149 Estimate the Difference between Population Proportions (2 of 3)
149 Estimate the Difference between Population Proportions (2 of 3)
Learning Objectives
- Construct a confidence interval to estimate the difference between two population proportions (or the size of a treatment effect) when conditions are met. Interpret the confidence interval in context.
- Interpret the meaning of a confidence level associated with a confidence interval and describe how the confidence level affects the margin of error.
- Given the description of a statistical study, evaluate whether conclusions are reasonable.
Confidence Interval for a Difference in Two Population Proportions: Beyond the Basics
For all confidence intervals, the margin of error is based on the standard error. We know from “Distributions of Differences in Sample Proportions” that the standard error for the sampling distribution of differences in sample proportions is:
[latex]\sqrt{\frac{{p}_{1}(1-{p}_{1})}{{n}_{1}}+\frac{{p}_{2}(1-{p}_{2})}{{n}_{2}}}[/latex]
Obviously, if we are trying to estimate the difference in population proportions, we will not know [latex]{p}_{1}[/latex] or [latex]{p}_{2}[/latex]. So we estimate these population proportions with our sample proportions. This is the same approach we used when had to estimate the standard error for the distribution of sample proportions in Inference for One Proportion. The estimated standard error becomes
[latex]\sqrt{\frac{{\stackrel{ˆ}{p}}_{1}(1-{\stackrel{ˆ}{p}}_{1})}{{n}_{1}}+\frac{{\stackrel{ˆ}{p}}_{2}(1-{\stackrel{ˆ}{p}}_{2})}{{n}_{2}}}[/latex]
This formula estimates the average error between a difference in sample proportions and the true difference in population proportions.
So a 95% confidence interval has the following formula:
[latex](\mathrm{difference}\text{}\mathrm{in}\text{}\mathrm{sample}\text{}\mathrm{proportions})\text{}±\text{}2(\mathrm{standard}\text{}\mathrm{error})[/latex]
[latex]({\stackrel{ˆ}{p}}_{1}-{\stackrel{ˆ}{p}}_{2})\text{}±\text{}2\sqrt{\frac{{\stackrel{ˆ}{p}}_{1}(1-{\stackrel{ˆ}{p}}_{1})}{{n}_{1}}+\frac{{\stackrel{ˆ}{p}}_{2}(1-{\stackrel{ˆ}{p}}_{2})}{{n}_{2}}}[/latex]
We can use this formula only if a normal model is a good fit for the sampling distribution. Recall that this is true only if the expected number of successes and failures in each sample is at least 10. For those who like formulas, these conditions translate into the following inequalities.
[latex]{n}_{1}{p}_{1}≥10\text{ }{n}_{1}(1-{p}_{1})≥10\text{ }{n}_{2}{p}_{2}≥10\text{ }{n}_{2}(1-{p}_{2})≥10[/latex]
We have to adjust these conditions because we do not know the population proportions [latex]{p}_{1}[/latex] and [latex]{p}_{2}[/latex]. We make the same adjustment we made in Inference for One Proportion. We require that the actual number of successes and failures in each sample is at least 10. For those who like formulas, these conditions translate into replacing [latex]{p}_{1}[/latex] and [latex]{p}_{2}[/latex] with the corresponding sample proportions. Luckily, this tweak works and the normal distribution still