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York University

Economics

ECON 3210

Razvan Sufana

Winter

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CHAPTER 5
Exercise Solutions
132 Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 133
EXERCISE 5.1
(a) y 1,x 2 3 0, 0
x* x* y*
i2 i3 i
0 1 0
1 2 1
2 1 2
2 0 2
1 1 1
2 1 2
0 1 1
1 1 0
1 0 1
(b) ii2 23, 3i 3 16, i i* 4, ix 10
** 2 * *
ii2 3 i * 33 2i i * *i ix 13 4 0
(c) b2 2 2 0.8125
i2 3i3 i2 * * 16 0
2
ii 3 2 i* 23 2i i* **i ix 46
b3 2 2 0.4
i2 3i3 i2 * * 16 0
3 32 1 2 1
(d) e 0.4, 0.9875, 0.025, 0.375, 1.4125, 0.025, 0.6, 0.4125, 0.1875
2
2 ei 3.8375
(e) 0.6396
NK 93
( (xi2 i2 3 3 i xi 2
(f) r23 2 2 2 2 0
xi2 2i x3i 3 i x x 2 3
2
0.6396
(g) se(b2b2 var6 )11 ( ( xx r 2 2 0.1999
i 2 3 2
(h) SSE i 3i8375 SST (y y) 16,
SSR SST SSE 12.16 25 R2 SSR 12.1625 0.7602
SST 16 Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 134
EXERCISE 5.2
(a) A 95% confidence interval for is
2
b2 (0.975,6) 2.8125 2.447 0.1999 (0.3233,1.3017)
(b) The null and alternative hypotheses are
H,0 211 1 2
cTlhelated t-value is
b2 1 0.8125 1
t 0.9377
se(b2 0.1999
At a 5% significance level, we reject H 0 if tt (0.975, 6).447 . Since 0.9377 2.447 ,
we do not reject H .
0 Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 135
EXERCISE 5.3
b1 0.0091
(a) (i) The t-statistic fo1is 0.476 .
se(b1) 0.0191
0.0276
(ii) The standard error forb2b2s se( ) 0.00418 .
6.6086
(iii) The estimate for 3 isb30.0002 ( 6.9624) 0.0014 .
(iv) To compute R 2, we need SSE and SST. From the output, SSE 5.752896 . To find
SST, we use the result
SST 0.0633
y N 1
which gives SST 1 518 (0.0633) 2 6.08246 . Thus,
R 1 SS 1 5.7290 0.054
SST 6.08246
SSE 5.752896
(v) The estimated error standard deviation is 0.061622
(NK ) 1519 4
(b) Thelue b20.0276 implies that if ln( TOTEXP) increases by 1 unit the alcohol share
will increase by 0.0276. The change in the alcohol share from a 1-unit change in total
expenditure depends on the level of total expenditure. Specifically,
d(WALC) d(TOTEXP) 0.0276 TOTEXP . A 1% increase in total expenditure leads to a
0.000276 increase in the alcohol share of expenditure.
vTlue b30.0014 suggests that if the age of the household head increases by 1 year
the share of alcohol expenditure of that household decreases by 0.0014.
vTlue b 0.0133 suggests that if the household has one more child the share of the
4
alcohol expenditure decreases by 0.0133.
(c) A 95% confidence interval for is
3
b3 0.975,1515)3 0.0014 1.96 0.0002 ( 0.0018, 0.0010)
This interval tells us that, if the age of the household head increases by 1 year, the share of
the alcohol expenditure is estimated to decrease by an amount between 0.0018 and 0.001. Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 136
Exercise 5.3 (Continued)
(d) The null and alternative hypotheses are H,0 4 1 4 .
b
cTlhelated t-value ist 4 4.075
se(b4
At a 5% significance level, we reject H if tt 1.96 . Since 4.075 1.96 ,
0 (0.975,1515)
we reject H 0 and conclude that the number of chil dren in the household influences the
budget proportion on alcohol. Having an additiona l child is likely to lead to a smaller
budget share for alcohol because of the non-al cohol expenditure demands of that child.
Also, perhaps households with more children pr efer to drink less, believing that drinking
may be a bad example for their children. Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 137
EXERCISE 5.4
(a) The regression results are:
WTRANS 0.0315 0.0414ln TOEXP 0.001AGE 0.0130 NK R 0.0247
e (0.0322) (0.0071) (0.0004) (0.0055)
(b) Tvealue b 0.0414 suggests that as lnTOTEXP increases by 1 unit the budget
2
proportion for transport increases by 0.0414. Alternatively, one can say that a 10%
increase in total expenditure will incree the budget proportion for transportation by
0.004. (See Chapter 4.3.3.) The positive sign 2fis according to our expectation because
as households become richer they tend to us e more luxurious forms of transport and the
proportion of the budget for transport increases.
vTlue b 0.0001 implies that as the age of the head of the household increases by 1
3
year the budget share for transport decreases by 0.0001. The expected sign fo3 is not
clear. For a given level of total expenditure a nd a given number of children, it is difficult
to predict the effect of age on transport share.
valuee b40.0130 implies that an additional ch ild decreases the budget share for
transport by 0.013. The negative sign means that adding children to a household increases
expenditure on other items (such as food and clothing) more than it does on transportation.
Alternatively, having more children may lead a household to turn to cheaper forms of
transport.
(c) The p-value for testing H :0 against the alternative H :0 where is the
0 3 1 3 3
coefficient of AGE is 0.869, suggesting that AGE could be excluded from the equation.
Similar tests for the coefficients of the other two variables yield p-values less than 0.05.
(d) The proportion of variation in the budget oportion allocated to transport explained by
this equation is 0.0247.
(e) For a one-child household:
WTRANS 00.0315 0.0414ln(TOTEXP 0) 0 0001AGE 0.013NK
0.0315 0.0414 ln(98.7) 0.0001 6 0.013 1
0.1420
For a two-child household:
WTRANS 00.0315 0.0414ln(TOTEXP 0) 0 0001AGE 0.013NK
0.0315 0.0414 ln(98.7) 0.0001 6 0.013 2
0.1290 Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 138
EXERCISE 5.5
(a) The estimated equation is
VALUE 2 8.4067 0.1834CR IME 22. 8109NITOX 6.3715ROOMS 0.0478AGE
(se) (5.3659) (0.0365) (4.1607) (0.3924) (0.0141)
1.3353DIST 0.2723ACCESS 0.0126TAX 1.1768PTRATIO
(0.2001) (0.0723) (0.0038) (0.1394)
The estimated equation suggests that as theper capita crime rate increases by 1 unit the
home value decreases by $183.4. The higher the level of air pollution the lower the value
of the home; a one unit increase in the nitric oxide concentration leads to a decline in value
of $22,811. Increasing the average number of rooms leads to an increase in the home
value; an increase in one room leads to an increase of $6,372. An increase in the
proportion of owner-occupied units built priorto 1940 leads to a decline in the home
value. The further the weighted distances to the five Boston employment centers the lower
the home value by $1,335 for every unit of weighted distance. The higher the tax rate per
$10,000 the lower the home value. Finally, the higher the pupil-teacher ratio, the lower the
home value.
(b) A 95% confidence interval for the coefficient of CRIME is
b2 (0.975,497)2 0.1834 1.965 0.0365 ( 0.255, 0.112) .
A 95% confidence interval for the coefficient of ACCESS is
bt b se( ) 0.2723 1.965 0.0723 (0.130,0.414)
7 (0.975,497)7
:t7 s(c) n We want to test Ha0:7r1oom room . The value of the t statistic is
brooms7 6.3715 7
t se(b ) 0.3924 1.6017
rooms
At 0.05 , we reject H 0 if the absolute calculated t is greater than 1.965. Since
1.6017 1.965, we do not reject H . The data is consistent with the hypothesis that
0
increasing the number of rooms by one increases the value of a house by $7000.
(d) We want to test H 0Hp1tratioaainst : ptratio 1. The value of the t statistic is
1.1768 1
t 1.2683
0.1394
At a significance level of 0.05 we rejectH 0 if the calculated t is less than the critical
value t(0.05,497).648 . Since 1.2683 1.648, we do not reject H 0. We cannot
conclude that reducing the pupil-teacher ratio by 10 will increase the value of a house by
more than $10,000. Chapter 5, Exercise Solutions, Principles of Econometrics, 4e 139
EXERCISE 5.6
In each case we use a two-tail test with a 5% si gnificance level. The critical values are given by
t 2.000 t 2.000
(0.025,60) and (0.975,60) . The rejection regiont 2ort 2.
(a) The value of the t statistic for testing the null h0 2thesiagainst the alternative
H :0 is
1 2
b2 3
t 1.5
se(b2) 4
Since 2 , we fail to rejec0 H and conclude that there is no sample evidence to
suggest that2 0.
(b) erting H0: 1 + 22 = 5 against the alternat1ve1H : 2 + 2 5, we use the statistic
b1 225
t
se 1 22
For the numerator of the t-value, we have52
1 2
The denominator is given by
se(1 2 ) 1va2( 2 b1 r2b ) 4 v12( ) 4 cov( , )
34442661.3
3
Therefore, t 0.9045
3.3166
Since 59.0 , we fail to rejec0 H . There is no sample evidence to suggest that
1 225 .
(c)torting H 0 123 against the alternati1 1423 , we use
4 (b)b b
t 1 2 3
se(1 2 3 )
6 4 1 3 Now, 4 (b)1 2 3 , and
se(b1 2 3 )b1 2 3( )
bbb1 2b3 2 3 1 31 2) 2cov( , ) 2cov( , ) 2cov( , )
4

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