Solution Notes, HW 9
1. Assume the experiment was run using a completely randomized design, (i.e. no blocks, and a model
with only an overall mean, treatment-related effects, and i.i.d. noise for each measurement). Compute
the least-squares estimates for each of the 15 factorial effects and their standard errors, and sums-ofsquares and degrees of freedom for an ANOVA partitioned with lines for treatments, residual, and
corrected total.
treatment
(1)
d
...
abc
abcd
1
48.87
31.52
...
57.06
50.67
replicate
2
3
53.78 54.37
35.04 34.88
...
...
50.53 51.43
45.38 56.48
• e.g. α̂ = [−48.84 − 53.78 − ... + 45.38 + 56.48]/48
• treatments:
– df = 15
– SS = 48 * α̂2 + ... 14 more of these
• residual:
– df = 16*(3-1)
– SS = (48.87 − ȳ(1) )2 + (53.78 − ȳ(1) )2 + (54.37 − ȳ(1) )2 + likewise for 15 other groups
• Var(...)
ˆ = σ 2 /48, stderr.(...)
ˆ =
q
M S(residual)/48
2. Assume the experiment was run using a randomized complete block design, each rep constituting
one block. How will the effect estimates differ from those in part 1? Compute sums-of-squares
and degrees of freedom for an ANOVA partitioned with lines for blocks, treatments, residual, and
corrected total.
treatment
(1)
d
...
abc
abcd
1
48.87
31.52
...
57.06
50.67
replicate
2
3
53.78 54.37
35.04 34.88
...
...
50.53 51.43
45.38 56.48
Treatments are orthogonal to blocks, so none of the estimates are affected. (Formally, H1 X2 is the
same for this design as for the previous one.) The treatment sum-of-squares from the previous table
is still correct, but now there’s an extra line:
• replicates:
– df = 2
– SS = 16(ȳblock 1 − ȳall )2 + 16(ȳblock 2 − ȳall )2 + 16(ȳblock 3 − ȳall )2
• residual
– 2 fewer degrees of freedom
– SS reduced by SS(replicates)
3. Assume that blocks of size 4 must be used. Select a generating relation to use (i.e. two factorial
effects and their generalized interaction). Use the same generating relation in each rep (i.e. don’t use
partial confounding), do not confound any main effects with blocks, and confound as few two-factor
interactions as possible. Assuming blocks are fixed, how will the effect estimates differ from those
in parts 1 and 2? Compute sums-of-squares and degrees of freedom for an ANOVA partitioned with
lines for blocks, treatments, residual, and corrected total.
• How about I = ABC = ACD ( = BD )
• Now 4 (blocks per rep) times 3 (reps) = 12 total blocks
• Effect estimates are the same, except that ABC, ACD, and BD are now confounded with blocks
and so are no longer estimable.
• treatments:
– df = 15 (from before) - 3 (lost to blocking)
ˆ 2 − 48 ∗ (αγδ)
ˆ 2 − 48 ∗ (βδ)
ˆ 2
– SS = previous −48 ∗ (αβγ)
• blocks (replaces previous “replicates” line):
– df = 12 - 1
– SS = 4(ȳblock 1 − ȳall )2 + ... + 4(ȳblock 12 − ȳall )2
• OR, leave previous “replicates” in, and add blocks-within-reps ...
– df = 3*(4-1)
– SS = 4(ȳblock 1 − ȳrep 1 )2 + 4(ȳblock 2 − ȳrep 1 )2 + 4(ȳblock 3 − ȳrep 1 )2 + 4(ȳblock 4 − ȳrep 1 )2 +
likewise for reps 2 and 3
• residual
– df is 47 (c.t.) - 11 (blocks) - 12 (treatments)
– SS is SS (c.t.) - SS (blocks) - SS (treatments)
4. Repeat the previous part, but under the assumption that the blocks have been selected randomly.
With this form of analysis, all factorial effects can be represented as “treatments” in the ANOVA,
but not all will be compared to the same denominator mean square. How will the effect estimates
differ from those in previous parts? Compute sums-of-squares and degrees of freedom for an ANOVA
partitioned with lines for confounded effects, unconfounded effects, and residual terms appropriate
for each of them.
• all 15 estimates as in part 1 ...
CAN actually work this two different ways, depending on interpretation.
#1 Suppose all 12 blocks are randomly drawn from the same “block bag”. That is, we regard two
blocks within a rep as having the same relationship as two blocks from different reps.
• whole-plot effects: ABC, ACD, BD.
–
–
–
–
df = 3
ˆ 2 + 48 ∗ (αγδ)
ˆ 2 + 48 ∗ (βδ)
ˆ 2
SS = 48 ∗ (αβγ)
whole-plot corrected total: 11 df “block” SS from last part
whole-plot residual = w-p c.t. - w-p treatments
• split-plot effects: all the rest
– df = 15 - 3
– SS = as in part 3
– split-plot residual df and SS as in part 3
#2 Suppose there may be systematic (unknown, nonrandom) difference among reps, but that within
a rep the 4 blocks represent randomly drawns from the “block bag”. Now, we regard two blocks
within a rep as potentially more similar than two blocks from different reps.
• need to reintroduce “replicate” as a fixed effect line in the ANOVA table
– df = 2, SS as before, this is still orthogonal to all effects
• whole-plot corrected total and residual
– df and SS are now reduced by these “replicate” quantities
– whole-plot noise is now WITHIN-REP variability among blocks, apart from that which is
associated with the 3 whole-plot effects
• split-plot analysis is not affected
5. Now assume that blocks are fixed, but select a different generating relation for each rep (i.e. use
partial confounding) in such a way that no main effect is used, and at least 2/3 information is available
for each interaction. Compute the least-squares estimates for each of the 15 factorial effects and their
standard errors, and sums-of-squares and degrees of freedom for blocks, partially confounded effects,
unconfounded effects, residual, and corrected total.
treatment
(1)
d
...
abc
abcd
1
48.87
31.52
...
57.06
50.67
replicate
2
3
53.78 54.37
35.04 34.88
...
...
50.53 51.43
45.38 56.48
• Suppose:
– Rep 1: I = ABC = ACD ( = BD )
– Rep 2: I = BCD = ABD ( = AC )
– Rep 3: I = AB = CD ( = ABCD )
ˆ = [−48.87 − 54.37 + 31.52 + 34.88... + 56.48]/32
• then e.g., (αβδ)
• likewise for other 8 effects involved in blocking scheme
• remaining 15 - 9 effects are estimated as if there were no blocking (i.e. all data, and N = 48)
• overall treatment SS is the sum of squared l.s. estimates, each multiplied by the number of data
values used in that estimate
• residual
– df = 47 (c.t.) - 11 (reps and blocks) - 15 (effects)
– SS also by subtraction
6. Pretend now that only 8 measurements could be made in the entire experiment. Do this by selecting
a single factorial effect, splitting replicate number 1 into two blocks, and ignoring one of the blocks.
What “strings” of factorial effects can be estimated from your fractional factorial? Compute these
estimates.
• Suppose I=+ABCD is selected
treatment
(1)
cd
bd
bc
ad
ac
ab
abcd
replicate
1
48.87
49.84
37.28
51.16
50.14
63.10
45.56
50.67
• estimable strings are A+BCD, B+ACD , C+ABD , ... 8 in all
• estimation:
– for example, α̂ = [−48.87 − 49.84 − ... + 50.14 + 63.10 + ...]/8
– but this actually estimates α + (βγδ) ...
7. Now suppose this “half-rep” extends throughout all 3 rep’s; you have only 8 treatments represented
in the design, but each of these give you 3 measurements in an unblocked design. What are the leastsquares estimates of the 7 estimable effect strings, and the sum-of-squares and degrees of freedom
for residual variation in an ANOVA.
• still using I = +ABCD,
data
treatment
(1)
cd
bd
bc
ad
ac
ab
abcd
48.87
49.84
37.28
51.16
50.14
63.10
45.56
50.67
53.78
47.19
46.81
49.53
44.70
58.01
50.02
45.38
54.37
51.28
44.87
52.26
53.37
66.53
52.21
56.48
• effect strings are orthogonal to each other, estimates are as usual:
– e.g., α̂ =[ - 48.87 - 53.78 - (all others at low level of A) +50.14 + 44.70 + (all others at low
level of A) ] / 24
– still really estimates α + (βγδ)
source
d.f. S.S.
treatments 7
24α̂2 + ... (6 more terms)
residual
diff diff
P
2
c.t.
23
all (y − ȳ)
8. Continuing the last part, suppose these 24 observations were obtained in a blocked experiment, with
each rep constituting one block. (Each block has the same 8 treatments.) What effect will this
have on the estimates of the 7 estimable effect strings? Compute the sum-of-squares and degrees of
freedom for residual variation, assuming fixed blocks and no block-by-treatment interaction.
• still using I = +ABCD,
treatment
(1)
cd
bd
bc
ad
ac
ab
abcd
1
48.87
49.84
37.28
51.16
50.14
63.10
45.56
50.67
replicate
2
3
53.78 54.37
47.19 51.28
46.81 44.87
49.53 52.26
44.70 53.37
58.01 66.53
50.02 52.21
45.38 56.48
• blocks are orthogonal to estimable effect strings, so estimates are constructed as if blocks aren’t
used
• ANOVA is as in the last problem, except for the addition of a line for blocks (and so reduction
of “residual” from before)
source
blocks
treatments
residual
c.t.
d.f.
2
7
diff
23
S.S.
P3
8(ȳblock i − ȳ)2
24α̂ + ... (6 more terms)
diff
P
2
all (y − ȳ)
i=1
2
9. Finally, consider a 25 problems (not the one discussed above) in which effects BCE and ADE are used
as “splitting” contrasts to divide treatments into groups of size 8. Select the group of 8 treatments
for which BCE is -1 for all treatments and ADE is +1 for all treatments for an unreplicated fractional
factorial experiment.
( Note that the full generating relation for this fraction is I = −BCE = +ADE = −ABCD)
• List the treatments (by name, e.g. “abd”) that are included in this fraction.
. a, d, be, ce, abc, bcd, abde, acde
• List the effect strings that contain main effects and are estimable based on this fraction.
. A − ABCE + DE − BCD, B − CE + ABDE − ACD, C − BE + ACDE − ABD, D −
BCDE + AE − ABC, E − BC + AD − ABCDE
• Double this experiment by adding a second fraction of the same size, generated by the same
factorial effects with different signs. Select the second fraction to maximize the resolution of the
resulting combined design. Write the full generating relation for this second fraction (including
signs and generalized interactions).
. I = +BCE = −ADE = −ABCD
• Taken together, these two fractions form a 1/2 fraction. List all effect strings that contain
main effects and can be estimated from this design. Suppose the two smaller fractions must be
treated as blocks in this overall design; which of the individual effects or effect strings (if any)
are confounded with blocks in this design?
. A − BCD, B − ACD, C − ABD, D − ABC, E − ABCDE; BCE − ADE is confounded
with blocks.