Supplementary Exercise 7.145 of IPS7e
-------------------------------------
(continuation of 7.143, although we use the supplied data obtained
after natural log transformation; in practice one would use the original
data to avoid the round-off errors.)

The statistical model is the same as in Exercise 7.143, except that the 
assumed normal distribution (and its parameters) are on log scale instead 
of original scale.

Minitab commands:

MTB > WOpen "H:\VHM\VHM801\Datasets\Minitab\Chapter 7\ex07_145.mtw".
Retrieving worksheet from file: 'H:\VHM\VHM801\Datasets\Minitab\Chapter
7\ex07_145.mtw'
Worksheet was saved on 09/10/2014

MTB > Describe  'loc';
SUBC>   Mean;
SUBC>   SEMean;
SUBC>   StDeviation;
SUBC>   QOne;
SUBC>   Median;
SUBC>   QThree;
SUBC>   Minimum;
SUBC>   Maximum;
SUBC>   Skewness;
SUBC>   Kurtosis;
SUBC>   N;
SUBC>   NMissing;
SUBC>   GBoxplot.
Descriptive Statistics: loc 

Variable   N  N*   Mean  SE Mean  StDev  Minimum     Q1  Median     Q3  Maximum
loc       31   0  3.338    0.109  0.609    2.090  2.880   3.410  3.860    4.360

Variable  Skewness  Kurtosis
loc          -0.12     -0.72
 
Boxplot of loc 

MTB > GSummary  'loc'.
Summary Report for loc 

MTB > Stem-and-Leaf 'loc';
SUBC>   Trim.
Stem-and-Leaf Display: loc 

Stem-and-leaf of loc  N  = 31
Leaf Unit = 0.10

 1   2  0
 3   2  23
 3   2
 4   2  7
 10  2  888899
 15  3  00001
 15  3
(5)  3  44455
 11  3  667
 8   3  89
 6   4  000
 3   4  233

MTB > PPlot 'loc';
SUBC>   Normal;
SUBC>   Symbol;
SUBC>   FitD;
SUBC>   Grid 2;
SUBC>   Grid 1;
SUBC>   MGrid 1.
Probability Plot of loc 
The P-value of the Anderson-Darling test of normality is 0.327

MTB > Onet 'loc'.
One-Sample T: loc 

Variable   N   Mean  StDev  SE Mean      95% CI
loc       31  3.338  0.609    0.109  (3.114, 3.561)

Answers to questions:
---------------------
(a)
The distribution of log-transformed oc-values is close to symmetrical
(the skewness is -0.12 so the slight left-skewness is of no
importance). The stemplot and histogram show a few "gaps" but the
normality test does not give any evidence against a normal distribution.
There does not appear to be any outlying observations.

(b)
The 95% CI is given above. As the data show no evidence against being
normally distributed, we may consider this interval as exact. There is
certainly no problem with using the t-procedure for these data.

(c)
The backtransformed mean is exp(3.33774)=28.2. It is lower than both the
estimated mean (33.4) and the estimated median (30.2) for the original
data, but it is a valid estimate of the median. If the assumed normal
distribution on the log-scale is valid, this estimate is better than the
median computed directly from the untransformed data.

The backtransformed endpoints of the 95% CI are exp(3.11430)=22.5 and
exp(3.56118)=35.2. This interval is considerably narrower than the 95%
CI for the median displayed in the Summary Report window (its
computation will be explained in session 8). Again, assuming the normal
distribution on log-scale to be valid, this is the best interval we can
get for the median.