Question

A medical researcher wants to construct a 99% confidence interval for the proportion of knee replacement surgeries that result in complications. An article in the Journal of Bone and Joint Surgery suggested that approximately 8% of such operations result in complications. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.082 ?

Answer 1

sorry I didn't mean to post

Answer 2

Answer: Sample size of 73 is needed so that the confidence interval will have a margin of error of 0.082

Explanation:

The formula to calculate the margin of error for population proportion is given by :-

`E=z_(\alpha/2)*\sqrt{(p(1-p))/(n)}`

Given : The proportion of operations result in complications :
`p= 0.08`

Significance level :
`\alpha = 1-0.99=0.01`

Critical value =
`z_(0.005)=2.576` [By standard normal distribution table]

Margin of error : E= 0.082

Substitute all the value in the above formula, we get

`0.082=2.576*\sqrt{(0.08(0.92))/(n)}\\\\\Rightarrow\ 0.03183=\sqrt{(0.0736)/(n)}`

Squaring both sides , we get

`0.0010131489=(0.0736)/(n)\\\\\Rightarrow\ n=(0.0736)/(0.0010131489)=72.644800779\approx73`

Hence, the required sample size = 73