A sample of 469 teenage births at a large city hospital is observed for the number of births resulting in twins. The 95% confidence interval for the proportion of all births among teenage girls resulting in twins was estimated to be (0.006, 0.026). What is the margin of error for this interval?
The confidence interval can be written as: \(x \pm m\) where m is margin of error, x is mid-point \(x = \frac{.006+.026}{2} = 0.016\) \(m = .026 - .016 = 0.01\)
