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Answer:
99% confidence interval is 0.07 < σ < 0.32
Step-by-step explanation:
Given that;
standard deviation s = 0.12 sec
s² = 0.12² = 0.0144
degree of freedom DF = n - 1 = 8 - 1 = 7
99% confidence interval
∝ = 1 - 99% = 1 - 0.99 = 0.01
now, we find x² critical values for ∝/2 = 0.005 and 1 - ∝/2 = ( 1 - 0.005) = 0.995, df = 7
The Lower critical value = 20.2777
The Upper critical value = 0.9893
Now, confidence interval is given by
√[ ( (n-1)×s² ) / ( ) ] < σ < √[ ( (n-1)×s² ) / (
) ]
so we substitute
√[ ( 7×0.0144 ) / ( 20.2777 ) ] < σ < √[ ( 7×0.0144 ) / ( 0.9893 ) ]
√[ ( 7×0.0144 ) / ( 20.2777 ) ] < σ < √[ ( 7×0.0144 ) / ( 0.9893 ) ]
√0.0049711 < σ < √0.10189
0.0705 < σ < 0.3192
Rounding to the nearest 2 decimal places
0.07 < σ < 0.32
Therefore; 99% confidence interval is 0.07 < σ < 0.32