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.

There was 504 pages in that chunk