Econ 299
Tutor Questions Chapter 6
University of Alberta
Econ 299 – Tutor Questions Chapter 6
Use the following data for the next 2 questions:
Year / 1999 / 2000 / 2001 / 2002 / 2003Car Sales (S) (Hundreds) / 6 / 4 / 5 / 7 / 3
Gas Price (P) (cents) / 60 / 65 / 68 / 72 / 75
1) Given the formula:
Car Salesi=β1+ β2Gas Pricei+єi
Test the following hypotheses:
a) If gas were free, 800 cars would be sold (α = 0.05)
Sbar =∑s/n
= (6+4+5+7+3)/5
=5
Pbar =∑p/n
=(60+65+68+72+75)/5
=68
β2hat=∑(P-Pbar)(S-Sbar)/(P-Pbar)2
=(6-5)(60-68)+(4-5)(65-68)+(5-5)(68-68)+(7-5)(72-68)+(3-5)(75-68)/
=(60-68)2+(65-68)2+(68-68)2+(72-68)2+(75-68)2
=-8+3+0+8-14/(64+9+0+16+49)
=-11/138
=-0.0800
β1hat=Sbar- β2hat(Pbar)
=5-0.080(68)
=10.4
Shat1 =10.4-0.08(60)=5.6
Shat2 =10.4-0.08(65)=5.2
Shat3 =10.4-0.08(68)=5.0
Shat4 =10.4-0.08(72)=4.6
Shat5 =10.4-0.08(75)=4.4
Sigmahat2 =∑ehat2/(n-2)
=(6-5.6)2+(4-5.2)2+(5-5)2+(7-4.6)2+(3-4.4)2/3
=(0.16+1.44+0+5.76+1.96)/3
=3.11
Varhat(β1hat)=sigmahat2 ∑p2/N∑(p-pbar)2
=3.1(23258)/5(138)
=104
Se(β1hat) = {Varhat(β1hat)}1/2
= (104)1/2
= 10.2
H0: β1=8
Ha: β1≠8
t*(n-2=3,5%)= 3.18 (from table)
test t = (β1hat- β1)/se(β1)
= (10.4-8)/10.2
= 2.4/10.2
= 0.235
-t*>t>t*, Do not reject H0 . If Gas where free, it is consistent with our data that 800 cars could be bought.
b) Gas prices have no effect on car sales (α = 0.01)
H0: β2=0
Ha: β2≠0
Varhat (β2hat) =sigmahat2 /∑(p-pbar)2
=3.1/(138)
=0.0225
Se(β2hat) = {Varhat(β2hat)}1/2
= (0.0255)1/2
= 0.160
t*(n-2=3, 1%)=5.84 (from table)
test t = (β2hat-β2)/se(β2)
= (-0.080-0)/0.160
= -0.500
-t>t>t*, Do not reject H0 . It is consistent with our data that gas prices could have no effect on Car sales.
Calculate and interpret the following confidence intervals:
c) 90% CI for β2hat
t*(n-2=3, 10%)=2.35 (from table)
CI=[ β2hat ± t*(se(β2hat))
=[-0.080±2.35(0.022) 1/2]
=[-0.429, 0.269]
In repreated samples, 90% of these confidence intervals will contain the true β2 value.
d) 95% CI for the mean of car sales
t*(n-1=4, 5%)=2.78 (mean has n-1 df)
sd(s) =∑(s-sbar)2/n-1
=[(6-5)2+(4-5)2+(5-5)2+(7-5)2+(3-5)2]/n-1
=(1+1+0+4+4)/4
=10/4
=2.50
Sd(sbar) = sd(s)/n1/2
=2.50/51/2
=1.12
CI=[sbar±t*(sd(sbar))
=[5±2.78(1.12)]
=[-1.88, 8.11]
In repreated samples, 95% of these confidence intervals will contain the true average car sales.
2) Indicate whether the following statements are true, false, or uncertain. Justify your claims.
a) My loglog model has an R2 of 0.43. My linear model has an R2 of 0.47. Therefore, the linear model is inferior.
False. R2 is a good measure of goodness of fit, but it cannot be used to compare between models.
b) There is a 95% chance that the actual value is contained in the 95% CI.
False. Given repeated Samples, 95% of confidence intervals will contain the actual parameter.
c) The null hypothesis is never accepted.
True. We either reject or do not reject the null hypothesis. It is never accepted.
d) 1-R2 is the portion of the variation not explained by the model.
True. R2 is the portion explained by the model, therefore 1-R2 is the portion not explained by the model.
3) Why is it that OLS is BLUE? (12 marks)
Best: OLS has the smallest variance of any linear unbiased estimator.
Linear: In estimating the coefficients, Y is never taken to a power more than one nor are there any non-linear functions of Y . Therefore OLS is linear in Y
Unbiased: The expected values of the estimated coefficients (βhat’s) are the actual coefficients (β’s)
Estimator: OLS estimates the coefficients (β’s)
4) Prove that the OLS estimate of β1 is unbiased. (12 marks)
(Hint: see text for proof that β2 is unbiased.)
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