Solution file for additional exercise 13.4
------------------------------------------
(continuation of additional exercise 10.4)

Planning (or post-hoc calculation) for a study on blood pH-values in mice, 
with the aim of comparing two strains. The dataset has 7 litters for each 
strain. The sample size calculations can be based on litter means, because 
the comparison of strains can be based solely on the litter means (compare 
the full model analysis and the analysis of litter means in exercise 10.4).

The estimated variance for litter means is 0.00131. This value we can
get either from the analysis of litter means, or from the full analysis
as MS(litter)/4 (because there are 4 mice in each litter). The standard
deviation (sigma) is 0.03626, without any rounding-off. The effect of interest 
is set at 0.05 units.

a)
MTB > Power;
SUBC>   TTwo;
SUBC>     Sample 7;
SUBC>     Difference .05;
SUBC>     Sigma .03626.
Power and Sample Size 

2-Sample t Test

Testing mean 1 = mean 2 (versus ?)
Calculating power for mean 1 = mean 2 + difference
a = 0.05  Assumed standard deviation = 0.03626

            Sample
Difference    Size     Power
      0.05       7  0.659106

The sample size is for each group.

Comments:
---------
The power was 0.66 to detect a difference of 0.05 units with 7 litters per strain.


b)
MTB > Power;
SUBC>   TTwo;
SUBC>     Difference .05;
SUBC>     Power .8;
SUBC>     Sigma .03626.
Power and Sample Size 

2-Sample t Test

Testing mean 1 = mean 2 (versus ?)
Calculating power for mean 1 = mean 2 + difference
a = 0.05  Assumed standard deviation = 0.03626

            Sample  Target
Difference    Size   Power  Actual Power
      0.05      10     0.8      0.830319

The sample size is for each group.

Comments:
---------
One would need 10 litters of each strain, with 4 mice per litter, to be 
reasonably confident to detect the difference of interest.


c) 
In order to compute necessary sample sizes for a design with only 2
mice per litter, we need to recompute the variance of a litter mean,
because it (naturally) depends on the number of replications. From the full
model,

y_ijk = mu + alpha_i + B_ij + eps_ijk,

we derive the formula

mean(y)_ij = mu + alpha_i + B_ij + mean(eps)_ij,

from which we compute the variance with 2 mice per litter as

Var(mean(y)_ij) = Var(B_ij) + Var(mean(eps)_ij)
                = sigma_B^2 + sigma_eps^2/2
                = 0.00072 + 0.00237/2 = 0.001905

and the standard deviation as:  sigma(mean(y)_ij) = 0.04365

MTB > Power;
SUBC>   TTwo;
SUBC>     Difference .05;
SUBC>     Power .8;
SUBC>     Sigma 0.04365.
Power and Sample Size 

2-Sample t Test

Testing mean 1 = mean 2 (versus not =)
Calculating power for mean 1 = mean 2 + difference
Alpha = 0.05  Assumed standard deviation = 0.04365

            Sample  Target
Difference    Size   Power  Actual Power
      0.05      13     0.8      0.800072

The sample size is for each group.

Comments:
---------
So we would need 13 litters with only 2 mice per litter, 
corresponding to 13*2=26 mice per strain. With 4 mice per litter, the
total number of mice was 10*4=40. Thus, less mice per litter would
require less mice in total. This is because there is more information in
replicating litters (both information about litter variation and mice
variation) than replicating mice (only information about mice variation).