The manufacturer of a certain type of product claims that the machine that fills packages of these products is set up in such a way that the average net weight of the packages is 32 grams with a variance of 0.015 square grams. A consumer agency took a random sample of 25 packages from the production line, obtained a sample variance of 0.029 square grams and constructed a 95% confidence interval for the population variance. It is known that the machine is stopped and adjusted if either both or one of the two limits of the confidence interval is not in the interval 0.008 to 0.030. Based on the agent’s sample information, does the machine need an adjustment? Assume that the product net weights in all packages are normally distributed.

clealtAfforcewug

clealtAfforcewug

Answered question

2022-11-18

The manufacturer of a certain type of product claims that the machine that fills packages of these products is set up in such a way that the average net weight of the packages is 32 grams with a variance of 0.015 square grams. A consumer agency took a random sample of 25 packages from the production line, obtained a sample variance of 0.029 square grams and constructed a 95% confidence interval for the population variance. It is known that the machine is stopped and adjusted if either both or one of the two limits of the confidence interval is not in the interval 0.008 to 0.030. Based on the agent’s sample information, does the machine need an adjustment? Assume that the product net weights in all packages are normally distributed.
I know the confidence interval is defined by:
[ x ¯ c s n , x ¯ + c s n ]
I'm just not sure which mean or standard deviation to use.

Answer & Explanation

mignonechatte00f

mignonechatte00f

Beginner2022-11-19Added 13 answers

Step 1
So your confidence interval is a confidence interval for the mean, but what we're looking for in this question is the confidence interval of the variance. In order to find that, we need to use the fact that for a normally distributed sample, the variance follows a chi-squared distribution.
Step 2
Because the sample size is 25, we know that 24 s 2 σ 2 follows a chi-squared distribution with 25 1 = 24 "degrees of freedom" (that is, ν = 24). Using a table (or a calculator), construct a 95% confidence interval about the mean of a ν = 24 chi-squared distribution, and solve for s 2 to construct the confidence interval for the variance, s 2 .
For this problem, the confidence interval you should have is
0.0177 < σ 2 < 0.0561

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