SOLUTION: In a survey of 203 households, a Food Marketing Institute found that 129 households spend more than $125 a week on groceries. Find the 80% confidence interval for the true proport

Question 1182834: In a survey of 203 households, a Food Marketing Institute found that 129 households spend more than $125 a week on groceries.
Find the 80% confidence interval for the true proportion of the households that spend more than $125 a week on groceries.
a. Answer as open-interval to 3 places
b. Express the same answer as a tri-linear inequality using decimals accurate to 3 places
c. Express the same answer using the point estimate and margin of error. Answer as decimals to 3 places
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
129/203 is the point estimate=0.6355
80% half-interval is z(0.90)* sqrt (p hat*(1- p hat)/n)
=1.282* sqrt (0.6355*0.3645/.203)
=0.0433
the interval is (0.592, 0.679)
0.592 < mean < 0.679
0.635 +/- 0.043
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