Question

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. In a clinical test with 2400 subjects, 720 showed improvement from the treatment. Find the margin of error for the 99% confidence interval used to estimate the population proportion.

Answer 1

Answer: 0.0241

Step-by-step explanation:

The formula we use to find the margin of error :

$E=z^*\sqrt{\dfrac{p(1-p)}{n}}$

, where z* = Critical value , n= Sample size  and p = Sample proportion.

As per given , we have

n= 2400

Sample proportion of subjects showed improvement from the treatment:

$p=\dfrac{720}{2400}=0.3$

Critical value for 99% confidence = z*= 2.576  (By z-table)

Now , the margin of error for the 99% confidence interval used to estimate the population proportion. :

$E=(2.576)\sqrt{\dfrac{0.3(1-0.3)}{2400}}$

$E=(2.576)\sqrt{0.0000875}$

$E=(2.576)(0.00935414346693)$

$E=0.0240962735708\approx0.0241$ [Round to the four decimal places]

Hence, the margin of error for the 99% confidence interval used to estimate the population proportion. =0.0241