Solution Notes for Homework Assignment 1
1. (3 points) Some relevant sentences from the manuscript are on page 267:
• “The experimental design was Completely Randomized Design with three replications using 11 insecticides and control.”
• “The 0.2 ml of each diluted insecticide was dropped inside the test tubes ... [and] gently rolled to produce uniformly insecticide-coated wells ...”
• “Ten adults were used for each treatment. One adult parasitoid was released into each insecticide-coated test tube.”
(a) “adult parasitoid,” together with a test tube.
(b) insecticides, along with a control.
(c) A brief description of the experimental protocol is given, including statements about uniformity of temperature and food source, but there is little specific detail (which is, unfortunately, common in many journal descriptions). It also sounds as though a standard procedure was used to coat each tube with insecticide, but randomization is not mentioned. There is replication in the experiment, apparently 30 units per treatment, but “3 replicates” of “10 adults” isn’t explained.
2. (2 points) Response was apparently binomial (percent of 30 units in each case); a generlized linear model with an appropriate link function could be used for analysis.
3. (5 points) Let x i
, i = 1 ...
12 be the 13-element row vector with first and ( i + 1) st elements 1, and all others 0. These are the unique rows of X , so any linear combination of the rows of X can be expressed as a linear combination of x i
’s, and vice versa. Let θ = ( α, τ
1
, ..., τ
12
)
0
.
l
0
θ is estimable iff l can be written as a linear combination of x i
’s.
(a) α = l
0
θ because with
{ l }
1
{ l }
1
= 1 and all other elements 0. If l =
= 1. But this implies P
13 i =2
{ l } i
P
12 i =1 w i x i
, then
= 1; not true for the l
P
12 i =1
. Hence α w i
= 1 is not estimable.
(b) α + τ
1
= l
0
θ where l = x
1
. Hence α + τ
1 is estimable.
(c) α + then
τ
1
+ τ
2
= l
0
θ where { l }
1
= { l }
2
= { l }
3
= 1 and all other elements 0. If l = P
12 i =1 w i x i
,
P
12 i =1 w i
= 1 because { l }
1
= 1. But this implies P
13 i =2
{ l } i
= 1; not true for the l .
Hence α + τ
1
+ τ
2 is not estimable.
(d) τ
1
− τ
2
= l
0
θ where l = x
1
− x
2
. Hence τ
1
− τ
2 is estimable.
(e) τ
1
+ τ
2
− 2 τ
3
= l
0
θ where l = x
1
+ x
2
− 2 x
3
. Hence τ
1
+ τ
2
− 2 τ
3 is estimable.
You could equivalently establish non-estimability for (a) and (c) by noting that Xd = 0 for d = ( − 1 , 1 , 1 , ..., 1)
0 and showing that c
0 d = 0 for the required c .