Question

A sample of 25 patients in a doctor's office showed that they had to wait an average of 35 minutes before they could see the doctor. The sample standard deviation is 8 minutes. Assume the population of waiting times is normally distributed. At 99% confidence, compute the margin of error.

Answer 1

Given that,

s = 8

n = 25

Degrees of freedom = df = n - 1 = 25- 1 = 24

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

_{t /2  df} = t_0.005,24= 2.7969

Margin of error = E = t_/2,df * (s /n)

= 2.7969* ( 8/ 25) = 4.47504