Question 1131776: In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2640 subjects randomly selected from an online group involved with ears. 1038 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys.
a) Find the best point estimate of the population proportion p.
= ____
(Round to three decimal places as needed.)
b) Identify the value of the margin of error E.
E = _____
(Round to three decimal places as needed.)
c) Construct the confidence interval.
= ____ < p < ____
(Round to three decimal places as needed.)
d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.
A.
There is a 90% chance that the true value of the population proportion will fall between the lower bound and the upper bound.
B.
One has 90% confidence that the sample proportion is equal to the population proportion.
C.
One has 90% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
D.
90% of sample proportions will fall between the lower bound and the upper bound.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The point estimate is the best guess and that is 1038/2640 or 0.393
the half-interval is z(0.95), for two sided confidence interval, * sqrt (p*(1-p)/n); the sqrt term is the standard error of the mean.
=1.645*sqrt(0.393*0.607/2640)
=0.0156
The CI is the point estimate+/- z*SE or 0.393+/-0.0156
(0.3774, 0.4086)
This means that one is 90% confident that the true value, unknown and unknowable is in this interval. That is C.
A is not true because the population proportion either does or doesn't fall in the interval, it isn't moving around.
B is not usually true ever except in small populations.
D refers to doing many sample proportions, and 90% of them will contain the population parameter. We don't know which 90 those are, however.
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