Out of 600 people sampled, 420 had kids.

Question

Out of 600 people sampled, 420 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

Answer 1

Use the formula p +/- (Z_critical)*sqrt([p*(1-p)]/(n-1)) where p = 420/600 =.7, n = 600, Z_critical is the z-score where the probability to the left of the score is equal to .99 (the proportion of the confidence) + .005(the portion left of the .99 center portion) = .995 So, if you look in a Z-table, the body of the table where the probability of .9950 shown to the left of the z-score has z equal to 2.575. So, Z_critical = 2.575

Plugging in the values into the formula, we have .7 +/- 2.575*sqrt([.7*.3]/599)

So, we are 99% confident that p within the population lies within this interval: (0.6518, 0.7482) rounded to 4 decimals.

I hope this helps! - Janet B.