Question

A doctor's office collected sample data for 25 patients, and found that they had to wait an average of 35 minutes with a sample standard deviation of 10 minutes, in order to see the doctor. (The wait times were also found to be normally distributed.) We are interested in calculating a 98% confidence interval for the average waiting time for the population of all patients.

PART 1: What would be the standard error of the mean waiting time (to 1 decimal place)?

Answer 1

Solution :

Given that,

=

s =

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

At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2_,df = t_0.01,90 = ( using student t table)

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

= * ( / )

=

The 98% confidence interval estimate of the population mean is,

- E < < + E

- < < +

  < <

( )