SOLUTION: An employment counselor found that in a sample of 100 unemployed workers, 65% were not interested in returning to work. Find the 95% confidence interval of the true proportion of

SOLUTION: An employment counselor found that in a sample of 100 unemployed workers, 65% were not interested in returning to work. Find the 95% confidence interval of the true proportion of

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Question 1155770: An employment counselor found that in a sample of 100 unemployed workers, 65% were not
interested in returning to work. Find the 95% confidence interval of the true proportion of
workers who do not wish to return to work.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The half-interval is +/- 1.96*sqrt(p*(1-p)/n. and the sample proportion is 0.65
.65+/-1.96 sqrt (.65*.35/100)=0.65+/-0.93
(0.557, 0.743)
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