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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.

Nov 6th, 2015

Thank you for the opportunity to help you with your question!

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.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Nov 6th, 2015

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Nov 6th, 2015
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Nov 6th, 2015
Dec 4th, 2016
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