Question

Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 333 with 163 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Answer 1

A sample is used to estimate the population proportion p.

We have 163 successes in a sample of 333.

Now, we find that

The sample proportion is p=163/333=0.489

The z-value for 99.9% confidence interval is z=3.291.

The sample size is n=333

Now, we know that the formula of confidence interval is given by

So, the required answer is (0.399,0.579).