SOLUTION: in a certain state, a survey of 500 workers showed that 45% belonged to a union. Find 90% confidence interval of the true proportion of workers who belonged to a union.

Question 204791: in a certain state, a survey of 500 workers showed that 45% belonged to a union. Find 90% confidence interval of the true proportion of workers who belonged to a union.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
in a certain state, a survey of 500 workers showed that 45% belonged to a union. Find 90% confidence interval of the true proportion of workers who belonged to a union.
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sample proportion: 0.45
E = 1.645[sqrt(0.45*0.55/500)] = 0.0366
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90% CI: 0.45 - 0.0366 < p < 0.45 + 0.0366
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Cheers,
Stan H.

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