Suppose that a random sample of 50 bottles

Amahle Sizakele

Amahle Sizakele

Answered question

2022-06-15

Suppose that a random sample of 50 bottles of a particular brand of cough medicine is selected and the alcohol content of each bottle measured. The sample mean alcohol content is 8.6 ml and the standard deviation is 2.99 ml. The standard deviation of the alcohol content is required, by law, to be less than 2.8 ml.

A 90% confidence interval for the population variance of the alcohol content is

( 6.60 ; 12.91 ).

Thus, we can conclude with 90% confidence, that

 

Answer & Explanation

nick1337

nick1337

Expert2023-05-21Added 777 answers

To determine the 90% confidence interval for the population variance of the alcohol content, we can use the Chi-square distribution. The formula for the confidence interval is:
((n1)s2χα/2,n12,(n1)s2χ1α/2,n12),
where n is the sample size, s is the sample standard deviation, and χα/2,n12 and χ1α/2,n12 represent the values from the Chi-square distribution with n-1 degrees of freedom corresponding to the lower and upper critical values, respectively.
Substituting the given values into the formula, we have:
((501)·2.992χ0.05/2,5012},(501)·2.992χ10.05/2,5012),
where χ0.05/2,5012 and χ10.05/2,5012 represent the lower and upper critical values from the Chi-square distribution with 49 degrees of freedom at the significance level of 0.05/2 (or 0.025) and 1-0.05/2 (or 0.975), respectively.
According to the given information, the 90% confidence interval for the population variance of the alcohol content is (6.60, 12.91).
Thus, we can conclude with 90% confidence that the population variance of the alcohol content falls within the interval (6.60, 12.91).

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