Question

Surgical complications: A medical researcher wants to construct a 98% confidence interval for the proportion of knee replacement surgeries that result in complications.

(a) An article in a medical journal suggested that approximately 20% 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 is needed to obtain a 98% confidence interval with a margin of error of 0.08 using the estimate 0.2 for p.

Answer 1

Solution :

Given that,

= 0.2

1 - = 1 - 0.2 = 0.8

margin of error = E = 0.08

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z/2 = 2.326 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (2.326 / 0.08)2 * 0.2 * 0.8

=135.25

Sample size = 135