SOLUTION: Oil Changes A service station advertises that
customers will have to wait no more than 30 minutes for
an oil change. A sample of 28 oil changes has a standard
deviation of 5.2 m
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-> SOLUTION: Oil Changes A service station advertises that
customers will have to wait no more than 30 minutes for
an oil change. A sample of 28 oil changes has a standard
deviation of 5.2 m
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Question 1173270: Oil Changes A service station advertises that
customers will have to wait no more than 30 minutes for
an oil change. A sample of 28 oil changes has a standard
deviation of 5.2 minutes. Find the 95% confidence
interval of the population standard deviation of the time
spent waiting for an oil change. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! This is a Chi Square with df=27
(n-1) s^2/ChiSq 0.025, df=27 < sigma ^2 < (n-1)s^2/ChiSq 0.975,df=27
26*27.04/43.195 < 26*27.04/14.573 Chi Square from the table
=16.28 < sigma ^2 < 48.24
That is the 95% CI of the variance
take the square root for the CI of the sd
4.03 < sigma < 6.95
or (4.03, 6.95) units minutes
