Question 1114655: You randomly select and weigh 30 samples of an allergy medicine and the sample standard deviation is 1.20 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance
(a) Complete the table:
Population Parameter Being Estimated
Point Estimate
Sample Size n =
Margin of Error
Distribution Used
Reason for Using that Distribution
Confidence Level
Confidence Interval
Units for the Confidence Interval (if applicable)
(b) Interpret the confidence interval (don’t forget the units):
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Problem asks to construct a 99% confidence interval for the population variance
:
The variance is a non-negative number. This means the domain of the probability distribution is not (−∞,∞), therefore the normal distribution cannot be the distribution of a variance. The correct PDF must have a domain of [0,∞), it can be shown that if the original population of data is normally distributed, then the expression (n−1)s^2/σ^2 has a chi-square distribution with n−1 degrees of freedom.
:
alpha(a) = 1 -(99/100) = 0.01
:
a/2 = 0.005
:
critical probability(p*) = 1 - 0.005 = 0.995
:
degrees of freedom(df) = 30 -1 = 29
:
chi^2 (0.995) distribution with df = 29 is 13.121
:
chi^2 (0.005) distribution with df = 29 is 52.336
:
now we evaluate (n-1)s^2/chi^2 for 13.121 and 52.336
:
for 13.121
:
(30-1)(1.20)^2/13.121 = 3.1827
:
for 52.336
:
(30-1)(1.20)^2/52.336 = 0.7979
:
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99% confidence interval for variance is [0.7979, 3.1827] milligrams
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