The purpose of this activity is to give you hands-on practice in following up a test for the population proportion p in which H_o has been rejected with a confidence interval, and getting a sense of how the confidence interval is a natural and informative supplement to the test in these cases.

Background:

Recall from a previous activity the results of a study on the safety of airplane drinking water that was conducted by the U.S. Environmental Protection Agency (EPA). A study found that out of a random sample of 316 airplanes tested, 40 had coliform bacteria in the drinking water drawn from restrooms and kitchens. As a benchmark comparison, in 2003 the EPA found that about 3.5% of the U.S. population have coliform bacteria-infected drinking water. The question of interest is whether, based on the results of this study, we can conclude that drinking water on airplanes is more contaminated than drinking water in general. Let p be the proportion of contaminated drinking water on airplanes.

In a previous activity we tested H_o: p = .035 vs. H_a: p > .035 and found that the data provided extremely strong evidence to reject H_o and conclude that the proportion of contaminated drinking water in airplanes is larger than the proportion of contaminated drinking water in general (which is .035).

Now that we've concluded that, all we know about p is that we have very strong evidence that it is higher than .035. However, we have no sense of its magnitude. It will make sense to follow up the test by estimating p with a 95% confidence interval.

Based on the data, find a 95% confidence interval for p and interpret it in context. Recall that the formula for that is:

 $\displaystyle \hat{{p}}\pm{2}\sqrt{{\frac{{\hat{{p}}{\left({1}-\hat{{p}}\right)}}}{{n}}}}$