Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 223 with 43% successes.

Question

Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 223 with 43% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Answer 1

Hi Anni,

Formula for these is:

CI= p^ +/- z*SE

p^=sample proportion estimate= 0.43

z*=z-critical value

SE=standard error

SE= sqrt(p^q^/n)

q^=1-p^=0.57

n=sample size=223

SE= sqrt(0.43*0.57)/223

SE= 0.033

Critical z-value for 99.9:

z* = 3.29

Go to z-table and look up z-score for 0.9995--this yields the 3.29. You now have all you need to compute confidence interval:

p^= 0.43

SE= 0.033

z* = 3.29

CI= p^ +/- (z*SE)

Remember to report it as: a < p < b when you finish computing the interval. I hope this helps.