Answer:
99% confidence interval is 0.07 < Ļ < 0.32
Step-by-step explanation:
Given that;
standard deviation s = 0.12 sec
sĀ² = 0.12Ā² = 0.0144
degree of freedom DF = n - 1 = 8 - 1 = 7
99% confidence interval
ā = 1 - 99% = 1 - 0.99 = 0.01
now, we find xĀ² critical values for ā/2 = 0.005 and 1 - ā/2 = ( 1 - 0.005) = 0.995, df = 7
The Lower critical value [tex]X^{2}_{\frac{\alpha }{2}},7[/tex] = 20.2777
The Upper critical value [tex]X^{2}_{1-{\frac{\alpha }{2}},7[/tex] = 0.9893
Now, confidence interval is given by
ā[ ( (n-1)ĆsĀ² ) / ( [tex]X^{2}_{\frac{\alpha }{2}},7[/tex] ) ] < Ļ < ā[ ( (n-1)ĆsĀ² ) / ( [tex]X^{2}_{1-{\frac{\alpha }{2}},7[/tex] ) ]
so we substitute
ā[ ( 7Ć0.0144 ) / ( 20.2777 ) ] < Ļ < ā[ ( 7Ć0.0144 ) / ( 0.9893 ) ]
ā[ ( 7Ć0.0144 ) / ( 20.2777 ) ] < Ļ < ā[ ( 7Ć0.0144 ) / ( 0.9893 ) ]
ā0.0049711 < Ļ < ā0.10189
0.0705 < Ļ < 0.3192
Rounding to the nearest 2 decimal places
0.07 < Ļ < 0.32
Therefore; 99% confidence interval is 0.07 < Ļ < 0.32
