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

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

Step-by-step explanation:

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

$E=z_{\alpha/2}\times\sqrt{\dfrac{p(1-p)}{n}}$

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

$0.082=2.576\times\sqrt{\dfrac{0.08(0.92)}{n}}\\\\\Rightarrow\ 0.03183=\sqrt{\dfrac{0.0736}{n}}$

Squaring both sides , we get

$0.0010131489=\dfrac{0.0736}{n}\\\\\Rightarrow\ n=\dfrac{0.0736}{0.0010131489}=72.644800779\approx73$

Hence, the required sample size = 73

Answer 2

Answer:

sorry I didn't mean to post