99% confidence interval for the mean length

Question

A quality control engineer is interested in the mean length of sheet insulation being cut automatically by machine. The desired length of the insulation is 12 ft. It is known that the standard deviation in the cutting length is 0.15 ft. A sample of 70 ut sheets yields a mean length of 12.14 ft. Find the 99% confidence interval for the mean length cut by machine.

Answer 1

The formula for a confidence interval is given by

μ ± zσ/√n

where μ is the mean, z is the z-score corresponding to a 99% confidence level, σ is the standard deviation and n is the sample size.

You are given the standard deviation of σ = .15.  You also know that the mean is μ = 12.14 and that the sample size is n = 70.  The z-score corresponding to a 99% confidence level is 2.57.  Then the computation is given by

μ ± zσ/√n

12.14 ± (2.57)(0.15)/√70

12.14 ± .046

Then the confidence interval for the mean length is [12.09, 12.19]

Hope this helps you!