Question

Suppose Peter, the receptionist at Dr. Mitchell’s office, is interested in the variation in waiting times in the office on Fridays. He knows that the waiting time overall has a standard deviation of 9.7 minutes, which equates to a variance of (9.7)2=94.09 minutes2. For Fridays only, he collects a random sample of 32 waiting times and calculates the standard deviation for the wait time to be 6.9 minutes, which translates to a variance of (6.9)2=47.61 minutes2.

Compute the 95%

confidence interval for the variance in wait times on Fridays. Express the results to three decimal places. Find the upper and lower limit.

Answer 1

Answer:-

Given that:-

Suppose Peter, the receptionist at Dr. Mitchell’s office, is interested in the variation in waiting times in the office on Fridays. He knows that the waiting time overall has a standard deviation of 9.7 minutes, which equates to a variance of (9.7)2=94.09 minutes2. For Fridays only, he collects a random sample of 32 waiting times and calculates the standard deviation for the wait time to be 6.9 minutes, which translates to a variance of (6.9)2=47.61 minutes2.

n=32

Sample variance S^{2}=47.61

95% confidence interval for variance

confidence interval formula

n-1=32-1=31

df=n-1=31

  

Critical values   

Confidence interval is given by

Upper limit:- 84.151

Lower limit:- 30.600