Find the 95% confidence interval for the variance of the weights of containers of motor oil if a sample of 14 containers has a variance of 3.2. Assume the variable is normally distributed.

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n = 14 with s^2 = 3.2

with level of test = 5% the Confidence Interval formula is CI = ( [ (n-1)*S^2]/ [CHI-SQR-Upper] , [(n-1)*S^2]/[CHI-SQR-Lower] )

I assume this is a TWO-TAILED TEST.

so CHI-SQR distribution has n-1 = 13 degrees of freedom

at the 5% level we should have: Upper chi-squared value is: 24.7356 with upper tail containing 2.5%

The Lower chi-squared value is 5.00874 at the lower end holding 97.5% of distribution.

CI = ( [13*3.2] / [24.7356] , [13*3.2] / [5.00874] ) = (1.681787, 8.305482)

We can approximate the 95% Confidence interval to (1.68, 8.31) to the nearest hundredth, for the variance of weights.

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