SOLUTION: In a city, out of 400 people sampled, 152 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.
Round your answers
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-> SOLUTION: In a city, out of 400 people sampled, 152 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.
Round your answers
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Question 1199321: In a city, out of 400 people sampled, 152 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.
Round your answers to three decimal places. Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! sample size is 400
p = 152/400 = .38
q = 1-p = .62
standard error = sqrt(p*q/n) = sqrt(.38*.62/400) = .0242693222.
two tailed 99% confidence interval has a critical z-score of plus or minus 2.5758 rounded to 4 decimal places.
use the z-score formula to find the critical raw scores.
z-score formula is z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard error.
when x = 2.5758, the formula becomes -2.5758 = (x-.38)/.0242693222.
solve for x to get x = -2.5758 * .0242693222 + .38 = .3174878799.
when x = 2.5758, the formula becomes 2.5758 = (x-.38)/.0242693222.
solve for x to get x = 2.5758 * .0242693222 + .38 = .4425129201
round your answer to 3 decimal places to get the 99% confidence interval is from .317 to .443.
