(a) An article in a medical journal suggested that approximately 14% 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 0.08

This question is incomplete, the complete question is;

Surgical complications: A medical researcher wants to construct a

99.5% confidence interval for the proportion of knee replacement surgeries that result in complications.

- An article in a medical journal suggested that approximately 14% 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 0.08?

a sample of operations needed is 149

Step-by-step explanation:

Given that;

confidence interval = 99.5% = 0.995 so

margin of error E = 0.08

p = 14% = 0.14

now we obtain the critical value of z t the 99.5 confidence interval

∝ = 1 - confidence interval

∝ = 1 - 0.995

∝ = 0.005

∝/2 = 0.0025

obtaining the area of probability in the right tail

Area of probability to the right is = 1 - 0.0025 = 0.9975

from probability table; critical value of t =2.81

using the formula

n = p( 1 - p ) [ (z_∝/2)/E)² ]

so we substitute  

n = 0.14 ( 1 - 0.14 ) [ 2.81 / 0.08 )²  

n = (0.14 × 0.86 ) × 1233.7656

n = 148.5453 ≈ 149

Therefore , a sample of operations needed is 149

idk

step-by-step explanation: