Solution file for GO Exercise 7.5
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Planning of a study on comparing the quality of video tapes of 6 different
brands. The outcome is a distortion measure, with an estimated variance
of 0.25, i.e. sigma=0.5.

a) We need to think a bit about how the question is to be interpreted.
First, what test is referred to? The possible interpretations are the
overall F-test for equality between brands and a test targeted at
comparing high and low cost brands (from a suitably designed contrast).
For this solution, I have chosen the first interpretation. Second, what 
information exactly is provided about the alternative hypothesis we will
work under? For this solution, I have chosen the simple interpretation
that true difference between high and low cost tapes equals 1 unit, and
that no substantial differences between tape brands exist.

With 4 tapes of each brand, a true difference of 1 unit between
the tapes of best and poorest quality, and a significance level of 0.01,
we compute the power using Minitab as follows:

MTB > Power;
SUBC>   OneWay 6;
SUBC>     Sample 4;
SUBC>     MaxDifference 1;
SUBC>     Sigma .5;
SUBC>     Alpha 0.01.
Power and Sample Size 

One-way ANOVA

a = 0.01  Assumed standard deviation = 0.5

Factors: 1  Number of levels: 6

   Maximum  Sample
Difference    Size     Power
         1       4  0.186276

The sample size is for each level.

Comments:
---------
This calculation is available in Stata only through the add-on command 
fpower, as follows:
  fpower , a(6) alpha(.01) delta(2)
(here, the value for delta is computed as: effect/sd  = 1/0.5 = 2).

If one wanted to use the Stata menu or the R function power.anova.test, one would 
need to know (hypothesize) the means of all groups, not only the two most 
extreme ones. Powers will be different depending on how the other group means
are set, but we can make the following assessments:

highest power: three means at 0, three means at 1
lowest power: one mean at 0, 4 means at 0.5, one mean at 1

This is because these two settings are the easiest and most difficult
ones, respectively, for the one-way ANOVA. The Stata listings below show
that the first of these has much higher power, and the second has
the same power as above.

. power oneway 0 0 0 1 1 1, varerror(.25) alpha(0.01) npergroup(4)
Estimated power for one-way ANOVA
F test for group effect
...
Estimated power:
        power =    0.7313

. power oneway 0 0.5 0.5 0.5 0.5 1, varerror(.25) alpha(0.01) npergroup(4)
Estimated power for one-way ANOVA
F test for group effect
...
Estimated power:
        power =    0.1863

Because the study involves four high cost tapes and two lost cost tapes,
it would perhaps make most sense to consider the following group means:

. power oneway 0 0 1 1 1 1, varerror(.25) alpha(0.01) npergroup(4)

Estimated power for one-way ANOVA
F test for group effect
...
Estimated power:
        power =    0.6587

See appendix for the same calculations using R.


b) 
95% CI for mean group differences should have a length of at most 2.
We give two solutions, for different interpretations of the question.
The simplest interpretation (i) is that the CI is based on a difference
between two of the 6 groups. Only slightly more complex is the situation
(ii) where the CI is obtained from a contrast based on the first 4 groups.

(i)
For (i), the constant A(n) equals sqrt(2/n), where n is the sample size in 
each of the groups. This is because the standard error of the mean 
difference of two groups is sigma*sqrt(2/n). Therefore, we must solve the equation
(for the margin of error)

1 >= 2*sigma*sqrt(2/n)  =>  n >= (2*sigma*sqrt(2)/1)^2 = 2

The degrees of freedom for error with 6 groups of 2 will be 6, so we
rerun the calculation with 2 replaced by t(6,0.975)=2.447.

1 >= 2.447*sigma*sqrt(2/n)  =>  n >= (2.447*sigma*sqrt(2)/1)^2 = 3

A sample size of 3 in each group should be sufficient to get 95%
confidence intervals for pairwise comparisons of length 2. Note that
with a residual standard deviation of 0.5, this is a fairly modest
requirement.

(ii)
For (ii), the relevant contrast for A-B is

  theta = 0.5*mu1 + 0.5*mu2 - 0.5*mu3 - 0.5*mu4.

Because each of the group means will have variance sigma^2/n, the 
contrast estimate will have variance 

  0.5^2 * sigma^2/n + ... + 0.5^2 * sigma^2/n (4 terms) = sigma^2/n,

and A(n)=sqrt(1/n). We can now use the same approach as above, and the
first calculation will give us: n >= (2*sigma)^2 = 1. The minimal group
size is 2, so we redo the calculation (as above) with 2.447: 
n >= (2.447*sigma)^2 = 1.5. Therefore, the minimal group size of 2 
will give a CI for the contrast for A-B with a length less than 2.

---

Appendix: output for a) from power calculation using R function:

> power.anova.test(groups=6, n=4, between.var=var(c(0,0,0,1,1,1)), within.var=0.25, sig.level=0.01)

     Balanced one-way analysis of variance power calculation 

         groups = 6
              n = 4
    between.var = 0.3
     within.var = 0.25
      sig.level = 0.01
          power = 0.7312544

NOTE: n is number in each group

> power.anova.test(groups=6, n=4, between.var=var(c(0,.5,.5,.5,.5,1)), within.var=0.25, sig.level=0.01)

     Balanced one-way analysis of variance power calculation 

         groups = 6
              n = 4
    between.var = 0.1
     within.var = 0.25
      sig.level = 0.01
          power = 0.1862755

NOTE: n is number in each group

> power.anova.test(groups=6, n=4, between.var=var(c(0,0,1,1,1,1)), within.var=0.25, sig.level=0.01)

     Balanced one-way analysis of variance power calculation 

         groups = 6
              n = 4
    between.var = 0.2666667
     within.var = 0.25
      sig.level = 0.01
          power = 0.6587254

NOTE: n is number in each group