Question

I am given a sample size of 469, 95% confidence interval of 0.006, 0.026. How do I find the sample proportion? I'd appreciate even just a formula.

Answer 1

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 sample
proportion of teenage births at this hospital resulting in twins?

Answer 2

Work backword to get p. You know the confidence interval is .0026 to .006, and if you know how the confidence interval is created, you know that it is p +/ z (s/sqrt(n)). (Or z may be t if you teacher taught it that way. Anyway, this means that the two ends of the confidence interval are the same distance from the middle, which is the sample proportion p you are looking for. The width of the interval is .006 - .0026 = 0.0034. If you divide that in half (0.0017), then your confidence interval would be calculated as p +/- .0017. Now work from either end to prove that you got it right. from the lower end of the confidence interval, .0026 + .0017 = 0.0043. And from the upper end, 0.006 - 0.0017 = 0.0043. So the sample proportion p is 0.0043. The confidence interval is 0.0043 +/ 0.0017, or 0.0036 to 0.006. You could also get the number from the sample that is the number of teenage births out of 469 that resulted in twins. 469 x 0.0043 = 2.