I've got a set of data and I want to answer the following:
Calculate an upper confidence bound with confidence level 95% for the population standard deviation of maximum pressure.
The set of data is
33.2, 41.8, 37.3, 40.2, 36.7, 39.1, 36.2, 41.8
36.0, 35.2, 36.7, 38.9, 35.8, 35.2, 40.1
We are learning how to use Minitab and we've been asked to solve this using Minitab. I only see an option to compute the one-sample t for the mean, and I'm unsure how to compute the one-sample t for standard deviation.
Edit: I just found the 1 variance computation and got this as the result:
95% Confidence Intervals
CI for CI for
Variable Method StDev Variance
C1 Chi-Square (1.88, 4.06) (3.54, 16.45)
Bonett (1.95, 3.91) (3.79, 15.28)
Which of these is the answer I'm looking for? I'm not sure what Chi-square and Bonnett are?
I believe you need to do a one-sided confidence interval. (Your question is somewhat ambiguous. If you want the upper-end of a two-sided CI, then your procedure gives 16.45 for the variance.)
The code generated by what I suppose to be the appropriate menu choices (STAT > Basic > One variance > data in C1, Omit test, Option > Less than) is as follows:
The above is from Minitab 16. (Bonnett's procedure does not assume the data are normal.) I will leave it to you to figure out exactly what distribution theory Minitab is using to get the one-sided chi-squared upper confidence bound.
In R, the computation for the upper-bound on the variance is:
Take the square root to get a bound for the SD.