You are told that a 95% confidence interval for a population proportion is (0.3775, 0.6225) What was the sample proportion that lead to this confidence interval?. Also, what is the size of the sample used.

The mean of the confidence interval is (0.3775 + 0.6225) / 2 = 0.5. Therefore, the standard deviation of the proportion would have been sqrt[0.5 * (1 - 0.5) / n], where n is the sample size. This expression simplifies to sqrt (0.25/n).

A 95% CI has a corresponding z = 1.96, so since the distance from 0.5 to 0.3775 (or 0.6225 to 0.5) is equal to 0.1225. Therefore, if we divide 0.1225 / 1.96 = 0.0625, we get the value of the SD, and this should be equal to sqrt (0.25/n).

0.0625 = sqrt (0.25/n)

n = 64

This means that the proportion was 0.5 and the sample size was 64.