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STOCKHOLM UNIVERSITY

Department of StatisticsPar Stockhammar

Solutions to the Exam in Econometrics, Part I, 2011-05-23

Problem 1

a) n = 88

yi = 25 835.05

xi = 16 462.34y2i = 8 500 750.6

x2i = 3 329 789.6

xiyi = 5 209 990.7

(xi x) (yi y) = 377 534.76

(yi y)2 = 917 854.51

(xi x)

2 = 250 144.32

u2i = 348 053.43

2 = (xi x) (yi y)(xi x)

2 =

377534

250144= 1.5093

1 = y 2x= 2583588

1.5093 16462

88 = 11.244

The estimate 1.5093 of2 means that an increase (decrease) in house size xi ofone square meter is associated on average with an increase (decrease) in houseprice of 1.5093 thousands of dollars, or 1 509.3 dollars. The estimate 11.2438of 1 means that the average house price when xi = 0 is 11.244 thousands ofdollars, or 11 244 dollars

b)2 = u2in2 = 34805386 = 4 047.13c)ESS= T SS RSS=

(yi y)

2u2i = 917 854.51 348 053.43 = 569

801.

r2 = ES S

T SS =

569801

917854 = 0.6208

Interpretation ofr2 = 0.6208: The value of 0.6208 indicates that 62.08 percentof the total sample (or observed) variation in yi (house prices) is explained bythe regressorxi (house size, measured in square meters).

d) The 95% confidence interval is given by:

2 t0.025(86) 2that is:

1.5093 1.988 0.1272

and the confidence interval is (1.256; 1.762).

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e) The test statistic is:

t=2 22

where

2 =

2(xi x)

2 =

4047.13

250144 = 0.1272

The observed tvalue becomes:

tobs= 1.5093 1

0.1272 = 4.004

The decision rule is: reject H0 if |tobs| > t0.025(86) = 1.988. Decision: Reject

H0.

Meaning of test outcome: Rejection of the null hypothesis2 = 1 against thealternative hypothesis 2= 1 means that the sample evidence favours that theslope is different from 1.

f) The estimate ofyi given xi = 100 is given by:

(yi |xi = 100) = 11.244 + 1.5093 100 = 162.174g) The 95% confidence interval for the estimate in f) is given by:

yi t0.025(86)2 1n

+ (x0 x)2

(xi x)2

which here becomes:

162.174 1.988

4047.13 188

+(100 187.07)2

250144

That is, the 95% confidence interval for the trueyi |xi = 100 is 162.17425.817or between 136 357 and 187 991 dollars.

h) Use the following test statistic to test whether the incremental contributionof the new variable is significant:

F = (R2new R

2old)/numberofnewregressors

(1 R2new)/(n numberofparametersinthe newmodel)

which here becomes:

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F = (0.7 0.6208)/1(1 0.7)/(88 3)

= 22.44> F(1, 85)f orallvalueson.

That is the contribution of the new variable is significant and should be kept inthe model.

Problem 2

a) False. Heteroscedasticity occurs when the variance of the disturbance termis not the same for all observations

b) False. The estimates are still unbiased but inconsistent.

c) False. R2 will probably increase even if we include irrelevant variables.

d) False.

e) True.

f) True.

g) False. It is a test for autocorrelation in the residuals.

h) True.

Problem 3

a) Anvand Durbin-Watsons test:H0: Residualerna ar inte autokorrelerade (oberoende)H1: Residualerna ar negativt autokorrelerade

Testvariabel:

d=

nt=2(ut ut1)2n

t=1u2t = 27.30789.7924 = 2.79Here: dU,0.05= 1.777, dL,0.05= 0.559, dU,0.01= 1.489 and dL,0.01= 0.345 whichmeans that 4 dU,0.05< dobs< 4 dL,0.05 and 4 dU,0.01< dobs< 4 dL,0.01,the test is inconclusive.

b) If positive residuals tends to be followed by negative residuals one periodin time later, then the residuals are negatively autocorrelated. This might be

solved using the first difference method or the Cochrane-Orcutt procedure (seechapter 12)

c)2 = u2tn3 = 9.79245 1.958. We test H0 : 2 = 1 vs H0 : 2 = 1 using thetest statistic:

2 =2 (n 3)

2 2(n 3)

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which here becomes 2obs=1.958 5

1 = 9.7924

and20.975(5) = 0.831211<

2obs<

20.025(5) = 12.8325

d) A 95% confidence interval for 2 is given by:

(n 3)22/2

2 (n 3) 221/2

or

5 1.95812.8325

2 5 1.9580.831211

That is, with 95% confidence the true value of2 lies between 0.763 and 11.778.

e) Use the Fstatistic:

Fi = R2xix2x3xk/(k 2)

(1 R2xix2x3xk)/(n k+ 1)=

0.84/1

(1 0.84)/6 = 31.5

Fi,obs= 31.5> F(1, 6) for all values on .

Problem 4

The OLS estimate,,is given by: =

(xi x) (yi y)

(xi x)2 =

(xi w2 (x w2)) (yi w1 (y w1))

(xi w2 (x w2))2

=

(xi x) (yi y)

(xi x)2

The OLS estimate of becomes:= y w1 (x w2)b) According to a) is the OLS estimate ofunbiased. This is however not the

case for:E() = Ey w1 (x w2)= E

y x+E(w1+w2) = w1+w2

That is, E() =.c) Measurement error inyi. The variance ofincreases by the value of 1(xix)2 .

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