SOLUTION: 91 adults selected randomly from a town,61 have health insurance. Find the 90% confidence interval for the true proportion of all adults in the town who have onsurance.

Question 1054503: 91 adults selected randomly from a town,61 have health insurance. Find the 90% confidence interval for the true proportion of all adults in the town who have onsurance.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
91 adults selected randomly from a town,61 have health insurance. Find the 90% confidence interval for the true proportion of all adults in the town who have insurance.
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sample proportion = p-hat = 61/91 = 0.67
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ME = z*sqrt(p*q/n) = 1.645*sqrt(0.67*0.33/91) = 1.645*0.05 = 0.08
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90% CI:: 0.67-0.08 < p < 0.67+0.08
90% CI:: 0.59 < p < 0.75
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Cheers,
Stan H.
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