calculating probalities using a confidence interval of sample size

Question

For a random sample sample of 150 households in a large metropolitan area, the number of households in which at least one adult is currently unemployed and seeking a full time job is 12. Using a 95% confidence interval, estimate the proportion of households in the area in which at least one adult is unemployed.

Answer 1

This is a proportion calculation. You are looking at the proportion of households out of 150, and 12/150 is, by my calculation, is 0.08. The formula for a proportion CI confidence interval is first, find σ_p which is sqrt((p*(1-p)/N) so you use p = 0.08, then 1-0.08 and N = 150. Check this yourself. I get 0.022. Now you need the z table. For a 95% confidence interval where the number can be either greater or less, you need two-tailed. That is, you go from area.025 to .975. Look up the z value for .9750. Find the value of .9750 on your z table and read off the 3 digit z value. Then your proportion CI is

.08 +/- .022*z and you have the interval of the proportions, from something less than .08 to something more.