The daily intakes of milk (in ounces) for ten randomly selected people were: 26.0 13.0 13.5 14.8 19.1 13.6 21.6 30.8 15.8 14.1 find a 99% confidence interval for the population standard deviation σ.

Re-arranging the numbers;
13.0, 13.5, 13.6, 14.1, 14.8, 15.8, 19.1, 21.6, 26.0, 30.8

Mean, x_bar = (13.0+13.5+13.6+14.1+14.8+15.8+19.1+21.6+26.0+30.8)/10 = 18.23

Standard deviation, sigma = Sqrt [{(13-18.23)^2+(13.5-18.23)^2+(13.6-18.23)^2+(14.1-18.23)^2+(14.8-18.23)^2+(15.8-18.23)^2+(19.1-18.23)^2+(21.6-18.23)^2+(26-18.23)^2+(30.8-18.23)^2}/10] = Sqrt [{27.3529+22.3729+21.4369+17.0569+11.7649+5.9049+0.7569+11.3569+60.3729+158.0049}/10] = Sqrt [336.381/10] = 5.8

Therefore,
x_bar = 18.23
sigma = 5.8
At 99% confidence, Z = 2.58

Confidence interval = x_bar +/- Z*sigma/Sqrt (n) = 18.23+/- 2.58*5.8/Sqrt (10) = 18.23+/-4.73 = [(18.23-4.73), (18.23+4.73)] = [13.5,22.96]

I’m not sure i’m sorry