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

4202 treated with the​ drug,

195 developed the adverse reaction of nausea. Construct a

90​%

confidence interval for the proportion of adverse reactions.

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

​(Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that,

n = 4202

x = 195

Point estimate = sample proportion = = x / n = 195/4202=0.0464

1 -   = 1- 0.0464 =0.9536

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 (((0.0464*0.9536) / 4202)

E = 0.0053

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.0464- 0.0053< p <0.0464+ 0.0053

0.0411< p < 0.0517

The 90% confidence interval for the population proportion p is : 0.0411,0.0517