STAT 510
Homework 4
Due Date: 11:00 A.M., Wednesday, February 10
1. Consider a completely randomized experiment in which a total of 10 rats were randomly assigned to
5 treatment groups with 2 rats in each treatment group. Suppose the different treatments correspond
to different doses of a drug in milliliters per gram of body weight as indicated in the following table.
Treatment
Dose of Drug (mL/g)
1
0
2
2
3
4
4
8
5
16
Suppose for i = 1, . . . , 5 and j = 1, 2, yij is the weight at the end of the study of the jth rat from the
ith treatment group. Furthermore, suppose
yij = µi + ij ,
where µ1 , . . . , µ5 are unknown parameters and the ij terms are iid N (0, σ 2 ) for some unknown
σ 2 > 0. Use the R code and partial output provided below to answer the following questions.
>
>
>
>
>
>
d=rep(c(0,2,4,8,16),each=2)
#y is the data vector with entries ordered to appropriately
#match the vector d.
dose=factor(d)
o1=lm(y˜dose)
summary(o1)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 351.000
6.576 53.372 4.37e-08 ***
dose2
-10.000
9.301 -1.075 0.331406
dose4
-6.000
9.301 -0.645 0.547277
dose8
-17.000
9.301 -1.828 0.127119
dose16
-70.500
9.301 -7.580 0.000634 ***
> anova(o1)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value
dose
6505.6
Residuals
432.5
> is.numeric(d)
[1] TRUE
> o2=lm(y˜d)
> anova(o2)
Pr(>F)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value
d
5899.6
Residuals
1038.5
Pr(>F)
> o3=lm(y˜d+dose)
> anova(o3)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value
Pr(>F)
d
0.0004245 ***
dose
0.1907591
Residuals
(a) Provide the numerical value of the BLUE of µ1 .
(b) Provide the numerical value of the BLUE of µ2 .
(c) Determine the standard error of the BLUE of µ2 .
(d) Conduct a test of H0 : µ1 = µ2 . Provide a test statistic, the distribution of that test statistic (be
very precise), a p-value, and a conclusion.
(e) Provide an F -statistic for testing H0 : µ3 = µ4 .
(f) Does a simple linear regression model with body weight as a response and dose as a quantitative
explanatory variable fit these data adequately? Provide a test statistic, its degrees of freedom, a
p-value, and a conclusion.
(g) Provide a matrix C and a vector d so that the null hypothesis of the test in part (f) may be written
as H0 : Cβ = d, where β = (µ1 , . . . , µ5 )0 .
(h) Fill in the missing entries in the ANOVA table produced by the R command anova(o3). (This
is the last R command in the provided code.)
2. Suppose X is an n × p matrix and B is a p × p non-singular matrix. Prove that
C (X) = C XB −1 .
3. Consider the simple linear regression model
yi = β0 + β1 xi + i
i = 1, . . . , n
where 1 , . . . , n are iid and have a normal distribution N 0, σ 2 and β0 ,β1 , and σ 2 > 0 are unknown
parameters. The model matrix for this linear model can be written as X = [1, x], where 1 is an n × 1
vector of ones and x = (x1 , . . . , xn )0 .
(a) We have learned that the least squares estimator of β in a general linear model with a full-rank
−1
matrix X is given by β̂ = (X 0 X) X 0 y. Simplify this expression for the special case of
simple linear regression to obtain an expression for the least squares estimators of β0 and β1 .
Express your final answers using summation notation.
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(b) There are some computational advantages to working with a model matrix whose columns are
orthogonal. For the simple linear regression problem consider the model matrix
W = [1, x − x̄· 1]
This design matrix is obtained by “centering” the explanatory variable around its mean x̄· =
P
n
−1
= W . It will follow that
i=1 xi /n. Find a matrix B so that XB
Xβ = XB −1 Bβ = W α
where Bβ = α.
(c) Derive expressions for the least squares estimators of α0 and α1 (where α = (α0 , α1 )0 from part
−1
(b)) using α̂ = (W 0 W ) W 0 y.
(d) Multiply α̂ from part (c) by B −1 from part (b) to obtain expressions for β̂0 and β̂1 .
(e) Show that your answer to part (a) matches your answer to part (d).
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