Solutions to Exercises in Chapter 6

Chapter 6 Solutions to Exercises

1

Solutions to Exercises in Chapter 6

6.1( )

22

2

ˆ1 0.71051t

t

eR

y y= − =

−∑

∑

6.2 To calculate 2R we need the term ( )2

ty y−∑ .

( )2 2 2 788.5155t ty y y T y− = − =∑ ∑

Therefore, 2 666.720.8455

788.5155

SSRR

SST= = =

6.3 From 2 2

2 ˆ ˆ( ) 1 1te T KR

SST SST

− σ= − = −∑

we have, 2

2 (1 ) 552.36(1 0.7911)ˆ 6.4104

(20 2)

SST R

T K

− −σ = = =− −

6.4 (a)

*

*ˆ 5.83 8.69 (1.23) (1.17) where

10

t tt

t

y x xx

= +=

(b)

*

*ˆ 0.583 0.0869 ˆ

ˆ (0.123) (0.0117) where 10

t tt

t

y x yy

= +=

(c)

* *

* *ˆ 0.583 0.869 ˆ

ˆ (0.123) (0.117) where and 10 10

t tt t

t t

y x y xy x

= += =

The values of 2R would remain the same in all cases.


2

6.5 ( )! ! ! ! !y y e y e y et t t t t− = −∑ ∑ ∑ ( )= + − = + −∑ ∑ ∑ ∑ ∑b b x e y e b e b x e y et t t t t t t1 2 1 2! ! ! ! !

Consider the terms !et∑ and x et t!∑ ( )!e y b b x y Tb b xt t t t t∑ ∑ ∑ ∑= − − = − − =1 2 1 2 0

( )x e x y b b x x y b x b xt t t t t t t t t!∑ ∑ ∑ ∑ ∑= − − = − − =1 2 1 22 0

The last expressions in each of these equations become zero from the normal equations thatare used to solve for the least squares estimators. Substituting !et∑ = 0 and x et t!∑ = 0 back

into the original equation, we obtain ( )! !y y et t− =∑ 0 .

6.6 (a) Least squares estimates of the equation are given by

% ! . .( . ) ( . )

∆wut

t = − +

14284 8 7309

1

2 0355 2 8038R2 = 0.393

(b) There will be no relationship between %∆w and (1/u) if β2 = 0. Thus we test H0: β2 = 0against H1: β2 ≠ 0. Given H0 is true, the value of the test statistic is

( )tb

b= = =2

2

8 7309

2 80383114

se

.

..

At the 5% significance level and with 15 degrees of freedom the critical value is tc =2.131. Since 3.114 > 2.131, the value of the test statistic falls in the rejection region; wereject H0 and conclude that there is a relationship between %∆w and (1/u).

(c) See Figure 6.1.

(d) The natural rate of unemployment is given by the solution to the equation

− + =142848 7309

0..

ut

That is, ut = =8 7309

14284611%

.

..

(e)( )d w

du u

% .∆= − 8 7309

2= −8.7309 when u = 1

= −0.9701 when u = 3

(f) The impact of a change in the unemployment rate on the rate of change in wages is givenby the slope of the function, ( )%d w du∆ . Thus, the impact is greatest as the rate of

unemployment u approaches zero and it is smallest as u approaches infinity.

(g) The parameter β1 gives the smallest possible percentage change in wages. As u increasesindefinitely the percentage change in wages will never be less than β1. When β1 isnegative, as our estimate suggests, the percentage decrease is never greater than −β1. Ourestimate suggests that any decrease is never greater than 1.43%.


3

Figure 6.1 Estimated Phillips' Curve

-2

0

2

4

6

8

0 1 2 3 4 5 6 7 8 9

u

%∆∆∆∆w

(h) A 95% interval estimate for β1 is given by

( ) ( )( ) ( )b t bc1 1 14284 2 131 2 0355 577 2 91± = − ± = −se . . . . , .

This wide interval suggests our point estimate for β1 is not very reliable. The smallestpossible wage change could reasonably lie anywhere between an increase of 2.91% and adecrease of 5.77%.

A 95% interval estimate for β2 is given by

( ) ( )( ) ( )b t bc2 2 8 7309 2131 2 8038 2 76 14 71± = ± =se . . . . , .

This interval is also a wide one, making it difficult for us to be precise about the likelyinfluence of the unemployment rate on the rate of change of wages.

6.7 (a) Knowledge of the relationship between the rate of change of wages and theunemployment rate is very important for government policy makers and unions.Typically, governments like to keep the lid on inflation, but the cost of doing so may beincreasing unemployment. Unions are always attempting to negotiate pay rises for theiremployees. However, they must be wary of doing so if wage rises mean that fewer oftheir workers can get jobs. An economic model for relating the rate of change in wages tounemployment was derived in Section 6.3.2. It is called the Phillips' curve and is givenby

%∆wu

= +

β β1 21

A corresponding statistical model is

%∆wu

ett

t= +

+β β1 2

1


4

where some statistical assumptions are needed for the random error et. We assume the et

are independent normally distributed random variables with zero mean and constantvariance σ2. To estimate this model, we use a set of 18 observations on wt and ut for theperiod 1949 to 1966. The estimation method used is the linear least squares methodregressing %∆wt on 1/ut. This method gives the estimated equation as

% !∆wt = − +

14284 8 7309

2 0355 2 8038

1. .

( . ) ( . ) ut

R2 = 0.393

where standard errors are in parentheses. We would expect β1 < 0 and β2 > 0. Thus, fromthe estimated equation, b1 and b2 give the expected signs. The standard error of b1 is2.0355 which is relatively large and leads to b1 being statistically insignificant. Theestimate for b2 is statistically significant, but, as we discovered in Exercise 6.2(h), itslarge standard error leads to a wide confidence interval that conveys little economicinformation. Some of the economic implications were discussed in the answers toExercise 6.2.

(b) From the Economic Report to the President we can construct the following table where ut

is the unemployment rate for all workers, wt is average gross hourly earnings in currentdollars for the total private non-agricultural sector, and 1 1% 100( ) /t t t tw w w w− −∆ = − .

Year wt ut %∆wt

1974 4.24 5.5 -1975 4.53 8.3 6.83961976 4.86 7.6 7.28481977 5.25 6.9 8.02471978 5.69 6.0 8.38101979 6.16 5.8 8.26011980 6.66 7.0 8.11691981 7.25 7.5 8.85891982 7.67 9.5 5.79311983 8.01 9.5 4.4329

(c) The estimated equation for the Phillips' curve using the data for the years 1975 to 1983 is

% !∆wt = 0 8311 47 85118884 13712

1. .

( . ) ( . )+

ut

R2 = 0.635

where standard errors are in parentheses. The estimate b1 = 0.8311 suggests that theannual wage increase will never drop below 0.83%, even as the unemployment ratebecomes very large. This outcome may be unrealistic; we may expect workers to accept adrop in real wages for very high unemployment rates. If so, b1 should be negative. Thevery high standard error of b1, and its consequent insignificance, do not rule out thepossibility of β1 being negative. The estimate b2 = 47.851 with se(b2) = 13.712 iseconomically feasible and significant. However, the resulting confidence interval is stillwide. It conveys little information, although the large value of b2 suggests wage changesare very responsive to the unemployment rate.


5

6.8 (a) The result ryp2 = R2 can be verified using your computer software. Let

sy2 = sample variance of the yt = 2039.3

sp2 = sample variance of the !yt = 646.70

syp = sample covariance of yt and !yt = 646.70.

Then, the squared sample correlation between yt and !yt is given by

( )r

s

s sRyp

yp

y p

22

2 2

2

2646 70

646 7 2039 30 3171= =

×= =

.

. ..

(b) Similarly, sx2 = 39294 and sxy = 5041, yielding

( )

rs

s sRxy

xy

x y

22

2 2

2

25041

39294 2039 30 3171= =

×= =

..

(c) We want to show that

( )( )

( )( )[ ]( ) ( )

RSSR

SST

y y

y y

x x y y

x x y yr

t

t

t t

t t

xy2

2

2

2

2 22= =

−

−=

− −

− −=∑

∑∑

∑∑!

Now,

( ) ( ) ( )!y y b b x b b x b x xt t t− = + − − = −∑ ∑ ∑21 2 1 2

222 2

( )( )[ ]

( )[ ]( ) ( )( )[ ]

( )=

− −

−⋅ − =

− −

−

∑

∑

∑∑

∑x x y y

x xx x

x x y y

x x

t t

t

tt t

t

2

2 2

2

2

2

Submitting this expression into that for R2 yields the desired result.

6.9 In the learning curve model ln( ) ln( )u qt t= +β β1 2 , β1 is the logarithm of the unit cost of

production for the first unit produced and β2 is the elasticity of unit cost with respect tocumulative production. A 1% change in the cumulative production leads to a % change in theunit cost equal to β2.


6

6.10 See Figure 6.3.

Figure 6.3 Graph of ln(x)

-2

-1

0

1

2

3

1 2 3 4 5 6 7 8 9 10

x

ln(x)

6.11 The results are summarised in the following table. Standard errors are in parentheses.Changing the units of y only changes both b1 and b2, making them one hundred times smaller.Changing the units of x only influences only b2, making it 100 times bigger. When both x andy are scaled b1 changes but not b2.

Regression b1 b2

(i) y on x 40.768 0.1283(22.14) (0.0305)

(ii) y on x* 40.768 12.83(22.14) (3.05)

(iii) y* on x 0.40768 0.001283(0.2214) (0.000305)

(iv) y* on x* 0.40768 0.1283(0.2214) (0.0305)

6.12 The estimated equations, their plots and the plots of the residuals are below.

The first equation is

2

ˆ 0.6776 0.0161 (0.0725) (0.0026) (9.3427) (6.2533) 0.4595

ty tset R

= +

=

The corresponding plots are


7

-1.0

-0.5

0.0

0.5

1.0

0.0

0.5

1.0

1.5

2.0

2.5

50 55 60 65 70 75 80 85 90 95

Residual Actual Fitted

The second equation is

2

ˆ 0.5287 0.1855ln( ) (0.1472) (0.0481) (3.5909) (3.8545) 0.2441

ty tset R

= +

=

The plots are

-1.0

-0.5

0.0

0.5

1.0

0.0

0.5

1.0

1.5

2.0

2.5

50 55 60 65 70 75 80 85 90 95


The third equation is

2

2

ˆ 0.7914 0.000355 (0.0482) (0.000046) (16.434) (7.7856) 0.5685

ty tset R

= +

=

The plots are

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.0

0.5

1.0

1.5

2.0

2.5

50 55 60 65 70 75 80 85 90 95



8

To choose the preferable equation we consider the following.

1. The signs of the response parameters 2 2 2, and β α γ . We expect them to be positive

because we expect yield to increase over time as technology improves. The signs ofthe estimates of 2 2 2, and β α γ are as expected.

2. 2R . The value of 2R for the third equation is the highest, namely 0.5685.

3. The plots of the fitted equations and their residuals. The upper parts of the figuresdisplay the fitted equation while the lower parts display the residuals. Considering theplots for the fitted equations, the one obtained from the third equation seems to fit theobservations best. In terms of the residuals, the first two equations haveconcentrations of positive residuals at each end of the sample. The third equationprovides more of a balance of positive and negative residuals through out the sample.

The third equation is preferable.

6.13 To test 0 :H normally distributed errors for the preferable equation, equation 3, we need

values of skewness and kurtosis. They are S = -0.08465 and k = 3.25607. The test statistic is

22 ( 3)

0.1886 4

T kJB S

−= + =

The hypothesis 0H is rejected if 2(2)JB > χ for a given significance level. Since 2

(2) 5.991χ =

at the 0.05 level of significance, we do not reject 0H . Therefore, there is insufficient evidence

to conclude that the normal distribution assumption is unreasonable.

Different software packages can use different estimators for skewness and kurtosis; and somereport excess kurtosis ( 3)k − as the ‘kurtosis’. For example, SAS yields 0.8741S = and

3 0.4211k − = , leading to a test value of 0.416.JB =

6.14 (a) Equation 1: 0ˆ 0.6776 0.0161(49) 1.4665y = + =

Equation 2: 0ˆ 0.5287 0.1855ln(49) 1.2506y = + =

Equation 3: 20ˆ 0.7914 0.0004(49) 1.7518y = + =

(b) Equation 1: 2

ˆ ˆ 0.0161td y

d t= β =

Equation 2: 2

ˆ 1ˆ 0.0038td y

d t t= α =

Equation 3: 2

ˆˆ2 0.0392td y

td t

= γ =


9

(c) Equation 1, at 0ˆ49, 1.4665t y= = and

2

ˆ ˆ 0.5379ˆ ˆ

t

t t

d y t t

d t y y= β =

Equation 2, at 0ˆ49, 1.2506t y= = and

2

ˆ 1ˆ 0.1483

ˆ ˆt

t t

d y t

d t y y= α =

Equation 3, at 0 0ˆ49, 1.7518t y= = so

22

ˆˆ 21.0965

ˆ ˆt

t t

d y tt

d t y y

γ= =

The slopes td y

d t and the elasticities t

t

d y t

d t y give the marginal change in yield and the

percentage change in yield, respectively, that can be expected from technologicalchange in the next year. The results show that the predicted effect of technologicalchange is quite sensitive to the choice of functional form.

6.15 (a) For households with 1 child

^

2

1.0099 0.1495ln( ) (0.0401) (0.0090) (25.19) (-16.70) 0.3203

wfood totexpset R

= −

=

For households with 2 children:

^

2

0.9535 0.1294ln( ) (0.0365) (0.0080) (26.10) (-16.16) 0.2206

wfood totexpset R

= −

=

In terms of 2β we would expect a negative value because as the total expenditure

increases the food share should decrease. Both estimations give the expected sign.The standard errors for 1 2 and b b from both estimations are relatively small resulting

in high values of t ratios and significant estimates.

(b) For households with 1 child, the average total expenditure is 94.848 and

[ ] [ ]1 2

1 2

ln( ) 1 1.0099 0.1495 ln(94.848) 1ˆ 0.5461

ln( ) 1.0099 0.1495ln(94.848)

b b totexp

b b totexp

+ + − +η = = =

+ −

For households with 2 children, the average total expenditure is 101.17 and

[ ] [ ]1 2

1 2

ln( ) 1 0.9535 0.1294 ln(101.17) 1ˆ 0.6366

ln( ) 0.9535 0.1294ln(101.17)

b b totexp

b b totexp

+ + − +η = = =

+ −

Both of the elasticities are less than one; therefore, food is a necessity.


10

0.0

0.2

0.4

0.6

0.8

3 4 5 6

X1

WF

OO

D1

-0.4

-0.2

0.0

0.2

0.4

3 4 5 6

X1

RE

SID

Figure 6.4a Figure 6.4b

(c) Figures 6.4a and 6.4b are the fitted curve and the residual plots for households with 1child. The function linear in wfood and ln(totexp) seems to be an appropriate one.However, the observations vary considerably around the fitted line, consistent withthe low 2R value. Also, the absolute magnitude of the residuals appears to decline asln(totexp) increases. In Chapter 11 we discover that such behavior suggests theexistence of heteroskedasticity.

Figures 6.5a and 6.5b are plots of the fitted equation and the residuals for householdswith 2 children. They are similar to the ones with 1 child, indicating that thefunctional form is a reasonable one.

The values of JB for testing 0 :H errors are normally distributed are 10.7941 and

6.3794 for households with 1 child and 2 children, respectively. Since both values aregreater than the critical 2

(2) 5.991χ = , we reject 0H . The p-values obtained are 0.005

and 0.041, respectively, confirming that 0H is rejected. We conclude that for both

cases the errors are not normally distributed. (Using SAS estimates for skewness andexcess kurtosis, the JB values are 11.06 and 6.65, respectively.)

0.0

0.2

0.4

0.6

0.8

3.5 4.0 4.5 5.0 5.5 6.0

X2

WF

OO

D2

-0.4

-0.2

0.0

0.2

0.4

3.5 4.0 4.5 5.0 5.5 6.0

X2

RE

SID



11

6.16 Because food

wfoodtotexp

= , the equation (ignoring the error terms) becomes

1 2 ln( )food

totexptotexp

= β +β (1)

Or 1 2( ) ( ) ln( )food totexp totexp totexp= β + β

And [ ]1 2 ln( ) 1( )

d foodtotexp

d totexp= β +β + (2)

The food expenditure elasticity is defined as

( )

( )

d food totexp

d totexp foodη =

From (1) and (2) we obtain

[ ]1 2

1 2

ln( ) 1

ln( )

totexp

totexp

β +β +η =

β + β


^

2

0.3095 0.0486ln( ) (0.0217) (0.0048) (14.30) (-10.05) 0.1458

wfuel totexpset R

= −

=


^

2

0.3009 0.0464ln( ) (0.0198) (0.0043) (15.22) (-10.71) 0.1105

wfuel totexpset R

= −

=

We expect fuel to be a necessary good and as total expenditure increases the fuelshare should decrease. That is, we expect the sign of 2β to be negative. Both

estimations give the expected sign. The standard errors for 1 2 and b b from both

estimations are relatively small resulting in high values of t ratios.


[ ] [ ]1 2

1 2

ln( ) 1 0.3095 0.0486 ln(94.848) 1ˆ 0.4494

ln( ) 0.3095 0.0486ln(94.848)

b b totexp

b b totexp

+ + − +η = = =

+ −


[ ] [ ]1 2

1 2

ln( ) 1 0.3009 0.0464 ln(101.17) 1ˆ 0.4647

ln( ) 0.3009 0.0464ln(101.17)

b b totexp

b b totexp

+ + − +η = = =

+ −

Both of the elasticities are less than one; therefore, fuel is a necessity.


12

0.0

0.1

0.2

0.3

0.4

3 4 5 6

X1

WF

UE

L1

-0.1

0.0

0.1

0.2

0.3

3 4 5 6

X1

RE

SID


(c) Figures 6.6a and 6.6b are the fitted curve and the residual plots for households with 1child. The plot of the actual observations and the fitted equaton is in Figure 6.6a. Thefunctional form is reasonable in the sense that it is difficult to suggest an alternativefunction which will be a better fit to the scatter of points. On the other hand, thefunction provides a poor explanation of variation in the budget share of fuel.Consistent with the low 2R , the observations vary widely around the fitted line. Alsothe variation of the observations around the fitted line decreases as the totalexpenditure increases. The positive residuals vary in magnitude from zero to 0.3whereas the range of the negative residuals is zero to –0.1, suggesting a highlyskewed error distribution.

Figures 6.7a and 6.7b are the fitted curve and the residual plots for households with 2children. They are similar to the ones with 1 child. While it is difficult to suggest afunctional form which would fit better than the linear-log one, the function is not agood one for explaining variation in the fuel budget share. Perhaps a multipleregression model with additional explanatory variables would be an improvement.

0.0

0.1

0.2

0.3

0.4

0.5

3.5 4.0 4.5 5.0 5.5 6.0

X2

WF

UE

L2

-0.2

0.0

0.2

0.4

0.6

3.5 4.0 4.5 5.0 5.5 6.0

X2

RE

SID



13

The values of JB for testing 0 :H errors are normally distributed are 874.4 and 8738.2

for households with 1 child and 2 children, respectively. Since both values are greaterthan the critical 2

(2) 5.991χ = , we reject 0H . The p-values are 0.0000 for both cases

confirming that 0H is rejected. We conclude that for both cases the errors are not

normally distributed.


^

2

0.0242 0.0205ln( ) (0.0303) (0.0068) (-0.798) (3.003) 0.0153

walc totexpset R

= − +

=


^

2

0.0459 0.0225ln( ) (0.0239) (0.0052) (-1.916) (4.281) 0.0195

walc totexpset R

= − +

=

We expect alcohol to be a non-necessary good and so as total expenditure increases,the alcohol share should increase. That is, we expect the sign of 2β to be positive.

Both estimations give the expected sign. The standard errors for 2b from both

estimations are relatively small resulting in high values of t ratios and significantestimates.


[ ] [ ]1 2

1 2

ln( ) 1 0.0242 0.0205 ln(94.848) 1ˆ 1.2966

ln( ) 0.0242 0.0205ln(94.848)

b b totexp

b b totexp

+ + − + +η = = =

+ − +


[ ] [ ]1 2

1 2

ln( ) 1 0.0459 0.0225 ln(101.17) 1ˆ 1.3881

ln( ) 0.0459 0.0225ln(101.17)

b b totexp

b b totexp

+ + − + +η = = =

+ − +

Both of the elasticities are greater than one; therefore, alcohol is a non-necessity.

(c) Figures 6.8a and 6.8b are the fitted curve and the residual plots for households with 1child. The functional form is reasonable in the sense that it is difficult to suggest analternative function which will be a better fit to the scatter of points. On the otherhand, the function provides a poor explanation of variation in the budget share ofalcohol. Consistent with the low 2R , the observations vary widely around the fittedline. The positive residuals vary in magnitude from zero to 0.3 whereas the range ofthe negative residuals is zero to –0.1, suggesting a highly skewed error distribution.


14

0.0

0.1

0.2

0.3

0.4

3 4 5 6

X1

WA

LC

1

-0.1

0.0

0.1

0.2

0.3

0.4

3 4 5 6

X1

RE

SID


Figures 6.9a and 6.9b are the fitted curve and the residual plots for households with 2children. They are similar to the ones with 1 child. While it is difficult to suggest afunctional form which would fit better than the linear-log one, the function is not agood one for explaining variation in the alcohol budget share. Perhaps, again, amultiple regression model with additional explanatory variables would be animprovement.

The values of JB for testing 0 :H errors are normally distributed are 346.1 and 1133.6

for households with 1 child and 2 children, respectively. Since both values are greaterthan the critical 2

(2) 5.991χ = , we reject 0H . This conclusion can also be reached

using the p-values. The p-values for this test are 0.0000 for both cases. We canconclude that for both cases the errors are not normally distributed.

0.0

0.1

0.2

0.3

0.4

0.5

3.5 4.0 4.5 5.0 5.5 6.0

X2

WA

LC

2

-0.1

0.0

0.1

0.2

0.3

0.4

3.5 4.0 4.5 5.0 5.5 6.0

X2

RE

SID



15

0.0

0.5

1.0

1.5

2.0

2.5

0.00 0.05 0.10 0.15

X

Y

-3

-2

-1

0

1

-10 -8 -6 -4 -2 0

LNX

LNY


0.0

0.5

1.0

1.5

2.0

2.5

-10 -8 -6 -4 -2 0

LNX

Y

-3

-2

-1

0

1

0.00 0.05 0.10 0.15

X

LNY

Figure 6.10c Figure 6.10d

6.19 (a) Figures 6.10a, 6.10b, 6.10c and 6.10d are the plots of observations of y against x,ln(y) against ln(x), y against ln(x) and ln(y) against x, respectively. The functionalforms to choose from are specified in part (b). They are all linear functions of y and xor the logarithms of them. The preferable form is the one where the plot of theobservations is best represented by a linear line. This is the plot of y and x in Figure6.10a. In all the other figures a line with some curvature would be a better fit. Thus,the chosen functional form is the equation 1 2t t ty x e= β +β + .

(b) (i)

2

ˆ 0.1367 15.6676 (0.0302) (0.6912) (4.5220) (22.667) 0.9753

t ty xset R

= +

=

(ii)^

2

ln( ) 0.8428 0.3666ln( ) (0.3356) (0.0552) (2.5114) (6.6418) 0.7724

t ty xset R

= +

=


16

(iii)

2

ˆ 1.5535 0.1872ln( ) (0.2762) (0.0454) (5.6244) (4.1203) 0.5663

t ty xset R

= +

=(iv)

^

2

ln( ) 1.7590 23.3066 (0.1556) (3.5589) (-11.303) (6.5488) 0.7674

t ty xset R

= − +

=

To check whether the level of arsenic in the water influences the level of arsenic inthe toenails we test each equation for whether 2 2 2 2, , and β α γ θ are zero. We reject an

0H if a t ratio is greater than the absolute critical value. At level of significance 0.05,

and 13 degrees of freedom, 0.025 2.16t = . The calculated t ratios for the hypotheses

from each equation are all higher than the critical value. We therefore reject 0H and

conclude that there is evidence to suggest that the level of arsenic in water doesinfluence the level of arsenic in the toenails.

(c) The plots of the residuals from equation (i) to (iv) are presented in Figures 6.11-6.14.The residuals in Figures 6.12b and 6.13b display a rough U-shape, while those inFigure 6.14b exhibit a rough inverted U-shape. Thus, in these cases, the residuals forsmall and large values of x or ln(x) have signs different to the residuals thatcorrespond to middle values of x or ln(x). Such patterns suggest inappropriatefunctional forms. Since the residuals for the linear function do not display such apattern, the plots support our choice of function in part (a).

0.0

0.5

1.0

1.5

2.0

2.5

0.00 0.05 0.10 0.15

X

Y

-0.2

-0.1

0.0

0.1

0.2

0.3

0.00 0.05 0.10 0.15

X

RE

SID



17

-3

-2

-1

0

1

-10 -8 -6 -4 -2 0

LNX

LNY

-1.0

-0.5

0.0

0.5

1.0

-10 -8 -6 -4 -2 0

LNX

RE

SID


0.0

0.5

1.0

1.5

2.0

2.5

-10 -8 -6 -4 -2 0

LNX

Y

-1.0

-0.5

0.0

0.5

1.0

1.5

-10 -8 -6 -4 -2 0

LNX

RE

SID


-3

-2

-1

0

1

0.00 0.05 0.10 0.15

X

LNY

-1.0

-0.5

0.0

0.5

1.0

0.00 0.05 0.10 0.15

X

RE

SID



18

6.20 The estimated equations are below.

^

211.952 0.0168

(1.081) (0.0207) 0.0021fibre age

R= +

=

^

22250.5 9.3693

(123.2) (2.362) 0.048calories age

R= −

=

^

2297.07 1.1130

(26.26) (0.5035) 0.0154chol age

R= −

=

^

297.508 0.4142

(6.682) (0.1281) 0.0324fat age

R= −

=

The number in parentheses are standard errors. Based on this set of data it is suggested that aspeople grow older they consume food with more fibre, less calories, less cholesterol and lessfat. The 2R for each equation is very low indicating that age is not a good predictor fordietary intake.

6.21 (a) The estimated equations for dishwasher shipments as a function of durable goodsexpenditure and as a function of private residential investment are below.

^

2154.43 22.856

(295.5) (2.548) 0.7702DISH DUR

R= − +

=

^

22067.1 92.058

(534.7) (10.95) 0.7465DISH RES

R= − +

=

The equation that gives the better predictor is the equation that has the higher 2R . Inthis case it is the equation with DUR as the explanatory variable. Hence, DUR is thebetter predictor.

(b)^

8686 154.43 22.856 154.43 22.856(208.2) 4604DISH DUR= − + = − + ≈

^

8686 2067.1 92.058 2067.1 92.058(67.2) 4119DISH RES= − + = − + ≈

Since the actual 86 3915DISH = , for this future observation RES is a better predictor

than DUR.

(c) The residual plots for the two equations using DUR and RES as the explanatoryvariables are in Figures 6.15a and 6.15b, respectively. The residuals appear to becorrelated. There is a tendency for positive residuals to follow positive residuals andnegative residuals to follow negative residuals.


19

-1000

-500

0

500

1000

1500

60 62 64 66 68 70 72 74 76 78 80 82 84

DISH Residuals

-1000

-500

0

500

1000

60 62 64 66 68 70 72 74 76 78 80 82 84

DISH Residuals

Figure 6.15a using DUR Figure 6.15b using RES

6.22 (a) The estimated equations for washing machine shipments as a function of durablegoods expenditure and as a function of private residential investment are below.

^

23361.6 10.110

(224.3) (1.934) 0.5324WASH DUR

R= +

=

^

22037.4 50.707

(252.6) (5.174) 0.8001WASH RES

R= +

=

The equation with the better predictor is the equation that has the higher 2R , namely,the one with RES as the explanatory variable. Hence, RES is the better predictor.

(b)^

8686 3361.6 10.110 3361.6 10.11(208.2) 5467WASH DUR= + = + ≈

^

8686 2037.4 50.707 2037.4 50.707(67.2) 5445WASH RES= + = + ≈

Since 86 5320WASH = , RES is also the better predictor for the future observation.

(c) The residual plots for the two equations using DUR and RES as the explanatoryvariables are in Figures 6.16a and 6.16b, respectively. The residuals appear to becorrelated. There is a tendency for positive residuals to follow positive residuals andnegative residuals to follow negative residuals.

-1000

-500

0

500

1000

60 62 64 66 68 70 72 74 76 78 80 82 84

WASH Residuals

-600

-400

-200

0

200

400

600

60 62 64 66 68 70 72 74 76 78 80 82 84

WASH Residuals

Figure 6.16a using DUR Figure 6.16b using RES

Solutions to Exercises in Chapter 6

Documents

Transcript of Solutions to Exercises in Chapter 6