Exercise 27.19 of PSLS 3e
-------------------------

Data: measurements of percent nitrogen in gas bubbles in 9 specimens of amber;
the interest is in comparing the nitrogen level to the present atmosphere, which
contains 78.1% nitrogen.

Model: a simple random sample (i.i.d. sample) from a distribution with 
unknown mean, median and standard deviation.

(a)
Estimation:
sample mean = 59.6, sample median = 63.3, sample standard deviation = 6.26

Examining the distribution with a dotplot and a stemplot shows a substantial
left skewness (skewness=-0.92). This indicates that a normal distribution 
assumption may not reasonable, in particular as the normal probability plot 
looks far from straight, and the A-D normality test is significant (P=0.024).
It therefore seems logical to look for alternatives to a t-test, which
for n<15 would require the distribution to be close to normal (according
to the PSLS and IPS guidelines, reviewed on page 6L-9).

(b)
Wilcoxon's signed rank test takes its most extreme outcome: W=0 for the
positive group, because no observations are above 78.1. The approximate
P-value in Minitab is given as P=0.009 for a two-sided alternative;
Stata gives the approximate P-value as P=0.008. We conclude there is evidence 
to reject H0: median=78.1, and as all values are below 78.1 we have evidence 
to state that the median is below 78.1. In conclusion, ancient air seems to
differ (be lower) in nitrogen content than present-day air.

The analysis above follows the wording in the question to use
Wilcoxon's signed rank test. However, because one of main concerns with
the distribution shape is its skewness, Wilcoxon's signed rank test 
is NOT the obvious choice because it assumes the distribution to be
symmetrical. In this instance it would be more logical to use a sign
test. Observing 9 out of 9 in a B(9,0.5) has probability
  p9=0.5^9 = 0.001953,
and therefore the *exact* two-sided P-value for the sign test is
  P=2*p9 = 0.0039.

Because the event observed is the same and most extreme for both the 
sign test and Wilcoxon's signed rank test, the two tests must give the 
same P-value in this case. Thus we realize that the exact P-value for
Wilcoxon's signed rank test must be 0.0039 (Stata also gives this value). 
The conclusion for both tests is the same: reject H0 and state evidence 
that the median is less than 78.1.


Minitab commands and listing:

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 WOpen "R:\Chapter 27\ex27_019.mtw".
 Describe 'pctN';
  Mean;
  StDeviation;
  QOne;
  Median;
  QThree;
  Minimum;
  Maximum;
  Skewness;
  Kurtosis;
  N.

Variable  N   Mean  StDev  Minimum     Q1  Median     Q3  Maximum  Skewness  Kurtosis
pctN      9  59.59   6.26    49.10  52.90   63.30  64.45    65.00     -0.92     -0.99

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 Dotplot 'pctN'.

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 Stem-and-Leaf 'pctN'.

Stem-and-leaf of pctN  N  = 9
Leaf Unit = 1.0

 1   4  9
 2   5  1
 2   5
 3   5  4
 3   5
 3   5
 4   6  0
(2)  6  33
 3   6  445

---

 PPlot 'pctN';
  Normal;
  Symbol;
  FitD;
  Grid 2;
  Grid 1;
  MGrid 1.

The P-value for the A-D normality test is 0.024.

---

 WTest 78.1 'pctN';
  Alternative 0.

Test of median = 78.10 versus median not = 78.10

         N for   Wilcoxon         Estimated
      N   Test  Statistic      P     Median
pctN  9      9        0.0  0.009      59.65

---

 STest 78.1 'pctN';
  Alternative 0.

Sign test of median =  78.10 versus not = 78.10

      N  Below  Equal  Above       P  Median
pctN  9      9      0      0  0.0039   63.30

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