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Solutions to Exercises in Chapter 5 29 (c) The 95 confidence 30 Stock/ Watson - Introduction to Econometrics - Second Edition 5 (a) The estimated gain

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Chapter 5 Regression with a Single Regressor: Hypothesis Tests and Confidence Intervals ▫ Solutions to Empirical Exercises 1 (a) AHE = 3 32 + 0 45 × Age

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Chapter 5 Regression with a Single Regressor:

Hypothesis Tests and Confidence Intervals

Solutions to Exercises

1 (a) The 95% confidence interval for

β is { 5 82 1 96 2 21},-. ±. ×. that is

1 (b) Calculate the t-statistic: 1 1

ˆ058226335ˆSE( ) 2 21

act t

The p-value for the test

010Hβ:= vs.

0Hβ:≠ is

-value 2 ( | |) 2 ( 2 6335) 2 0 0042 0 0084 act pt=Φ- =Φ-. =×. =. . 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 1 1

ˆ( 5.6)0220.10ˆSE( ) 2 21

act t

The p-value for the test

01 :5.6Hβ=- vs. 11 :5.6Hβ≠- is -value 2 ( | |) 2 ( 0.10) 0.92 act pt=Φ- =Φ- = The p-value is larger than 0.10, so we cannot reject the null hypothesis at the 10%, 5% or 1% significance level. Because

15.6β=- is not rejected at the 5% level, this value is contained in

the 95% confidence interval. (d) The 99% confidence interval for 0 is {520.4 2.58 20.4},±× that is, 0

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

(b) The hypothesis testing for the gender gap is 01

0Hβ:= vs.

0.Hβ:≠ With a t-statistic

1 1

ˆ02.125.89ˆ()036

act tSE the p-value for the test is -value 2 ( | |) 2 ( 5.89) 2 0 0000 0 000 act pt=Φ- =Φ- =×. =. (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.

Solutions to Exercises in Chapter 5 29

β is {2 12 1 96 0 36},.±.×. that is,

1 (d) The sample average wage of women is 0

ˆ12 52/hour.$

β=. The sample average wage of men is

ˆˆ$12.52 $2.12 $14.64/hour.

(e) The binary variable regression model relating wages to gender can be written as either 01i

Wage Male uββ=+ +,

or 01i

Wage Female vγγ=+ +.

In the first regression equation,

Male equals 1 for men and 0 for women;

β is the population

mean of wages for women and 01 ββ+ is the population mean of wages for men. In the second regression equation,

Female equals 1 for women and 0 for men;

γ is the population mean of

wages for men and 01 γγ+ is the population mean of wages for women. We have the following relationship for the coefficients in the two regression equations: 001 01 0

Given the coefficient estimates

0 and 1 , we have 001 10 01

ˆˆˆ14.64

ˆˆ212

Due to the relationship among coefficient estimates, for each individual observation, the OLS residual is the same under the two regression equations: . iiuv=Thus the sum of squared residuals, 2 1 n i i SSRu is the same under the two regressions. This implies that both 1 2 1SSR n SER = and 2 1 SSR TSS

R=- are unchanged.

In summary, in regressing Wages on

,Female we will get

14.64 2 12Wages Female=-. ,

006 SER 4.2R=. , = .

4. (a) -3.13 + 1.47 × 16 = $20.39 per hour

(b) The wage is expected to increase from $14.51 to $17.45 or by $2.94 per hour. (c) The increase in wages for college education is 1

× 4. Thus, the counselor's assertion is that

1 = 10/4 = 2.50. The t-statistic for this null hypothesis is

1.47 2.50

0.07

14.71,t

=- which has a p-value of 0.00. Thus, the counselor's assertion can be rejected at the 1% significance level. A

95% confidence for

30 Stock/Watson - Introduction to Econometrics - Second Edition

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

± 2.58 × 2.5 = 13.9 ± 6.45.

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.

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 1 = 0; but this null hypothesis was rejected at the 5% level in part (a). (d) 1 would be rejected at the 5% level in 5% of the samples; 95% of the confidence intervals would contain the value 1 = 0.

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 t

28
distribution. (b) The t-statistic is

61.5 55

7.4 0.88, act t == 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; t act is less than this critical value, so that the null hypothesis is not rejected at the 5% level.

9. (a)

11 12 nXn

YY Yβ= +++? so that it is linear function of Y

1 , Y 2 , ..., Y n (b) E(Y i |X 1 , ..., X n 1 X i , thus 1121
11 1

11(|,,) ( )|,,)

11 nnn n

EX X E YY YX XXn

Xnβ

10. Let n

0 denote the number of observation with X = 0 and n 1 denote the number of observations with

X = 1; note that

11 n i i Xn 1 |;Xnn= 1 1 11 n ii ni XY Y 21011
222
1111
() 1 ; nnnnnn ii nnnii

XX XnX n n

11 001

n i i nY nY Y so that 01 10n n nn YYY=+

Solutions to Exercises in Chapter 5 31

From the least squares formula

1111
122
1110
0 1 11010
00 nnn ii i iii iii nn ii ii

XXYY XYY XYYn

XX XX nnn

nnnnYY Y Y Y YYnnnn and 01011

01 0110 00

ˆˆ()nnnnnYX Y YYY YYnn nn

11. Using the results from 5.10,

0 m

Yβ= and

1 wm

YYβ=- From Chapter 3, SE( )

m m S m n

Y= and

SE( ) .

mw mw ss wm nn

YY-= + Plugging in the numbers

ˆ523.1

β= and

ˆˆSE( ) 6.22; 38.0

ββ==- and

ˆSE( ) 7.65.

12. Equation (4.22) gives

0 2

ˆ222

var( )where 1 ii x ii ii

HuHXEXnEH

Using the facts that ( | ) 0

EuX= and var

2 ii u uXσ= (homoskedasticity), we have 22
2 000 xx ii i ii i i i i ii x i

EHu Eu Xu Eu EXEuXEX EX

EX and 2 2 2 2 2222
22
2 22
2 2 2 2| x ii i ii i xx iii ii ii x xiiii i i

EHu Eu XuEX

Eu Xu Xu

EX EX

Eu EXEu X

EXEX 22
2 2 22 2
2 2 2 2 2 2 1. iii x xiuxu u i i x u i

EXEu X

EX EXEX EX

Because ( ) 0,

EHu= var

2 ii ii

Hu E Hu=,so

32 Stock/Watson - Introduction to Econometrics - Second Edition

2 22
2 var( ) [( ) ] 1 x ii ii u i

Hu E HuEX

We can also get

22
22
222
2 222
22
2 112
12 1 xxx ii ii iii xx xi ii i

EH E X E X XEX EX EX

EX EXEX

Thus 0quotesdbs_dbs20.pdfusesText_26

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[PDF] Solutions to Exercises

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Chapter 5

Regression with a Single Regressor:

Hypothesis Tests and Confidence Intervals

Solutions to Exercises

1 (a) The 95% confidence interval for

β is { 5 82 1 96 2 21},-. ±. ×. that is

ˆ058226335ˆSE( ) 2 21

The p-value for the test

010Hβ:= vs.

0Hβ:≠ is

ˆ( 5.6)0220.10ˆSE( ) 2 21

The p-value for the test

15.6β=- is not rejected at the 5% level, this value is contained in

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

0Hβ:= vs.

0.Hβ:≠ With a t-statistic

ˆ02.125.89ˆ()036

1% significance level.

Solutions to Exercises in Chapter 5 29

β is {2 12 1 96 0 36},.±.×. that is,

ˆ12 52/hour.$

β=. The sample average wage of men is

ˆˆ$12.52 $2.12 $14.64/hour.

Wage Male uββ=+ +,

Wage Female vγγ=+ +.

In the first regression equation,

Male equals 1 for men and 0 for women;

β is the population

Female equals 1 for women and 0 for men;

γ is the population mean of

Given the coefficient estimates

ˆˆˆ14.64

ˆˆ212

R=- are unchanged.

In summary, in regressing Wages on

14.64 2 12Wages Female=-. ,

006 SER 4.2R=. , = .

4. (a) -3.13 + 1.47 × 16 = $20.39 per hour

× 4. Thus, the counselor's assertion is that

1.47 2.50

14.71,t

95% confidence for

30 Stock/Watson - Introduction to Econometrics - Second Edition

5 (a) The estimated gain from being in a small class is 13.9 points. This is equal to approximately 1/5

± 2.58 × 2.5 = 13.9 ± 6.45.

6. (a) The question asks whether the variability in test scores in large classes is the same as the

7. (a) The t-statistic is

2.13= with a p-value of 0.03; since the p-value is less than 0.05, the null

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 t

61.5 55

20.5. Thus, the null hypothesis is not rejected at the 5% level.

9. (a)

YY Yβ= +++? so that it is linear function of Y

11(|,,) ( )|,,)

EX X E YY YX XXn

Xnβ

10. Let n

X = 1; note that

XX XnX n n

11 001

Solutions to Exercises in Chapter 5 31

From the least squares formula

XXYY XYY XYYn

XX XX nnn

01 0110 00

ˆˆ()nnnnnYX Y YYY YYnn nn

11. Using the results from 5.10,

Yβ= and

YYβ=- From Chapter 3, SE( )

Y= and

SE( ) .

YY-= + Plugging in the numbers