Solution file for additional exercise 13.3
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Planning of a study on the effect of a social marketing campaign for
smoke-free homes, with the aim of comparing two regions (presence and
absence of campaign). The sample size calculation is based on comparing
the two regions after the campaign wrt. their proportions of 
pre-contemplatory homes, or on comparing pre- and post-campaign values
within the same region. Both of these situations would be two independent
samples (note that it is not the same households being interviewed before
and after the campaign). Effect sizes of interest are 0.1 and 0.01, 
with an estimated non-campaign proportion of 0.55. The output from Minitab's 
power menu (see appendices for Stata and R listings) gives the results.

MTB > Power;
SUBC>   PTwo;
SUBC>     PCompare .45 .54;
SUBC>     Power .8;
SUBC>     PBaseline .55.
Power and Sample Size 

Test for Two Proportions

Testing comparison p = baseline p (versus not =)
Calculating power for baseline p = 0.55
Alpha = 0.05

              Sample  Target
Comparison p    Size   Power  Actual Power
        0.45     392     0.8      0.800741
        0.54   38926     0.8      0.800008

The sample size is for each group.

Comments:
---------
To detect a difference of 0.1 requires about 400 families. To detect a
difference of 0.01 is almost impossible.

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Appendix: output from power calculation using the Stata menu:

. power twoproportions .55 (.45 .54), test(chi2)

Performing iteration ...

Estimated sample sizes for a two-sample proportions test
Pearson's chi-squared test 
Ho: p2 = p1  versus  Ha: p2 != p1

  +-----------------------------------------------------------------+
  |   alpha   power       N      N1      N2   delta      p1      p2 |
  |-----------------------------------------------------------------|
  |     .05      .8     784     392     392     -.1     .55     .45 |
  |     .05      .8   77852   38926   38926    -.01     .55     .54 |
  +-----------------------------------------------------------------+

Same thing with a continuity correction (presumed to give more accurate
values):

. power twoproportions .55 (.45 .54), test(chi2) continuity

Performing iteration ...

Estimated sample sizes for a two-sample proportions test
Pearson's chi-squared test 
Ho: p2 = p1  versus  Ha: p2 != p1

  +-----------------------------------------------------------------+
  |   alpha   power       N      N1      N2   delta      p1      p2 |
  |-----------------------------------------------------------------|
  |     .05      .8     824     412     412     -.1     .55     .45 |
  |     .05      .8   78250   39125   39125    -.01     .55     .54 |
  +-----------------------------------------------------------------+

We could also try the same based on a likelihood-ratio chi-square test
(instead of the Pearson chi-square test), effectively replacing the option
test(chi2) with test(lrchi2). The results are however in this case exactly 
the same as with the Pearson chi-square test without continuity correction.

The menu for Fisher's exact test only allows calculation of power, so
one would need to use a trial-and-error approach to get a sample size
corresponding to a certain power.

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Appendix: output from power calculation using R function:

> power.prop.test(p1=.55, p2=.45, sig.level=0.05, power=0.8, alternative="two.sided")

     Two-sample comparison of proportions power calculation 

              n = 391.263
             p1 = 0.55
             p2 = 0.45
      sig.level = 0.05
          power = 0.8
    alternative = two.sided

NOTE: n is number in *each* group

> power.prop.test(p1=.55, p2=.54, sig.level=0.05, power=0.8, alternative="two.sided")

     Two-sample comparison of proportions power calculation 

              n = 38925.34
             p1 = 0.55
             p2 = 0.54
      sig.level = 0.05
          power = 0.8
    alternative = two.sided

NOTE: n is number in *each* group