It was reported that in a survey of 4790 American youngsters

It was reported that in a survey of 4790 American youngsters aged 6 to 19, 18% were seriously overweight (a body mass index of at least 30; this index is a measure of weight relative to height). Calculate a confidence interval using a 99% confidence level for the proportion of all American youngsters who are seriously overweight. (Round your answers to three decimal places.)

Solution

sample size, n=4753 sample proportion, p = 0.15 we use normal approximation, for this we check that both np and n(1-p) > 5. Since n*p = 712.95 > 5 and n*(1-p) = 4040.05 > 5, we can take binomial random variable as normally distributed, with mean = p = 0.15 and std error = root( p * (1-p) /n ) = 0.005179 For constructing Confidence interval, Margin of Error (ME) = z x SD = 0.0133 99% confidence interval is given by: Sample Mean +/- (Margin of Error) 0.15 +/- 0.0133 = (0.1367 , 0.1633)