Sample Midterm with solutions -3775-CORRIGE
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University of Ottawa *
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Course
3775
Subject
Mathematics
Date
Jan 9, 2024
Type
Pages
3
Uploaded by MegaWorld9539
MAT 3375
Midterm test
solutions
Octobre 22
, 2019
Professor M. Alvo
Time: 80 minutes
Student number
:
Given name
:
Family name
:
This is an open book exam. Calculators are permitted. Answer all questions.
The test is out of 20.
1.
(10 points) The Tri-City O
ffi
ce Equipment Corporation sells an im-
ported copier on a franchise basis and performs preventive maintenance and
repair service on this copier.
The data below have been collected from 45
recent calls on users to perform routine preventive maintenance service; for
each call, X is the number of copiers serviced and Y is the total number of
minutes spent by the service person. Assume a first-order regression model
is appropriate.
Y
i
=
β
0
+
β
1
X
i
+
"
i
a. Complete the analysis of variance table below.
Solution: The SS add to the total as do the df. MS=SS/df.
b.
Compute the test statistic to test the null hypothesis that
β
1
= 0
against the alternative that
β
1
6
= 0. What is its distribution?
1
<un jeu de données
avait été fourni>
cette solution continue dans la table à la
page suivante
How would you carry out the test
with significance level alpha?
En bleu, des commentaires de ma part.
En rouge, des ajouts aux solutions.
F
=
MS
(
R
)
/MSE
= 325
.
97
/
0
.
3365 = 968
.
7
> F
↵
(1
,
43)
.
Here
F
↵
(1
,
43)
is the upper 100(1
-
↵
)% point of an F distribution with 1 and 43 degrees of
freedom.
c. Interpret
β
0
in your estimated regression function.
β
0
is the intercept in the regression model. It indicated the value of Y
when X=0 and is a start up value.
d. Obtain a point estimate of the mean service time when X = 5 copiers
are serviced.
ˆ
Y
=
b
0
+ 5
b
1
e. What are the usual assumptions made in linear model?
We assume
Y
i
=
β
0
+
β
1
X
i
+
"
i
where the
{
"
i
}
are i.i.d. normal (0
,
σ
2
) error terms.
Analysis of Variance Table
Source
SS
df
MS
Time
325.97
1
325.97
Error
14.47
43
0.3365
Total
340.44
44
2. (5 points) The plot below refers to the rate of crime against Number
of the population having at least a high school education for several counties
in the US. Carry out a test on whether or not Education is important as a
predictor by using the extra sum of squares principle.
Make the usual as-
sumptions for the linear regression model. Use the output data from Rstudio
below.
Model
SSE
df
Rate˜0
522098
84
Rate˜Education
2665
82
a. State the full and reduced models.
Solution: Full model:
Y
i
=
β
0
+
β
1
X
i
+
"
i
Reduced model:
Y
i
=
β
0
+
"
i
b.
Obtain the test statistic F* for the general linear test and state its
distribution.
F
⇤
=
SSE
(
R
)
-
SSE
(
F
)
df
R
-
df
F
SSE
(
F
)
df
F
2
solution (a) continue ici…….
F
968.7
<une visualization
avait été fournie>
Reduced.
Full
SS(“Reduced”) .. c’est
la norme carrée du
vecteur
(yi) - \bar y (1n)
qui montre la carence
du modèle “réduit”,
c’est à dire le
modèle sans \beta_1
“Full”..
c’est ce
qu’on a appelé
résidu, c’est à
dire la norme
carrée du vecteur
(yi) - (\hat yi), i.e.
l’erreur du grand
modèle
SS(Reduced) - SS(Full) = ce qu’on a appelé SS(Regr).
Faire le test avec niveau de confiance alpha veux dire rejeter l’hypothèse si la condition soulignée est vraie.
Text
y_i = epsilon_i
(on reconnait ça d’après le titre “Rate~0”
et aussi le fait qu’il y a 2 df, c’est à dire
2 paramètres à estimer, pour passer au
Full model.)
=
522098
-
2665
84
-
82
2665
82
= 7990
.
8
F
⇤
has an F distribution with 2 and 82 degrees of freedom
3. (5 points)The summary data as well as the ANOVA table that follow
show assessed value for property tax purposes (Y in thousand dollars) and
sales price (X, in thousand dollars) for a sample of 15 parcels of land for
industrial development sold recently in
arm’s length
transactions in a tax
district. The following R commands was used
>
fit= lm(formula = Sales ˜ Assessed)
>
summary(fit)
a. Obtain the regression equation. What is the estimate of the variance?
Solution:
ˆ
Y
=
-
5
.
812 + 2
.
542
X
b. Construct 95% confidence intervals for the coe
ffi
cients. Express your
answers in terms of the student coe
ffi
cients and specify the appropriate de-
grees of freedom.
Coe
ffi
cients
Estimate
S.E.
t-value
Intercept
-5.812
3.065
-1.896
Assessed
2.542
0.2245
11.3222
ANOVA
Source
SS
df
MS
F
Assessed
401.81
1
401.81
128.18
Residuals
40.75
13
3.13
CI for
β
0
:
-
5
.
812
±
t
↵
/
2
(3
.
065)
CI for
β
1
: 2
.
542
±
t
↵
/
2
(0
.
2245)
where is the
t
↵
/
2
is the upper 100(1
-
↵
/
2)% point of a Student-t distri-
bution with 13 degrees of freedom.
3
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