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This text provides information about Desklib, an online library for study material. It includes solved assignments, essays, dissertations and more. The article also highlights the usefulness of Hotelling T2 distribution, Q-Q plots, and confidence intervals for linear combinations of linear variables. It also explains how to calculate the test statistic and confidence interval for a given model.

Revision Questions 1
Exam Revision November 2017
(Student Name)
Course
Professor
Institution
City and State of the University
Date

Revision Questions 2
Question 1
1.
i)
X~ N p ( U P∗1 , ε P∗P )
Let, Am∗ p be a constant matrix
Let Y m∗1 = Am∗ p XP∗1
yi = ai∗1
[ Y 1
⋮
Y nm ] = [a11 … a1 p
⋮ ¿ ¿ …¿amp ¿ ] [ X1
⋮
X P ]
This implies that
yi = ai1
[ y
⋮
ym ] = [a11 X1+ a12 X 2 … a1 p X p
⋮ ¿ ¿ …¿amp X p ¿ ]
Taking expectations on both sides
[ E( y1 )
⋮
E( ym ) ] = ¿
= [ E( y1 )
⋮
E( ym ) ] = [ a11 u1 +a22 u2+¿ … ¿¿ ⋮ ¿¿ ⋮ ¿ am , 1 u1 +am ,2 u2+¿ …¿amp up ¿ ]
This is the
E(Y) = AE(x) = Au
Where A is mxp vector 2
U is px1 vector

Revision Questions 3
Similarly,
Var(Y) = Var[Ax]
= A*Var (x) A'
Var(Y) = A∑ A'
Y = AX ~ Nm(AU , A∑ A')
Here Y is a mx1 vector with mean vector of AU and variance and covariance vector A ∑ A'
ii)
If A=
[ 1
n … 1
n
⋮ … ⋮
1
n … 1
n ]1 xp
[ x1
⋮
xn ] n∗1
= ∑
i=1
n
Xi
n
The mean vector of X is using
[ 1
n … 1
n
⋮ … ⋮
1
n … 1
n ]1 xn
[ u1
⋮
un ]u∗1
Hence
U' = ∑
i=1
n
ui
n
Variance of X is ε '[variance]

Revision Questions 4
ε '=
[ 1
n … 1
n
⋮ … ⋮
1
n … 1
n ] [ σ1
2 … σ n
2
⋮ … ⋮
σ1 n
2 … σ n
2 ] [ 1
n
⋮
1
n ] =
[ 1
n … 1
n
⋮ … ⋮
1
n … 1
n ] [ σ1
2
n +
∑
1=1
n
σ1 i
2
n … σ 1n
2
n +
∑
1=1
n
σ1 n
2
n
⋮ … ⋮
σn
2
n +
∑
1=1
n
σ ¿
2
n … σn
2
n +
∑
1=1
n
σn
2
n
]
ε '=∑
1=1
n
σi
2
n2
+ 2
n2 ∑
i ≠ j
n
σ1 i
ε '= ∑
1=1
n
σi
2
n2
+ 2∑
i ≠ j
n
σ1 i
n2
X ~ N ( μ' , ε' )
Then X is the univariate normal distribution with mean = u' and variance = ε '
2.
i) Hotelling T 2 distribution is a multivariate distribution which is proportional to an F-
distribution. The distribution is a generalization of students t-statistics used in
multivariate hypothesis testing.
ii)
Hotelling T-squared distribution has many applications. The distribution can be used to
test the hypothesis that pregnant mothers take the required nutritional diet during their 9 –
months of pregnancy. If an organization wants to establish whether the pregnant mothers take the
required nutritional diet during their 9 – months of pregnancy, a sample of pregnant women is
taken and the mean of their nutrient intake is calculated and tests were done as below:
H0 : μ=μ0 (women take in all the required food nutrients)
H1 : μ ≠ μ0 (women do not at least one of the required food nutrients)

Revision Questions 5
Assumptions made for the univariate case
a) Only one single nutritional component is measured
b) Homoskedasticity (Equality of variance) is assumed
c) The variables are independent (Independence assumption )
d) Normality of subjects is assumed
Since this is a univariate test, the test statistics are calculated as below:
t=
x−μ0
√ s2
n
~ tn−1
The null hypothesis follows the t-distribution.
Decision rule: Reject the null hypothesis id the observed/calculated t value is greater than the t
critical value
|t|> tn−1 , α
2
Summary
In question 2 part(ii) , I have given a case study or example where Hotelling t squared
distribution is practically applied then I have constructed possible hypothesis from the case
study . I also stated all the assumptions of Hotelling t squared , test statistic and he distribution it
follows.
3.
Usefulness of a Q-Q plot
i. Graphic representation of Q-Q plot assists in testing for normality
ii. The presence or absence of outliers in Q-Q plot assists in knowing the how
symmetrical a data sequence is to the normal distribution
Construction of Q-Q plot

Revision Questions 6
i. Arrange values in an ascending order
ii. Draw a normal distribution curve into n+1 segments
iii. Find the z-value for every segment
4.
Simultaneous (1- α)*100% T 2 the confidence interval is used to calculate a confidence interval
for linear combinations of linear variables.
y j ± √ p(n−1)
n− p F p ,n− p , α
√ SY
2
n
Bonferroni (1- α )*100% confidence interval is used to calculate confidence interval s or
individual variables.
y j
± tn −1 , α
2 p √ SY
2
n
Number 2
1.
Random variables T 1, T 2, T 3, T 4 are dependent given ,
X= ( T 1
⋮
T n
) ~ N
( μ1
⋮
μn
, ϵ
) y = ( μ1
⋮
μn
)
H0: μ = μ1+μ2+ μ3
3 ⟹ 3 μ1- μ2- μ3-μ4
Thus
A = (3,-1,-1,-1)
i)
When n is small, we cannot use the normality test , so
Let X1 → X2 be a 4-dimensional sample

Revision Questions 7
Let Y 1= AX1, i= I 1(n)
E( Y i) = A.E( Xi) = A. μ = 0
We can estimate, ^σ y
2 = 1
n−1 ∑
i=1
m
( yi− y)2
= 1
n−1 ∑
i=1
m
( A Xi −A X )2
= 1
n−1 ∑
i=1
m
( A Xi −A X )(A Xi− A X )'
= 1
n−1 ∑
i=1
m
A (Xi −X )(Xi− X )' A'
= A [ ∑
i=1
m
A ( Xi−X )(Xi −X )'
] A'
According to standard t-test ;
T =
Y
( ^σ y
√ n ) ~ tn−1
(T=
Y
( ^σ y
√n ) = √n A X
^σ y
)
The technique is to transform Xis to univariates by the use standard t-test
ii)
When n is large, the same technique as in part (i) above is used but with an additional result.
For large n, S= 1
n−1 ∑
i=1
m
( X i−X )( Xi −X )' ⟶ ε almost surely
Through continuous mapping theorem,
^σ y= √ AS A' ⟶ √ A ∑ A' almost surely

Revision Questions 8
Using continuous mapping theorem which states that if x⟶a with probability p and g() is a
function with finite discontinuities then g' ¿) x⟶ g(a) almost surely
A X ~ N(0, A ∑ A'
n )= √ n A X
√ A ∑ A' ~ N(0,1) and √ A ∑ A'
√ AS A' ⟶ 1 almost surely
So, ( √n A Y
√ A ∑ A' ) ( √ A ∑ A'
√ AS A' ) = √n A X
√ AS A' ~ N(0,1)
2.
We use t-distribution in this question
Since this is a univariate test, the test statistics is calculated as below:
t=
x−μ0
√ s2
n
~ tn−1
The null hypothesis follows the t-distribution.
Decision rule: Reject the null hypothesis id the observed/calculated t value is greater than the t
critical value
|t|> tn−1 , α
2
Number 3
1.
Y 1 = e1
' X = -0.383X1 +-0.924 X2
Y 2 = e1
' X = -0.924 X1 +-0.383 X2
2.
Principal component Y 1
λ1
λ1 + λ2
= 5.828
5.828+0.172 = 0.9713

Revision Questions 9
Principal component Y 3
λ2
λ1 + λ2
= 0.172
5.828+0.172 = 0.0287
3.
No correlation between X and Y 2 . The principal component Y 2 explains 2.87% of the total
population variance.
Question 4
1.
The equation can be written down in a matrix form as;
Y = X β + ε
Where
Y = ( y1 , y2 … . yn)'
X= X = [1 X21 X31
⋮ ⋮ ⋮
1 X2 n X3 n ] β = [ β0
⋮
β2 ] ε = [ε0
⋮
ε2 ]
Thus the least estimators of β are
^β = (X' X )−1 X'Y , ^Y = X ^β
^εreduced = (y- ^y) ⟹ ∑ εi
2 = ( y− ^y)1(y- ^y)
The degrees of freedom for error sum of squares are -3
Test statistic
F =
(error ∑ of squares for reduced model−error ∑ of squares for full model)
( Error ∑ of squares for full model
n−4 )

Revision Questions 10
F=
( y − ^y)1 ( y − ^y)
( X β +ε
n−4 )
2.
Suppose β2=β3= 0
Then,
y1 = β0 + β2 X1 i + ε i
X = ¿
Then
^β = (X' X )−1 X'Y
^Y = X ^β
^εreduced = (y- ^y) ⟹ ∑ εi
2 = ( y− ^y)1(y- ^y) …..
The test statistic, F is calculated as
F=
(∑ εi
2 reduced−∑ ε i
2 full)
2
((∑ εi
2 full)
n−4 )
3.
The confidence interval for β2−β3 is calculated as:
{√ ^σ2
√ a' (X ' X )−1 a}tn−4 ± a' ^β
Here
a=[0 0 1−1]'
x is a matrix design under full model
^σ 2 is ∑ εi
2 full
n−4

Revision Questions 11
^β is estimated in full model
tn−4 is critical value of t distribution . At 95% confidence interval, tn−4 is calculated as;
P(t> tn−4) = 0.025