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Question

Out of 500 people sampled, 180 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

< p <

Answer 1

p +/- z * √ (pq/n)

p = 180/500

q = 1-p

n = 500

As n>30 we use a z-value rather than a t-value

for 99% confidence interval we need the area to the left of the positive Z value to be 0.995

giving z = 2.58

So you can calculate the confidence interval now