Question

A researcher wants to estimate the proportion of depressed individuals taking a new anti-depressant drug who find relief. A random sample of 225 individuals who had been taking the drug is questioned; 173 of them found relief from depression. Based upon this, compute a 99% 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. What is the lower limit of the 99% confidence interval? What is the upper limit of the 99% confidence interval?

Answer 1

Solution :

Given that,

n = 225

x = 173

Point estimate = sample proportion = = x / n = 173/225=0.769

1 -   = 1- 0.769 =0.231

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

  Margin of error = E = Z/2   * (((( * (1 - )) / n)

= 2.576* (((0.769*0.231) / 225)

E = 0.07

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.769-0.07 < p < 0.769+0.07

0.70< p < 0.84

The 99% confidence interval for the population proportion p is : lower limit=0.70,upper limit= 0.84