SOLUTION: Out of 600 people sampled, 342 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.
Give your answers as decimals
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-> SOLUTION: Out of 600 people sampled, 342 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.
Give your answers as decimals
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Question 1198839: Out of 600 people sampled, 342 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.
Give your answers as decimals, to three places
You can put this solution on YOUR website! the proportion of people that have kids is 342 / 600 = .57
the proportion of people that don't have kids is .43
p = .57
q = 1 -.57 = .43
n = total number of people = 600
standard deviation is sqrt(p * q / n) = .57 * .43 / 600) = .0202113829.
z-score is used.
formula for z-score is:
z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation
the critical z-score at 90% two tailed confidence interval is equal to plus or minus 1.644853626.
low side z-score formula becomes -1.644853626 = (x - .57) / .0202113829 = .5367552335.
high side z-score formula becomes 1.644853626 = (x - .57) / .0202113829 = .6032447665.
the 90% confidence interval is from .537 to .603 rounded to 3 decimal places.
