Question 1197542: We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 288 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county.
Express your answer in tri-inequality form. Give your answers as decimals, to three places.
< p <
Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.
p =
±
Answer by math_tutor2020(3816)   (Show Source):
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At 99% confidence, the z critical value is roughly z = 2.576
Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value. Look at the bottom row labeled "Z" and above the 99% confidence level.
Alternatively, you can use the NormInv function on a spreadsheet to calculate the z critical value.
x = 288 = number of people who have children
n = 400 = sample size
phat = sample proportion of people who have children
phat = x/n
phat = 288/400
phat = 0.72
E = margin of error
E = z*sqrt(phat*(1-phat)/n)
E = 2.576*sqrt(0.72*(1-0.72)/400)
E = 0.058
L = lower boundary of the confidence interval
L = phat - E
L = 0.72 - 0.058
L = 0.662
U = upper boundary of the confidence interval
U = phat + E
U = 0.72 + 0.058
U = 0.778
The confidence interval in the format L < p < U is 0.662 < p < 0.778
When written in the format phat ± E, we get 0.72 ± 0.058
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