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?
Answer: a sample of operations needed is 149
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