}/\sqrt{200}=.031$ and $\hat{p_2}=.66$ with standard error $\mbox{SE}_2=\sqrt{(.66)(1-.66)}/\sqrt{200}=.033$. Then the difference .74-.66=.08 will have standard error $\sqrt{.031^2 + .033^2}=.045.$

We now state a confidence interval for the difference between two proportions.

  

The SE for the .08 change in retention rates is .045, so the .08 estimate is likely to be off by some amount close to .045. However, the 95% margin of error is approximately 2 SE's, or .090. A 95% confidence interval for the difference in proportions p₁-p₂ is $.08 \pm .09$ or $(-.01, \;.17)$.  

Coverting to percentages, the difference between retention rates for 1989 and 1999 is 8% with a 95% margin of error of 9%. A 95% confidence interval for the true difference is $(-1\%,\; 17\%)$.