Question

A researcher wants to estimate the proportion of depressed individuals taking a new anti-depressant drug who find relief. A random sample of 250 individuals who had been taking the drug is questioned; 180 of them found relief from depression. Based upon this, compute a 95% confidence interval for the proportion of all depressed individuals taking the drug who find relief. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. ( If necessary, consult a list of formulas.)

What is the lower limit of the 95% confidence interval?

What is the upper limit of the 95% confidence interval?

Answer #1

Sample proportion = 180 / 250 = 0.72

95% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.72 - 1.96 * sqrt [ 0.72 * ( 1 - 0.72) / 250] < p < 0.72 + 1.96 * sqrt [ 0.72 * ( 1 - 0.72) / 250]

0.664 < p < 0.776

**Lower limit = 0.66**

**Upper limit = 0.78**

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