Statistics Homework

Question

Out of 200 people sampled, 74 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

 < p < 

Answer 1

A proportion is a mean, and given the large sample sizes in this problem, we can treat it like another other confidence interval problem.

1.) calculate the sample mean, = p = 74/200 = .37

2.) calculate the sample standard deviation, = sqrt(p*q) = sqrt((.37)*(1 - .37)) = .4828

3.) We are interested int he sampling distribution of the mean (SE), so we need to calculate the standard error of the mean, = SD/sqrt(N) = .4828/sqrt(200) = .03414

4.) We want a 90% confidence interval, so we need to find the two-tailed critical value for Z. You may use a Z table or use Excel's norm.s.inv(p) function. The norm.s.inv function will give the Z score at the p <= .90. To get the required two-tailed Z score, we simply run norm.s.inv(.95), which gets the upper Z score = 1.645

5.) construct the confidence interval: p +- Z_crit*SE