A manufacturer produces, packages, and sells packs of their product targeted to weigh 51 grams. A quality control manager working for the company was concerned that the variation in the actual weights of the targeted 51-gram packs was larger than acceptable. That is, he was concerned that some packs weighed significantly less than 51-grams and some weighed significantly more than 51 grams. In an attempt to estimate the standard deviation of the weights of all of the 51-gram packs the manufacturer makes, he took a random sample of 26 packs off of the factory line. The random sample yielded the sample mean 50.97 grams and the standard deviation of 0.621 grams. Use the random sample to derive a 90% confidence interval for the variance (standard deviation) of the actual weights of the packs.

Note: if the normality plot is not provided you may assume that the actual weights of the packs are normally distributed.

1. Procedure: Select an answer One proportion Z procedure One mean T procedure One mean Z procedure One variance χ² procedure
2. Assumptions: (select everything that applies)
   • Normal population
   • The number of positive and negative responses are both greater than 10
   • Population standard deviation is known
   • Simple random sample
   • Sample size is greater than 30
   • Population standard deviation is unknown
3. Unknown parameter: Select an answer μ, population mean σ², population variance p, population proportion