Supplementary Exercise 7.68 of IPS7e
------------------------------------
(same data as in Supplementary Exercise 7.58 of IPS7e)

Data: 34 difference scores of preschool children in tests for
spatial-temporal reasoning after and before attending piano lessons.

Model: the 34 observations are a simple random sample (i.i.d. sample)
from a distribution corresponding to a population of difference scores
for children that attend piano lessons. As we'll use a nonparametric
method, the population parameter of interest is the median.

Estimation: estimated median=4.0 (from Exercise 7.58)

95% CI for median: (3,5) - from Minitab listing below (use interval with
confidence level 95% obtained by non-linear interpolation (NLI)).

Hypothesis H0: median=0 (no improvement by the piano lessons),
alternative hypothesis Ha: median>0 (some improvement)

The hypotheses may also be stated in terms of the probability p 
of an improvement (that is, the (non-zero) difference after-before
is positive): H0: p=0.5 vs. Ha: p>0.5.
  
Test: sign test using 28 positive and 4 negative differences,
P=P(Y>=28) where Y follows B(32,0.5), and from Minitab/Stata analyses for
the sign test we get an exact P-value less than 0.00005
(the actual value calculated in the binomial distribution is 0.0000097).

We may also compute an approximate P-value using either the z-test for 1
proportion, or the normal approximation to B(32,0.5). These approaches are 
really only of interest if we don't have access to software, or in lucky
cases a table for the binomial distribution in question, to allow us to 
compute the exact P-value.

The approximation of B(32,0.5) by a normal distribution gives:
  P ~= 1-P(Z<(28-0.5-16)/sqrt(32*0.5*0.5)) = 1-P(Z<4.066) = P(Z<-4.066)
     = 0.000024
This computation involves a continuity-correction in the binomial
distribution. The approximation is barely applicable because its condition
is not met (32*0.5*(1-0.5)=8<10), but the weaker condition given in the
textbooks are met (32*0.5=16 and 32*(1-0.5)=16 >10).

The classical z-test for 1 proportion gives:
  P ~= P(z> (28/32-0.5)/sqrt(0.5*0.5/32)) = P(z> 4.243) = P(z<-4.243) = 0.000011.
The conditions for use of this test are met (32*0.5=16 and 32*(1-0.5)=16>10).

Conclusion: the test gives very strong evidence of a positive median
in the distribution of differences. We can be pretty sure that the scores
are more likely to be higher than lower on the second measurement (after
the piano lessons).

As a final note, our previous analyses in 7.58 and 7.59 did not
identify any substantial problems with the normal distribution
assumption. Therefore, the need for a non-parametric analysis is not
obvious, and it is no surprise that our conclusions are very similar to
those of the normal distribution analysis.

---
Minitab commands and output:

MTB > WOpen "H:\VHM\VHM801\Datasets\Minitab\Chapter 7\ex07_058.mtw".
Retrieving worksheet from file: 'H:\VHM\VHM801\Datasets\Minitab\Chapter
7\ex07_058.mtw'
Worksheet was saved on 09/10/2014

MTB > SInterval 95.0 'change'.
Sign CI: change 

Sign confidence interval for median

                                 Confidence
                      Achieved    Interval
         N  Median  Confidence  Lower  Upper  Position
change  34   4.000      0.9424  3.000  5.000        12
                        0.9500  3.000  5.000       NLI
                        0.9757  3.000  5.000        11

MTB > STest 0.0 'change';
SUBC>   Alternative 1.
Sign Test for Median: change 

Sign test of median =  0.00000 versus > 0.00000

         N  Below  Equal  Above       P  Median
change  34      4      2     28  0.0000   4.000

MTB > CDF 4;
SUBC>   Binomial 32 .5.
Cumulative Distribution Function 

Binomial with n = 32 and p = 0.5

x  P( X <= x )
4    0.0000097

MTB > POne 32 28;
SUBC>   Test .5;
SUBC>   Confidence 95.0;
SUBC>   Alternative 1.
Test and CI for One Proportion 

Test of p = 0.5 vs p > 0.5
                                             Exact
Sample   X   N  Sample p  95% Lower Bound  P-Value
1       28  32  0.875000         0.736403    0.000