We wish to estimate what percent of adult residents in a certain
county are parents. Out of 400 adult residents sampled, 128 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 = ____ ± ____
Solution :
Given that,
Point estimate = sample proportion =
= x / n = 128 / 400 = 0.320
Z/2
= 2.576
Margin of error = E = Z
/ 2 *
((
* (1 -
)) / n)
= 2.576 * (((0.320
* 0.680) / 400)
= 0.060
A 99% confidence interval for population proportion p is ,
- E < p <
+ E
0.320 - 0.060 < p < 0.320 + 0.060
0.260 < p < 0.380
p = 0.320 0.060
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