Supplementary Exercise 7.66 of IPS7e
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Data: test scores of 20 Spanish teachers attending a summer institute. Scores
were obtained before and after the institute, so we choose to analyse the gain
in score between the two (comparable but not identical) tests.

Model: the 20 observations are a simple random sample (i.i.d. sample)
from a distribution with mean mu and standard devation sigma, both of
which are unknown parameters. The mean mu corresponds to the difference
in mean scores after minus before the summer institute. In other words,
the two original samples (pretest and posttest scores) are paired
samples, and we analyze the differences.

Descriptive analysis using Minitab:

---

 Describe 'gain';
  Mean;
  SEMean;
  StDeviation;
  QOne;
  Median;
  QThree;
  Minimum;
  Maximum;
  Skewness;
  Kurtosis;
  N.

Variable   N   Mean  SE Mean  StDev  Minimum      Q1  Median     Q3  Maximum  Skewness
gain      20  1.450    0.716  3.203   -5.000  -1.000   2.000  3.750    7.000     -0.50

Variable  Kurtosis
gain         -0.39

---

 GSummary 'gain'.

---

 Stem-and-Leaf 'gain'.

Stem-and-leaf of gain  N  = 20
Leaf Unit = 1.0

 2   -0  54
 4   -0  32
 6   -0  11
 8   0   11
(7)  0   2223333
 5   0   4455
 1   0   7

---

 PPlot 'gain';
  Normal;
  Symbol;
  FitD;
  Grid 2;
  Grid 1;
  MGrid 1.
Probability Plot of gain 
The P-value of the Anderson-Darling test of normality is 0.257

Comments:
---------
The distribution of gains looks unimodal and fairly symmetrical with a 
minor left-skewness (skewness=-0.50). There are no suspected outliers.
The A-D normality test is far from significant. Despite the minor
skewness, the guidelines for use of t-procedures (7L-3) suggest that we 
can safely use t-procedures for these data.

Next, we turn to the statistical inference. We compute a 95% confidence
interval using Minitab, as well as a statistical test. The null hypothesis 
should correspond to no effect of the institute,
       H0: mu=0
and the alternative hypothesis could correspond to an improvement over the
course of the institute (it is hard to imagine that the Spanish level
could have declined),
       Ha: mu>0

---

 Onet 'gain';
  Confidence 95.0;
  Alternative 0.

Variable   N   Mean  StDev  SE Mean       95% CI
gain      20  1.450  3.203    0.716  (-0.049, 2.949)

---

 Onet 'gain';
  Test 0;
  Confidence 95.0;
  Alternative 1.

Test of mu = 0 vs > 0
                                     95% Lower
Variable   N   Mean  StDev  SE Mean      Bound     T      P
gain      20  1.450  3.203    0.716      0.211  2.02  0.029

Comments:
---------
The observed t-test value is 2.02, which is weakly significant in a
t-distribution with df=19. There is (weak) evidence to indicate an
improvement in the test scores after the summer institute. Note that
with a two-sided alternative hypothesis, the test would not have reached
significance (P=2*0.029=0.058, or 0.057 without round-off error).
For this reason, the confidence interval includes 0 despite that P<0.05;
recall that testing by confidence interval only works for tests with a
two-sided alternative.

The confidence interval indicates the likely improment of test scores as
somewhere in the range of (0,3) units. Whether that is an interesting level 
of improvement probably depends on the actual tests. Given that the maximal 
score was 36, an average improvement of 1.5 units does not seem as much, 
but it depends on the initial level and the difficulty of the test.