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Question

Out of 400 people sampled, 176 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

I am 99% confident that the proportion of people who have kids is between ________  and _________ .

Answer 1

Let p-hat be the proportion of people who have kids. Then p-hat = X/n, where n is the sample size and X is the number of people who have kids.

p-hat = 176/400 = 0.44

The confidence level of 99% has the z-value of 2.58.

0.44 ± 2.58√(0.44)(0.56)/400 = [0.376 , 0.504]

I am 99% confident that the proportion of people who have kids is between 0.376 and 0.504.

Answer 2

sample proportion (phat) = 176/400 = 0.44.

sample size = 400

CI for true population proportion = phat +/- z-critical for 99% CI * square root(phat* (1-phat)/sample size)

z-critical for 99% is found by looking up z value for probability of 0.995 (since CI is the middle 99% of area), which is 2.575.

So CI = 0.44 +/- 2.575 * square root(0.44 * 0.56/400) = 0.44 +/- 0.064 = (0.376,0.504)