Supplementary exercise 6.33 of IPS7e
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Time spent studying statistics in an introductory stats class.
Let X_1,...,X_25 denote the times for 25 students.

We assume that the observations are i.i.d. (independent and identically
distributed) with mean mu and standard deviation sigma.
We also assume (quite unrealistically) that sigma is known to be 35 min.
The average of X's (Xmean) is given at 80 min.

(a) With a confidence level of 95%, our zstar-value is 1.96:

95% CI: Xmean +- zstar*sigma/sqrt(n) = 80 +- 1.96*35/sqrt(25)
        = 80 +- 13.72 = 80 +- 13.7 = (66.3, 93.7)

(b) No. The confidence interval is for the population mean, not for values 
of individual students. We can be 95% confident that the population mean 
study time is between 66.3 and 93.7. The population includes all students 
taking the class in question, and perhaps including also students taking 
similar classes elsewhere with the same textbook and schedule.

The individual study times are much more variable than the mean. If we assumed
the study times to follow a normal distribution N(80,35), then a 95% range, or 
interval (but not a confidence interval), based on this distribution would be 
80 +- 1.96*35 = 80 +- 68.6 = (11.4, 148.6).

Note that when the sample size n increases, the confidence intervals shrinks
but the interval for individual values does not change.

Minitab commands and output:

 OneZ 25 80;
  Sigma 35.


The assumed standard deviation = 35

 N   Mean  SE Mean      95% CI
25  80.00     7.00  (66.28, 93.72)

MTB > DPlot;
SUBC>   Scale 1;
SUBC>     LDisplay 1 0 0 0;
SUBC>   Distribution;
SUBC>     Normal 80 35;
SUBC>   Shade 3;
SUBC>     ShType 1;
SUBC>     ShValue 0.05.
Distribution Plot 

The distribution plot shows the interval (11.4, 148.6).