Question

In: Statistics and Probability

Given below is simple random sample data for wait times, in minutes, for a call center....

Given below is simple random sample data for wait times, in minutes, for a call center. At the 98% confidence level, calculate the confidence interval estimate for the variance in wait time for the population of all calls at the call center. Assume the population is normally distributed. 12.1 11.5 13.4 16.2 11.3 12.2 11.3

Expert Solution

Solution:

sample size n = 7

12.1 ,11.5, 13.4, 16.2,11.3 ,12.2 ,11.3

Using calculator , we find the sample variance.

sample variance = s² = 3.0990476190476

df = n - 1 = 7 - 1 = 6

Our aim is to construct 98% confidence interval for the population variance ²

c = 98% = 0.98

= 1 - c = 1 - 0.98 = 0.02

/ 2 = 0.01

1 - ( / 2) = 0.99

Now , using chi square table ,

= = 16.81

= = 0.87

The 98% confidence interval for ² is,

( 7 - 1) *3.0990476190476/ 16.81< ² < (7 -1 ) * 3.0990476190476/ 0.87

1.11 < ² < 21.32

orchestra answered 1 year ago

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