Question

Use the sample data and confidence level given below to complete parts​ (a) through​ (d).

A drug is used to help prevent blood clots in certain patients. In clinical​ trials, among

4924

patients treated with the​ drug,

170

developed the adverse reaction of nausea. Construct a

99

​%

confidence interval for the proportion of adverse reactions.

​a) Find the best point estimate of the population proportion p.

Answer 1

Solution :

Given that,

n = 4924

x = 170

(a)

Point estimate = sample proportion = = x / n = 170/4924=0.035

1 -   = 1- 0.035 =0.965

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* (((0.035*0.965) / 4924)

= 0.0067

the 99% confidence interval for the proportion of adverse reactions

- E < p < + E

0.035 - 0.0067 < p < 0.035+0.0067

0.0283< p < 0.0417

(0.0283 , 0.0417)