Question

A sample of 113 hypertensive people were given an anti-hypertensive drug, and the drug was found to be effective in 66 of those people. (By effective, we mean that their diastolic blood pressure is lowered by at least 10 mm Hg as judged from a repeat measurement taken 1 month after taking the drug.)

a) Find a 94% confidence interval for the true proportion of the sampled population for which the drug is effective.

b) Using the results from the above mentioned survey, how many people should be sampled to estimate the true proportion of hypertensive people for which the drug is effective to within 2% with 97% confidence?

c) If no previous estimate of the sample proportion is available, how large of a sample should be used in (b)?

Answer 1

Solution :

Given that,

a) n = 113

x = 66

Point estimate = sample proportion = = x / n = 66 / 113 = 0.584

1 - = 1 - 0.584 = 0.416

At 94% confidence level

= 1 - 94%

=1 - 0.94 =0.06

/2 = 0.03

Z/2 = Z_0.03 = 1.881

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

= 1.881 (((0.584 * 0.416) / 113)

= 0.087

A 94% confidence interval for population proportion p is ,

± E  

= 0.584 ± 0.087

= ( 0.497, 0.671 )

b) margin of error = E = 2% = 0.02

At 97% confidence level

= 1 - 97%

=1 - 0.97 =0.03

/2 = 0.015

Z/2 = Z_0.015 = 2.17

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.17 / 0.02)² * 0.584 * 0.416

= 2859.99

sample size = n = 2860

c) =  1 - =   0.5

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.17 / 0.02)² * 0.5 * 0.5

= 2943.06

sample size = n = 2944