Question 1079575: In a sample of 1000 people,630 said they were most likely to sleep when at home sick with a cold or flu.Determine the proportion of people who sleep when at home sick with a cold?
b) in order to calculate a confidence interval for p it is neccessary to be able to approximate probability distribution of the number of people who sleep when at home sick with a cold or flu ising a normal distribution.calculate the following np and n(1-p)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The point estimate is 630/1000=0.63=p
np=630
1-p=0.37
n(1-p)=370
the 95% CI is 0.63 +/_ 1.96* sqrt (0.63)(0.37)/1000
=0.0299 or 0.03
(0.60, 0.66)
The variance is np(1-p)=233.1; take the sqrt, multiply by 1.96, and the interval width was 29.92, using 630 as the mean.
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