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.
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.
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?