Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 247 with 85% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places

Step-by-step explanation:

The confidence interval for the population proportion is given by:-

, where = sample proportion, n =sample space , z* = critical z-value.

As per given, we have

Critical value for 99.9% confidence = 2.576

The required confidence interval will be:

The required confidence interval:

step-by-step explanation:

a) justify

step-by-step explanation:

justifying is just part of everyday life. justify kind of means proof so i think the answer is a) justify.