Question

A large candy manufacturer produces, packages and sells packs of candy targeted to weigh 50 grams. A quality control manager working for the company was concerned that the variation in the actual weights of the targeted 50-gram packs was larger than acceptable. That is, he was concerned that some packs weighed significantly less than 50 grams and some weighed significantly more than 50 grams. In an attempt to estimate σ, the standard deviation of the weights of all of the 50-gram packs the manufacturer makes, he took a random sample of n 15 packs off of the factory line. The sample yielded a sample variance of 4.2 grams2. Assume the weight of a random selected candy pack is approximately normally distributed. Use the sample to derive a 95% confidence interval for σ.

Answer 1

To derive a 95% confidence interval for σ, the standard deviation of the weights of all the 50-gram packs, we can use the chi-square distribution. The confidence interval is (7.267, 11.955) grams.

Step-by-step explanation:

To derive a 95% confidence interval for σ, the standard deviation of the weights of all the 50-gram packs, we can use the chi-square distribution. The sample size is 15, so the degrees of freedom are n-1 = 15-1 = 14. We need to find the lower and upper bounds of the chi-square distribution that contains 95% of the area. Using the chi-square table or calculator, with 14 degrees of freedom and a cumulative probability of 0.025 on each tail, the lower bound is 5.035 and the upper bound is 25.482.

The confidence interval for σ is (sqrt((n-1)s^2)/sqrt(χ^2_upper), sqrt((n-1)s^2)/sqrt(χ^2_lower)), where s^2 is the sample variance and χ^2_lower and χ^2_upper are the lower and upper bounds of the chi-square distribution, respectively. Substituting the values, the confidence interval for σ is (sqrt((14)(4.2))/sqrt(25.482), sqrt((14)(4.2))/sqrt(5.035)), which simplifies to (7.267, 11.955) grams.