ECON107 ASSIGNMENT 1 Answer Key Question 1 (50 marks)

1. ∑Xt2 = 32+22+12+(-1)2+02 = 15

∑XtYt = (5)(3)+(2)(2)+(3)(1)+(2)(-1)+(-2)(0) = 20

∑Xt = 3+2+1+(-1)+0 = 5

∑Yt = 5+2+3+2+(-2) = 10

𝑋 = Σ𝑋𝑡5

= 1

𝑌 = Σ𝑌𝑡5

= 2

2. 𝛽1� = nΣ𝑋𝑡𝑌𝑡 − Σ𝑋𝑡Σ𝑌𝑡nΣ𝑋𝑡2 – (Σ𝑋𝑡)2

R = (5)(20)−(5)(10)(5)(15)−(52)

= 1

𝛽0� = 𝑌 − 𝛽1�.𝑋 = 1

3.

-2

-1

0

1

2

3

4

5

-2 0 2 4

Yt

Xt

(1,2)

4. 𝛽1: An unit increase in X leads to an average of 𝛽1 unit increase in Y

𝛽0: an average value of Y when X is 0.

5. See above. 6. ∑ 𝑒𝑡5

𝑡=1 = ∑ (𝑌𝑡5𝑡=1 − 𝑌𝑡� )

= ∑ (𝑌𝑡5𝑡=1 − 𝛽0� − 𝛽1�.𝑋𝑡)

= 1+(-1)+1+2+(-3) = 0

Standard error of estimate, 𝜎� = �∑ (𝑒𝑡)25𝑡=1𝑛−2

= �12+(−1)2+12+22+(−3)2

5−2

= 2.309

Coefficient of determination, r2 = 𝐸𝑆𝑆𝑇𝑆𝑆

= ∑ (𝑌𝑡�−5𝑡=1 𝑌)2

∑ (𝑌𝑡−5𝑡=1 𝑌)2

= 1026

= 0.385

38.5% of the total variation in Yt is explained by the regression model (or by X).

7. H0: 𝛽1=0 H1: 𝛽1≠0

t = 𝛽1�−𝛽1𝑆(𝛽1�)

= (𝛽1�−𝛽1)�∑𝑋𝑡2

𝜎� ~ t(n-k-1) ,

where 𝑆(𝛽1�) = 𝜎�

�∑𝑋𝑡2 = 0.7303. So t=1.3693

Reject H0 if |T| > t0.05 = 2.35336

∴ Do not reject H0

Question 2 (20 marks)

1. 𝛽1� = nΣ𝑋𝑖𝑌𝑖 − Σ𝑋𝑖Σ𝑌𝑖nΣ𝑋𝑖2 – (Σ𝑋𝑖)2

R = Σ(𝑋𝑖−𝑋)(𝑌𝑖−𝑌)Σ(𝑋𝑖−𝑋)2

Add 1 to the dependent variable: new 𝛽1� = Σ(𝑋𝑖−𝑋)(𝑌𝑖+1−(𝑌+1))Σ(𝑋𝑖−𝑋)2

= Σ(𝑋𝑖−𝑋)(𝑌𝑖−𝑌)Σ(𝑋𝑖−𝑋)2

∴ There is no change in 𝛽1�.

𝛽0� = 𝑌 − 𝛽1�.𝑋

Add 1 to the dependent variable: new 𝛽0� = 𝑌 + 1 − 𝛽1�.𝑋

∴ 𝛽0� will increase by 1 unit.

2. 𝛽1� = nΣ𝑋𝑖𝑌𝑖 − Σ𝑋𝑖Σ𝑌𝑖nΣ𝑋𝑖2 – (Σ𝑋𝑖)2

R = Σ(𝑋𝑖−𝑋)(𝑌𝑖−𝑌)Σ(𝑋𝑖−𝑋)2

Add 1 to the independent variable: new 𝛽1� = Σ(𝑋𝑖+1−(𝑋+1))(𝑌𝑖−𝑌)Σ(𝑋𝑖+1−(𝑋+1))2

= Σ(𝑋𝑖−𝑋)(𝑌𝑖−𝑌)Σ(𝑋𝑖−𝑋)2

∴ There is no change in 𝛽1�.

𝛽0� = 𝑌 − 𝛽1�.𝑋

Add 1 to the independent variable: new 𝛽0� = 𝑌 − 𝛽1�(𝑋 + 1)

= 𝑌 − 𝛽1�.𝑋 − 𝛽1�

∴ 𝛽0� will increase by 𝛽1� unit.

Question 3 (30 marks)

𝛽1� =Σ(𝑋𝑖 − 𝑋)(𝑌𝑖 − 𝑌)

Σ(𝑋𝑖 − 𝑋)2

Since there is no variation in X, Σ(𝑋𝑖 − 𝑋)2 = 0. 𝛽1� is then undefined. As a result, residuals are not defined. So are RSS and ESS. Therefore, r2 is naturally undefined as well.

∴ The person is not justified as the regression model is not valid.

If observations were lined up at 45° from origin, 𝑋𝑖 = 𝑌𝑖

𝛽1� = Σ�𝑋𝑖−𝑋��𝑌𝑖−𝑌�

Σ�𝑋𝑖−𝑋�2

= Σ�𝑋𝑖−𝑋��𝑋𝑖−𝑋�

Σ�𝑋𝑖−𝑋�2 = 1

R2 = 𝐸𝑆𝑆𝑇𝑆𝑆

= ∑(𝑌𝚤�−𝑌)2

∑(𝑌𝑖−𝑌)2

= ∑(𝛽0�+𝛽1� .𝑋𝑖−𝑌)2

∑(𝑌𝑖−𝑌)2 (𝛽0�=0 as intercept is at origin)

= ∑(𝑋𝑖−𝑌)2

∑(𝑌𝑖−𝑌)2

= ∑(𝑌𝑖−𝑌)2

∑(𝑌𝑖−𝑌)2

= 1

∴ Regression line will be a ‘perfect fit’.