Find a 98% confidence interval for the population standard deviation σ

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Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data.

The amounts (in ounces) of juice in eight randomly selected juice bottles are:
     15.0  15.9  15.2  15.7
     15.0  15.8  15.4  15.4
Find a 98% confidence interval for the population standard deviation σ.    
Oct 20th, 2017

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

To find 98% confidence interval for the population standard deviation σ, we need the following:

s^2; n-1, Chi^2(right),Chi^2(left) to plug in the following formula:

(Sqrt((n-1)s^2/Chi^2(right)), Sqrt((n-1)s^2/Chi^2(right))

s^2=.1221; n=1 =7,Chi^2(right)(.01,7)=18.48  Chi^2(left)= 1.24

98% confidence interval= (0.2088 < SD < 0.9167)

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 10th, 2015

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