Solution file for Problem 16.2 (GO)
-----------------------------------

Planning of an experiment: 
measurements viscosity in batches of ice cream, produced under different conditions:
recipe (A/B), temperature (low/high), pressure (low/high).

Two batches can be run each day for a total of 8 days. Temperature and
pressure must remain constant during (both batches of) the day. Day to
day variability is expected because a fresh supply of milk is used every
day.

(a) The factors recipe, temperature and pressure are all treatment-type 
factors. Among these temperature and pressure can only be applied at the
day level. Therefore day becomes the experimental unit for these two
factors. Together the temperature and pressure comprise 4 combinations,
and with 8 days available there'll be 2 replicates of each factor
combination in a completely randomized design. The recipes can be
changed between the two batches run on the same day. Therefore we arrive
at a split-plot design:
  * whole plots: days
  * whole-plot factors: temperature and pressure
  * split plots: batches
  * split-plot factor: recipe

The skeleton ANOVA takes the form:

Source     DF
---------------
Temp        1
Pres        1
T*P         1
Days        4
Recipe      1
R*T         1
R*P         1
R*T*P       1
Error       4
----------------
Total      15

The degrees of freedom for Days (whole plots) may be obtained as the
residual degrees of freedom in the completely randomized design for
whole plots (8 observations). Alternatively, use the ANOVA table in the
notes (p.15) with a=4 and c=2.

The degrees of freedom for Error (split plots) may be obtained by
subtraction from the total. Alternatively, use the ANOVA table in the
notes with a=4, b=2 and c=2.

Randomization takes place first at the whole plot level. The 4 factor
combinations of temperature and pressure are randomly assigned to the 8
days with two replications of each combination. On each day, it is
randomly decided which of the two recipes should be used first.


(b) Simultaneous 95% confidence intervals means that the overall error
rate for all confidence intervals is 5%. As the question is about CIs
for combinations of temperature and pressure, there are 4 combinations
to choose from and 6=4*3/2 pairs of comparisons. This means that we can
do Bonferroni CIs with a correction for 6 comparisons.

The intervals will be based on the mean viscocities for the four
different combinations across days and recipes. As the data comprise 16
observations, each mean will be over 4 observations. The standard error
for each difference will be computed as
  sqrt(MS_D*2/4) = sqrt(MS_D/2)
where MS_D is the mean square in the ANOVA table for days nested within
treatment combinations. The degrees of freedom for MS_D equals 4, so the
Bonferroni corrected percentile corresponds to a tail probability of 
1-0.025/6=0.0042, or a t-percentile of 4.85.

The same result is obtained if the data are averaged across recipes to
the day level, and then analyzed by a 2-way ANOVA with replications, or
a 1-way ANOVA with replications for the combined temperature*pressure
factor with 4 levels.