Answers to Homework #5 - Statistics

7. If β = (β 1 , β 2 , β 3 , β 4 ), then we can make
⎛
⎞
1 0 1 0
1 0 1 0
1 0 0 1
1 0 0 1
X =
.
0 1 0 1
0 1 0 1
⎜
⎟
⎝ 0 1 1 0 ⎠
0 1 1 0
(b) There are three dimensions in the row space. We can use rows to find estimable
linear combinations! Two are: β 1 +β 3 , and β 1 +β 4 . We can not estimate β 1 because
(1, 0, 0, 0) is not in the row space. Similarly, we can’t estimate β 1 + β 2 .
(c) We can simply use the first three dummies in the model, so that ˜β = (β 1 , β 2 , β 3 ),
and
⎛ ⎞
1 0 1
1 0 1
1 0 0
1 0 0
˜X =
.
0 1 0
0 1 0
⎜ ⎟
⎝ 0 1 1 ⎠
0 1 1
(d) β 1 is the expected tumor size after two weeks of Treatment A, low dose; β 2 is the
expected tumor size after two weeks of Treatment B, low dose; β 3 is the expected
increase in tumor size after two weeks of treatment using high dose compared to
low dose, for either A or B.
8. Let x = (1, 2, 3, 2, 3, 4) ′ and d a = (1, 1, 1, 0, 0, 0) and 1 = (1, 1, 1, 1, 1, 1). Then we
want to project the response y onto the linear space L(1, d a , x). We know that we
will have “confounding” because the pot B plants are given more fertilizer. Let’s
compute the VIF for fertilizer. To do this, we get the R 2 for the model using x
as the response and d a (and 1) as predictors. We get ˆx = (2, 2, 2, 3, 3, 3) ′ , so the
SSE = 4. Also, SST = 5.5, and R 2 = 3/11. So V = 11/8.
9. Note that the design matrix is not full rank; we expect to get a zero eigenvalue. We
solve
⎛
⎞
4 − λ 2 2
⎜
⎟
det ⎝ 2 2 − λ 0 ⎠ = λ(λ − 2)(λ − 6).
2 0 2 − λ