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

43704370

patients treated with the drug,

139139

developed the adverse reaction of nausea. Construct a

9090 %

confidence interval for the proportion of adverse reactions.

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

nothing

Answer 1

Solution :

Given that,

n = 437

x = 139

Point estimate = sample proportion = = x / n = 139 / 437 = 0.318

1 - = 1 - 0.318 = 0.682

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z _0.05 = 1.645

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

= 1.645 * (((0.318 * 0.682) / 437)

= 0.037

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.318 - 0.037 < p < 0.318 + 0.037

0.281 < p < 0.355

The 90% confidence interval for the population proportion p is : (0.281 , 0.355)