Stock Watson 3U Exercise Solutions Chapter 5 Instructors PDF

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! ! ! ! Introduction!to!Econometrics!(3rd!Updated!Edition)! ! ! by! ! ! James!H.!Stock!and!Mark!W.!Watson! ! ! ! ! !

Solutions!to!End7of7Chapter!Exercises:!Chapter!5*" ! ! (This version August 17, 2014)

! ! ! ! ! ! ! ! ! ! ! *Limited!distribution:!For!Instructors!Only.!!Answers!to!all!odd=numbered! questions!are!provided!to!students!on!the!textbook!website.!!!If!you!find!errors!in! the!solutions,[email protected].!!

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 1 _____________________________________________________________________________________________________

! 5.1 (a) The 95% confidence interval for β1 is {−5.82 ± 1.96 × 2.21}, that is

−10. 152 ≤ β1 ≤ − 1. 4884. (b) Calculate the t-statistic: t act =

βˆ 1 − 0 −5.82 = = −2.6335. SE(βˆ 1) 2. 21

The p-value for the test H 0 : β1 = 0 vs. H 1 : β1 ≠ 0 is

p-value = 2 Φ( −| t act |) = 2Φ ( −2.6335) = 2× 0. 0042 = 0. 0084. The p-value is less than 0.01, so we can reject the null hypothesis at the 5% significance level, and also at the 1% significance level. (c) The t-statistic is tact =

βˆ 1 − ( −5.6) 0.22 = = 0.10 SE (βˆ 1) 2. 21

The p-value for the test H0 : β1 = − 5.6 vs. H1 : β1 ≠ − 5.6 is

p-value = 2 Φ ( −| t act |) = 2Φ (−0.10) = 0.92 The p-value is larger than 0.10, so we cannot reject the null hypothesis at the 10%, 5% or 1% significance level. Because β1 = − 5.6 is not rejected at the 5% level, this value is contained in the 95% confidence interval. (d) The 99% confidence interval for b0 is {520.4 ± 2.58 × 20.4}, that is,

467.7 ≤ β0 ≤ 573.0.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 2 _____________________________________________________________________________________________________

5.2. (a) The estimated gender gap equals $2.12/hour.

(b) The hypothesis testing for the gender gap is H 0 : β 1 = 0 vs. H1 : β 1 ≠ 0. With a tˆ −0 2.12 statistic tact = β 1 = = 5.89, the p-value for the test is SE (βˆ 1) 0. 36

p-value = 2Φ(−|t act |) = 2Φ (−5.89) = 2 × 0.0000 = 0.000 (to four decimal

places)The p-value is less than 0.01, so we can reject the null hypothesis that there is no gender gap at a 1% significance level.

(c) The 95% confidence interval for the gender gap β1 is {2.12 ±1.96 × 0.36}, that is, 1.41 ≤ β 1 ≤ 2.83.

(d) The sample average wage of women is βˆ0 = $12. 52/hour. The sample average wage of men is βˆ 0 + βˆ1 = $12.52 + $2.12 = $14.64/hour.

(e) The binary variable regression model relating wages to gender can be written as either Wage = β0 + β1 Male + ui , or Wage = γ 0 + γ 1Female + vi . In the first regression equation, Male equals 1 for men and 0 for women; β0 is the population mean of wages for women and β0 + β1 is the population mean of wages for men. In the second regression equation, Female equals 1 for women and 0 for men; γ 0 is the population mean of wages for men and γ 0 + γ1 is the population mean of wages for women. We have the following relationship for the coefficients in the two regression equations:

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 3 _____________________________________________________________________________________________________

5.2 (continued)

γ0 = β0 + β1 , γ0 + γ1 = β0 . Given the coefficient estimates βˆ 0 and βˆ 1, we have

γˆ0 = βˆ0 + βˆ1 = 14.64, γˆ1 = βˆ 0 − γˆ 0 = − βˆ1 = − 2. 12.

Due to the relationship among coefficient estimates, for each individual observation, the OLS residual is the same under the two regression equations: n

2 uˆi = vˆi . Thus the sum of squared residuals, SSR = ∑ i =1 uˆ i , is the same under the 1

2 SSR R 2 = 1 − TSS two regressions. This implies that both SER = ( SSR are n −1 ) and

unchanged.

In summary, in regressing Wages on Female, we will get

! = 14.64 − 2.12Female, Wages

R2 = 0 .06, SER = 4.2.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 4 _____________________________________________________________________________________________________

5.3. The 99% confidence interval is 1.5 × {3.94 ± 2.58 × 0.31) or 4.71 lbs ≤ WeightGain ≤ 7.11 lbs.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 5 _____________________________________________________________________________________________________

5.4. (a) −7.29 + 1.93 × 16 = $23.59 per hour

(b) The wage is expected to increase by 1.93×2 = $3.86 per hour.

(c) The increase in wages for college education is β1 × 4. Thus, the counselor’s assertion is that β1 = 10/4 = 2.50. The t-statistic for this null hypothesis is

t = 1.93−2.50 = −7.13, which has a p-value of 0.00. Thus, the counselor’s assertion 0.08 can be rejected at the 1% significance level. A 95% confidence for β 1 × 4 is 4 × (1.93 ± 1.96 × 0.08) or $7.09 ≤ Gain ≤ $8.35.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 6 _____________________________________________________________________________________________________

5. 5 (a) The estimated gain from being in a small class is 13.9 points. This is equal to approximately 1/5 of the standard deviation in test scores, a moderate increase.

(b) The t-statistic is t act = 13.9 = 5.56, which has a p-value of 0.00. Thus the null 2.5 hypothesis is rejected at the 5% (and 1%) level.

(c) 13.9 ± 2.58 × 2.5 = 13.9 ± 6.45.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 7 _____________________________________________________________________________________________________

5.6. (a) The question asks whether the variability in test scores in large classes is the same as the variability in small classes. It is hard to say. On the one hand, teachers in small classes might able so spend more time bringing all of the students along, reducing the poor performance of particularly unprepared students. On the other hand, most of the variability in test scores might be beyond the control of the teacher. (b) The formula in 5.3 is valid for heteroskesdasticity or homoskedasticity; thus inferences are valid in either case.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 8 _____________________________________________________________________________________________________

5.7. (a) The t-statistic is

3.2 1.5

= 2.13 with a p-value of 0.03; since the p-value is less than

0.05, the null hypothesis is rejected at the 5% level. (b) 3.2 ± 1.96 × 1.5 = 3.2 ± 2.94 (c) Yes. If Y and X are independent, then β1 = 0; but this null hypothesis was rejected at the 5% level in part (a). (d) β 1 would be rejected at the 5% level in 5% of the samples; 95% of the confidence intervals would contain the value β 1 = 0.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 9 _____________________________________________________________________________________________________

5.8. (a) 43.2 ± 2.05 × 10.2 or 43.2 ± 20.91, where 2.05 is the 5% two-sided critical value from the t28 distribution. (b) The t-statistic is t act = 61.57.4−55 = 0.88, which is less (in absolute value) than the critical value of 20.5. Thus, the null hypothesis is not rejected at the 5% level. (c) The one sided 5% critical value is 1.70; tact is less than this critical value, so that the null hypothesis is not rejected at the 5% level.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 10 _____________________________________________________________________________________________________

5.9. (a) β =

1 n

(Y1 + Y2 +!+ Yn ) so that it is linear function of Y1, Y2, …, Yn. X

(b) E(Yi|X1, …, Xn) = β 1Xi, thus

E(β |X1 ,…, X n ) = E =

1 1 (Y + Y +!+ Yn )|X 1 ,…, X n ) Xn 1 2

1 1 β ( X +!+ X n ) = β1 Xn 1 1

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 11 _____________________________________________________________________________________________________

5.10. Let n0 denote the number of observation with X = 0 and n1 denote the number of observations with X = 1; note that

∑

n i =1

∑

n

i=1

Xi = n1; X = n1| n; 2

n

(

)

( X i − X )2 = ∑i =1 X i2 − nX 2 = n1 − nn1 = n1 1 − nn1 =

n 1n 0 n

1 n1

∑

n i =1

Xi Yi = Y1;

that Y = nn1 Y1 + nn0 Y0 From the least squares formula n n X i (Yi − Y ) ∑ i=1 X iYi − Yn1 ∑ n ( X − X )(Yi − Y ) ∑ i=1 = = βˆ1 = i=1 n i n 2 2 n1n0 |n ∑ i=1( X i − X ) ∑ i=1 ( X i − X )

=

n ⎞ n n n⎛ (Y1 − Y ) = ⎜ Y − 1 Y1 − 0 Y0 ⎟ = Y1 − Y0 , n0 n0 ⎝ n n ⎠

n ⎞ n n + n0 ⎛n and βˆ0 = Y − βˆ1 X = ⎜ 0 Y0 + 1 Y1 ⎟ − ( Y1 − Y0 ) 1 = 1 Y0 = Y0 n ⎠ n n ⎝ n

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n

; n1Y1 + n0Y0 = ∑ i= 1Yi , so

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 12 _____________________________________________________________________________________________________

5.11. Using the results from 5.10, βˆ 0 = Ym and βˆ1 = Yw − Ym . From Chapter 3,

SE ( Ym ) =

Sm nm

and SE (Yw − Ym ) =

sm2 nm

s2 + nww . Plugging in the numbers βˆ 0 = 523.1 and

SE ( βˆ 0 ) = 6.22; βˆ1 = −38.0 and SE ( βˆ1 ) = 7.65.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 13 _____________________________________________________________________________________________________

5.12. Equation (4.22) gives

σ β2ˆ = 0

var (Hi ui ) n ⎡⎣ E ( Hi )⎤⎦ 2

, where Hi = 1 −

2

µX E ( X i2 )

Xi .

Using the facts that E (ui |Xi ) = 0 and var(u i| X i) = σ 2u (homoskedasticity), we have ⎛ ⎜

E ( Hi ui ) = E ⎜⎜ ui − ⎜⎜ ⎝

=0 −

⎞ ⎟

µx E(X

µx E ( X i2 )

2 i

)

Xi ui ⎟⎟ = E( ui) − ⎟⎟ ⎠

µx E ( Xi2 )

E[ Xi E( ui| Xi)]

×0 = 0,

and ⎧⎛ ⎪⎜ ⎪⎪⎜ ⎨⎜ i ⎪⎜ ⎪⎜ ⎪⎩⎝

E [(Hi ui ) ] = E u − 2

⎧ ⎪ ⎪ 2 ⎨ i ⎪ ⎪ ⎩⎪

µX E ⎛⎜ Xi2 ⎞⎟ ⎝

⎞ ⎟ ⎟ i i⎟ ⎟⎟ ⎠

Xu

⎠

2⎫ ⎪ ⎪⎪ ⎬ ⎪ ⎪ ⎭⎪ 2

⎫

⎪ ⎡ µ ⎤ 2 2 ⎪ X = E u − 2 ⎛ 2 ⎞ X u + ⎢ ⎛ 2 ⎞ ⎥ X i u i ⎬⎪ E ⎜⎝ Xi ⎟⎠ ⎢⎣ E ⎜⎝ X i ⎟⎠ ⎥⎦ ⎪ ⎭⎪

µX

2 i i

2

⎡ µ ⎤ ⎡ 2 ⎛ 2 ⎞⎤ ⎡ ⎤ µ X E⎢ X i E ⎜⎝ ui |X i ⎟⎠⎥ = E u − 2 ⎛ X 2 ⎞E ⎢⎢ X i E ⎝⎛⎜u2i |X i ⎠⎞⎟ ⎥⎥ + ⎢ ⎥ 2 ⎢ ⎥ ⎛ 2 ⎞ ⎜⎜ i ⎟⎟ ⎝ ⎠

E ⎜⎝ Xi

⎟ ⎠

⎣

⎦

⎢⎣ E ⎛⎜⎝ Xi ⎞⎟⎠ ⎥⎦

2

⎡ µ ⎤ ⎛ 2⎞ X µX σ u2 + ⎢ E ⎜⎜ X i ⎟⎟σ u2 ⎥ 2 ⎞ ⎛ E ⎛⎝⎜ X i ⎞⎠⎟ E X ⎢⎣ ⎜⎝ i ⎟⎠ ⎥⎦ ⎝ ⎠ ⎛ µ2 ⎞ = ⎜ 1 − ⎛ X 2 ⎞ ⎟ σu2. ⎜ E ⎜⎝ X i ⎟⎠ ⎟ ⎝ ⎠

µX = σ u2 − 2 2

Because E (Hi ui ) = 0, var ( H iu i ) = E[( H iu i) 2],so

⎛ µX2 ⎞ 2 ⎟σ . var ( Hi ui ) = E[( Hi ui ) 2 ] = ⎜1 − ⎜ E ( X 2i ) ⎟ u ⎝ ⎠ Also (continued next page)

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 14 _____________________________________________________________________________________________________

5.12 (continued)

E (H

2 i

)=E

⎧⎛ ⎪⎜ ⎪⎜ ⎨⎜ ⎪⎜ ⎪⎜ ⎩⎪ ⎝

1−

µX E ( X i2 )

X

⎞2 ⎫⎪ ⎟ ⎟ ⎪ i ⎟ ⎬ ⎟⎟ ⎪ ⎪ ⎠ ⎭⎪

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩⎪

2

2

2 ⎡ µ ⎤ µ2 X E X i2 = 1− µ X . =1− 2 X 2 + ⎢ ⎥ 2 2

E ( Xi

)

( )

⎢⎣ E ( X i ) ⎥⎦

E ( Xi

)

Thus

σ β2ˆ

0

⎛ µ2X ⎞ 2 ⎜1 − ⎟σ ⎜ E ( X i2 ) ⎟ u var (Hi ui ) σ u2 ⎝ ⎠ = = = 2 ⎡ nE ( H 2 )2 ⎤ ⎛ ⎛ ⎞ 2 µ X2 ⎞ µ i X ⎜ ⎟ n 1 ⎣⎢ ⎦⎥ n ⎜1 − − ⎟ ⎜ E ( X i2 ) ⎟ ⎜ E ( X i2 ) ⎟ ⎝ ⎠ ⎝ ⎠ E ( X i ) σu E ( X i2 )σ u2 = = . 2 2 nσ X2 n [E ( X i − µ X ) ] 2

2

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⎫

⎡ µ ⎤ 2 ⎪⎪ X X = E 1− 2 + ⎢ ⎥ Xi ⎬ i 2 ⎪ E ( X 2i ) ⎢⎣ E ( X i ) ⎥⎦ ⎪ ⎭⎪

µX

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 15 _____________________________________________________________________________________________________

5.13. (a) Yes, this follows from the assumptions in KC 4.3. (b) Yes, this follows from the assumptions in KC 4.3 and conditional homoskedasticity (c) They would be unchanged for the reasons specified in the answers to those questions. (d) (a) is unchanged; (b) is no longer true as the errors are not conditionally homosckesdastic.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 16 _____________________________________________________________________________________________________

5.14. (a) From Exercise (4.11), βˆ = ∑ ai Yi where a i =

Xi n

∑ j =1 X 2j

. Since the weights depend

only on X i but not on Yi , βˆ is a linear function of Y.

(b) E(βˆ|X 1 ,…, X n ) = β +

n ∑ i=1 X i E(ui |X 1 ,…, X n ) = β since E (ui | X1 , K , X n ) = 0 ∑ni=1 X j2

∑ n X 2Var (ui |X1 ,…, X n ) σ2 = (c) Var (βˆ |X 1 ,…, X n ) = i=1 i 2 n X j2 ∑ i=1 ⎡ ∑ ni=1 X 2j ⎤ ⎣ ⎦

(d) This follows the proof in the appendix.

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Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 5 17 _____________________________________________________________________________________________________

5.15. Because the samples are independent, βˆm ,1 and βˆw ,1 are independent. Thus

var ( βˆm,1 − βˆw,1 ) = var ( βˆm,1 ) + var( βˆw,1 ). Var ( βˆm ,1) is consistently estimated as [ SE( βˆm,1 )] 2 and Var ( βˆ w,1 ) is consistently estimated as [SE ( βˆw,1 )]2 , so that var( βˆ m,1 − βˆ w,1 ) is consistently estimated by [ SE( βˆm ,1 )] 2 + [ SE ( βˆw ,1 )]2 , and the result follows by noting the SE is the square root of the estimated variance.

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