Question

A survey of 950 U.S. adults found that 32% of people said that they would get no work done on Cyber Monday since they would spend all day shopping online. Find the 99% confidence interval of the true proportion. Round intermediate answers to at least five decimal places. Round your final answers to at least three decimal places.

Answer 1

Given that, n=950
p^= 32% = 0.32
(1–α)%=99%
(1–α)=0.99
α=0.01
α/2=0.005
Zα/2=2.58 ....... from standard normal table.
Margin of error=E=Zα/2 ×√{[p^(1-p^)]/n}
=2.58 ×√{[0.32(1-0.32)]/950}
=2.58 × 0.01513
= 0.03903....upto 5 decimal place.
Margin of error=E=0.03903
99% Confidence Interval for true population proportion p is given as,
p^ ± Margin of error
=(0.32-0.03903,0.32+0.03903)
=(0.28097,0.35903)
=(0.281,0.359) ....upto 3 decimal place.
Lower limit =0.281
Upper limit=0.359

99% confidence interval for the true population proportion lies between (0.281,0.359)