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 2: What t value would be used to calculate the interval estimate (to 3 decimal places)?

Answer 1

Solution :

Given that,

t _/2,df = 2.492

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

= 2.492 * (10 / 25)

Margin of error = E = 5.0

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

- E < < + E

35 - 5.0 < < 35 + 5.0

27 < < 37

(30 , 40)