Question

In a group of 35 people, 30% have newborn babies at a local hospital. Assume the weight of newborn babies has approximately normal distribution. Find a 95% Confidence Interval for the mean weight of all newborn babies at this hospital.

Answer 1

Solution:
Given that, n=35
p^=30%=0.30
(1–α)%=95%
α=0.05
α/2=0.025
Zα/2=1.96 ....... from standard normal table.
Margin of error=E=Zα/2 ×√{[p^(1-p^)]/n}
=1.96 ×√{[0.3(1-0.3)]/35}
=0.1518
Margin of error=E=0.1518
95% Confidence Interval for the mean weight of all newborn babies at this hospital is given as,
p^ ± Margin of error=(0.3-0.1518,0.3+0.1518)
=(0.1482,0.4518)
Lower limit =0.1482
Upper limit=0.4518
95% Confidence Interval for the mean weight of all newborn babies at this hospital is (0.1482,0.4518)