Question

The widths of 79 randomly selected window blinds were found to have a standard deviation of 2.37. Calculate the 95% confidence interval for the population standard deviation of the widths of all window blinds in this factory. Round your answers to 2 decimal places.

1Lower Bound ？

2Upper Bound？

Answer 1

Chi square critical values at 0.05 level with 78 df = 55.466 , 104.316

95% confidence interval for is

Sqrt [ (n-1) S² / /2 ] < < Sqrt [ (n-1) S² / 1-/2 ]

sqrt [ ( 79 - 1) * 2.37² / 104.316 ] < < sqrt [ ( 79 - 1) * 2.37² / 55.466 ]

2.05 < < 2.81

Lower bound = 2.05

Upper bound = 2.81