The thicknesses of 20 randomly selected linoleum tiles were found to have a standard deviation of 2.66 Construct the 95% confidence interval for the population standard deviation of the thicknesses of all linoleum tiles in this factory. Round your answers to two decimal places.
df = n - 1 = 20 - 1 = 19
Chi square critical values at 0.05 level with 19 df = 8.907 , 32.852
95% confidence interval for is
Sqrt [ (n-1) S2 / /2 ] < < Sqrt [ (n-1) S2 / 1-/2 ]
Sqrt [ (20 - 1 ) 2.662 / 32.852 ] < < Sqrt [ (20 - 1 ) 2.662 / 8.907 ]
2.02 < < 3.89
95% CI is ( 2.02 , 3.89)
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