For a 95% confidence interval we use the formula
p ± z*√(p(1-p) / n)
The sample size is 100, so n = 100.
First, we need to establish the sample proportion, which is the number of people who have kids in the sample to the total sample size.
p = (#people with kids) / (sample size n)
p = 60 / 100 = 0.60
The critical value z* for 95% confidence interval is 1.960.
Summarize:
n = 100
z* = 1.960
p = 0.6
(1-p) = 1-0.6 = 0.4
so we are ready to plug into original formula!
p ± z*√(p(1-p) / n)
=0.6 ± 1.960 √((0.6 × 0.4) / 100)
=0.6 ± 1.960 √(0.24 / 100)
=0.6 ± 1.960 √(0.0024)
=0.6 ± (1.960 × 0.048989795)
=0.6 ± (0.096019998)
round to 2 sig figs
=0.6 ± 0.096
which gives the upper confidence limit of
0.6 + 0.096
= 0.696 or 69.6 %
and a lower confidence limit of
0.6 - 0.096
= 0.503 or 50.3%
ANSWER:
0.503 < p < 0.696
OR
50.3% < p < 69.6%
