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ECON 2202 - LONG ANSWER QUESTIONS
TOPIC 5
Q5.1..
- Use
2
0
n
r
r
t
and α=0.10 to test significance of correlation (=linear relationship) between account
balance and number of transactions; r=-0.23, n=50, df=n-2.
2. H
0
: ρ = 0; H A
: ρ ≠ 0
- Reject H 0
if t 0
t α/2, n-
= t 0.05,
= 1.6759 or t 0
< - t α/
-t
α/
=-1.6759 t
α/
2 2
n
r
r
t
- Since –t α/
< t 0
< t α/
(-1.6759< - 1.6374 < 1.6759) do not reject the null hypothesis. A significant linear
relationship does not exist between account balance and number of transactions during the last month, at
the 10% significance level.
You might make a type II error, which occurs when you accept a null hypothesis which is false.
Q5.2..
a.
2
2
1
n x x
n xy x y
b
;
0 1
b y bx
b. ŷ = b
0
1
x = 1.5 + 1.6 x. Slope says that for every 1 unit increase in x, y increases by 1.6.
c. 1.
1
1 / 2 b
b t s
, 95% CI,
n
x
x
MSE
x n x
MSE
x x
MSE
s
b
2
2
2 2 2
1
; MSE=SSE/(n-k-
54 1. 5 * 14 1. 6 * 19. 5 1.
ˆ
0 1
2 2 2
SSE e y y y b y b xy
1
b
/ 2 , 1 0. 025 , 2
t t
nk
2
2
2 2 2
1
n
x
x
MSE
x n x
MSE
x x
MSE
s
b
1
1 / 2 b
b t s
=1.6±4.3027*0.8485. So the 95% CI for β 1
is (-2.0508, 5.2508).
X Y XY X
2
Y
2
Total 5.0 14 19.5 7.50 54
α/2=0.
d..
- Use
1
1
0
b
s
b
t
, α=0.05, n=4, k=
- Ho: β 1
=0 Ha: β 1
- Reject Ho if t 0
t α,/2, n-k-1,
= t 0.025, 2
=4.3027 or t 0
<- t α/2,
1
1
0
b
s
b
t
- Since - t
α,/2,
< t
0
< t
α,/
(-4.3027 < 1.8857 <4.3027) do not reject H
0
. Conclude the slope is
insignificant at the 5% significance level.
e. The alternative two tests are test of significance of the overall regression, and test of significance of
the correlation coefficient between x and y. Their respective test statistics are: F 0
= MSR/MSE and
2
n
r
r
t
.
Q5.3..
a. Relationship appears to be random or triangular.
Scatter Plot of x versus y
0
2
4
6
8
10
12
0 20 40 60 80 100
x
y
α/2=0.
1
1
0
b
s
b
t
- Since - t α,/2,
< t 0
< t α,/
(-3.4995< -0.5758 <3.4995) do not reject H 0
. Conclude the slope is
insignificant at the 5% significance level.
e. Since we didn’t reject H 0
, we could have made a Type II error (“accepting” H 0
when H 0
is false).
f. When x=10, then ŷ = b 0
x = 8.4401-0.0359 *10 =8.0811.
When x=xbar = ∑x/n=525/9=58.3333, then ŷ = b 0
x = 8.4401-0.0359 *58.3333 =6.
Q5.4..
Net Sales from 1988 to 1998
$
$100,
$200,
$300,
$400,
$500,
$600,
$700,
1986 1988 1990 1992 1994 1996 1998 2000
Working capital from 1988 to 1998
$
$50,
$100,
$150,
$200,
1986 1988 1990 1992 1994 1996 1998 2000
Net Sales versus Working Capital (1988 to 1998)
$
$100,
$200,
$300,
$400,
$500,
$600,
$700,
$0 $50,000 $100,000 $150,000 $200,
a. It makes sense to plot sales as the y variable (dependent variable) and working capital as the
independent variable. This scatter plot shows that there is a relatively strong, positive linear relationship
between the two variables. Note that if you plot each variable separately against time (time is always
independent), you also see a strong positive linear relationship between working capital and time, and
between net sales and time.
b. r = 0.9707. This measures the degree of linear relationship between Working Capital and Net Sales.
c..
- Use
2
0
n
r
r
t
to test +ve correlation between Working Capital and Net Sales, n=11, α=0.05.
2: H
0
: ρ ≤0; H
A
: ρ > 0 (The claim is the equivalent of a positive linear relationship).
3: t-distribution with α = 0.05 and df = n–2 = 11–2 = 9. t
0.05, 9
= 1.8331. Draw distribution and show the
rejection region on the left-tail. Mark the critical value -t 0.05, 9
Reject the null hypothesis if the t 0
< t α
= -1.8331, otherwise fail to reject.
2 2
0
n
r
r
t
5: Since t 0
t α
(12.1189 > -1.8331), do not reject the null hypothesis. Conclude that a positive linear
correlation exists between working capital and net sales at the 5% significance level.
2
2
/ 2
x x
x x
n
y t s
p
, 95% CI, x p
= 8, k=1, n=18, 1.
n k
SSE
s
2. ŷ = 200 + 150x = 200 + 150*8 = 1400
/ 2 , 1 0. 025 , 16
t t
nk
2
2
2
x x
x x
n
s
p
1400 2. 1199 *1.
1
ˆ
2
2
/ 2
x x
x x
n
y t s
p
. The 95% CI for y when x p
= 8 is is
d. The estimates differ for two reasons: the different values of x, and the fact the first prediction
interval is for the average or expected value of y, while the second is for a precise value (which will,
all else equal, have a larger standard error).
Q5.7..
a.
2 2
n x x x n y y y
n xy x y
r
xy
b. The correlation coefficient in a. tells us that there is a negative linear correlation between the dependent
and independent variables of the problem, though it is not very strong.
c..
- Test significance of correlation using
2
0
n
r
r
t
, α=0.05, n=
- Ho: ρ=0, Ha: ρ≠0.
- Reject Ho if t 0
t α/2,,n-
= -t 0.025,
= 2.0687 or t 0
< -t α/
t α/
= -2.0687 - t α/
2 2
0
n
r
r
t
- Since t 0
< - t α/
(-3 < -2.0687), reject Ho. This means the correlation is significantly different than
zero, and the two variables have a significant linear relationship.
d..
2
1
n x x x
n xy x y
b
0 1
b y bx
Population Model: y = β
0
1
x + ε
Sample Model : ŷ = b 0
x = 2.9 – 0.3 x
e. First way is to test significance of slope coefficient.
α/2=0.
- Use
1
1
0
b
s
b
t
, α=0.05, n=25, k=
- Ho: β 1
=0 Ha: β 1
- Reject Ho if t 0
t α/2, n-k-1,
= t 0.025, 23
=2.0687 or t 0
<- t α/2,
t α/
= -2.0687 - t α/
Second way is to test significance of overall regression.
- Use
MSE
MSR
F
0
, α=0.05, n=25, k=
- Ho: regression is not significant, Ha: regression is significant.
- Reject Ho if F 0
> F
α, (k, n-k-1)
= F
0.05, (1,2)
F
α,
Q5.8. This question will be taken up in class.
Q5.9..
- Use
2
2
/ 2
x x
x x
n
y t s
p
, n=20, k=1, s
ε
= sqrt (SSE/(n-k-1) = sqrt (800.25/18), x
p
= 80, α=0.10.
ˆ 9784 345. 50 * 9784 345. 50 * 80 17 , 856
p
y x
- 7341
/ 2 , 1 0. 05 , 18
t t
nk
2
2
2
x x
x x
n
s
p
17 , 856 1. 7341 * 1. 5076 17 , 856 2.
( )
1
ˆ
2
2
/ 2
x x
x x
n
y t s
p
. So the 90% CI for
the average value of y when x p
= 80 is (-17,858.6143, -17,853.3857)
Interpretation: there is a 90% probability the average value of y will be in (-17,858.6143, -17,853.3857) if x is
predicted to take the value 80.
α/2=0.
α=0.
Q5.11. For these solutions, you may get slightly different yet correct answers depending on whether you
use rounded values in calculations, or unrounded values.
a.
0 1
2
2
1
b y b x
n x x
n xy x y
b
b. Calculate SST and SSE.
SSE 2200 1. 1429 * 126 0. 7143 * 2648 164.
2200 9 *( 126 / 9 ) 436;
0 1
2
2 2 2
y b y b xy
SST y n y
c. 4.
n k
SSE
s MSE
; and
3448 9 *( 162 / 9 )
- 8487
( )
2
2 2 2
1
x nx
MSE
x x
MSE
s
b
d. Perform a test of the hypothesis H 0
: β 1
1. Use
1
1
0
b
s
b
t
, α=0.05 (the default value if none is given), n=9, k=1x
2. Ho: β 1
=10 Ha: β 1
3. Reject Ho if t 0
t α,/2, n-k-1,
= t 0.025, 7
=2.3646 or t 0
<- t α/2,
-t
α/
=-2.3646 t
α/
1
1 1
0
b
s
b
t
5. Since t
0
< - t
α,/
(-44.1716 < 2.3646) reject H
0
. Conclude the slope is not equal to 10 at the
5% significance level.
e.
ˆ 1. 1429 0. 7143 * 13 10.
0 1
p p
y b bx
α/2=0.
Q5.12..
a. OLS Assumptions:
Error terms are independent and normally distributed with zero mean and constant variance or ε
~ iid N(0,σ
2
equation is linear in the coefficients
independent variable is independent of error terms
b..
2
2
/ 2
1
ˆ
x x
x x
n
y t s
p
, x p
=17, n=42, and a 95% CI for the expected value
of y.
8 10 * 17 178
ˆ y
- t
α/2, n-k-
= t
0.025, 40
800
17 8
42
1
- 23
1
2
2
2
x x
x x
n
s
p
178 2. 0211 * 3. 9714 178 8.
1
ˆ
2
2
/ 2
x x
x x
n
y t s
p
so the
95% prediction interval for the expected value of y is (169.9735, 186.0265).
c..
2
2
/ 2
1
1
ˆ
x x
x x
n
y t s
p
, x p
=17, n=42, and a 95% CI for the value of y.
y ˆ 8 10 * 17 178
- t α/2, n-k-
= t 0.025, 40
800
17 8
42
1
- 23 1
1
1
2
2
2
x x
x x
n
s
p
178 2. 0211 * 11. 9115 178 24.
1
ˆ
2
2
/ 2
x x
x x
n
y t s
p
so the 95%
prediction interval for y given x p
=17 is (153.9256, 202.0744).
