伍德里奇计量经济学导论第四版

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied,

or distributed without the prior consent of the publisher.

28

CHAPTER 4

TEACHING NOTES

At the start of this chapter is good time to remind students that a specific error distribution played no role in the results of Chapter 3. That is because only the first two moments were derived under the full set of Gauss-Markov assumptions. Nevertheless, normality is needed to obtain exact normal sampling distributions (conditional on the explanatory variables). I

emphasize that the full set of CLM assumptions are used in this chapter, but that in Chapter 5 we relax the normality assumption and still perform approximately valid inference. One could argue that the classical linear model results could be skipped entirely, and that only large-sample analysis is needed. But, from a practical perspective, students still need to know where the t distribution comes from because virtually all regression packages report t statistics and obtain p -values off of the t distribution. I then find it very easy to cover Chapter 5 quickly, by just saying we can drop normality and still use t statistics and the associated p -values as being

approximately valid. Besides, occasionally students will have to analyze smaller data sets, especially if they do their own small surveys for a term project.

It is crucial to emphasize that we test hypotheses about unknown population parameters. I tell

my students that they will be punished if they write something like H 0:1

?β = 0 on an exam or, even worse, H 0: .632 = 0.

One useful feature of Chapter 4 is its illustration of how to rewrite a population model so that it contains the parameter of interest in testing a single restriction. I find this is easier, both theoretically and practically, than computing variances that can, in some cases, depend on

numerous covariance terms. The example of testing equality of the return to two- and four-year colleges illustrates the basic method, and shows that the respecified model can have a useful interpretation. Of course, some statistical packages now provide a standard error for linear combinations of estimates with a simple command, and that should be taught, too.

One can use an F test for single linear restrictions on multiple parameters, but this is less transparent than a t test and does not immediately produce the standard error needed for a

confidence interval or for testing a one-sided alternative. The trick of rewriting the population model is useful in several instances, including obtaining confidence intervals for predictions in Chapter 6, as well as for obtaining confidence intervals for marginal effects in models with interactions (also in Chapter 6).

The major league baseball player salary example illustrates the difference between individual and joint significance when explanatory variables (rbisyr and hrunsyr in this case) are highly correlated. I tend to emphasize the R -squared form of the F statistic because, in practice, it is applicable a large percentage of the time, and it is much more readily computed. I do regret that this example is biased toward students in countries where baseball is played. Still, it is one of the better examples of multicollinearity that I have come across, and students of all backgrounds seem to get the point.

课后答案网 w w w .k h d a w .c o m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 29SOLUTIONS TO PROBLEMS

4.1 (i) H 0:3β = 0. H 1:3β > 0.

(ii) The proportionate effect on n salary

is .00024(50) = .012. To obtain the percentage effect, we multiply this by 100: 1.2%. Therefore, a 50 point ceteris paribus increase in ros is predicted to increase salary by only 1.2%. Practically speaking, this is a very small effect for such a large change in ros .

(iii) The 10% critical value for a one-tailed test, using df = ∞, is obtained from Table G.2 as

1.28

2. The t statistic on ros is .00024/.00054 ≈ .44, which is well below the critical value. Therefore, we fail to reject H 0 at the 10% significance level.

(iv) Based on this sample, the estimated ros coefficient appears to be different from zero only because of sampling variation. On the other hand, including ros may not be causing any harm; it depends on how correlated it is with the other independent variables (although these are very significant even with ros in the equation).

4.2 (i) and (iii) generally cause the t statistics not to have a t distribution under H 0.

Homoskedasticity is one of the CLM assumptions. An important omitted variable violates Assumption MLR.3. The CLM assumptions contain no mention of the sample correlations among independent variables, except to rule out the case where the correlation is one.

4.3 (i) While the standard error on hrsemp has not changed, the magnitude of the coefficient has increased by half. The t statistic on hrsemp has gone from about –1.47 to –2.21, so now the coefficient is statistically less than zero at the 5% level. (From Table G.2 the 5% critical value with 40 df is –1.684. The 1% critical value is –2.423, so the p -value is between .01 and .0

5.)

(ii) If we add and subtract 2βlog(employ ) from the right-hand-side and collect terms, we

have

log(scrap ) = 0β + 1βhrsemp + [2βlog(sales) – 2βlog(employ )]

+ [2βlog(employ ) + 3βlog(employ )] + u

= 0β + 1βhrsemp + 2βlog(sales /employ )

+ (2β + 3β)log(employ ) + u ,

where the second equality follows from the fact that log(sales /employ ) = log(sales ) –

log(employ ). Defining 3θ ≡ 2β + 3β gives the result.

课后答案网 w w w .k h d a w .c o m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 30 (iii) No. We are interested in the coefficient on log(employ ), which has a t statistic of .2, which is very small. Therefore, we conclude that the size of the firm, as measured by employees, does not matter, once we control for training and sales per employee (in a logarithmic functional form).

(iv) The null hypothesis in the model from part (ii) is H 0:2β = –1. The t statistic is [–.951 –

(–1)]/.37 = (1 – .951)/.37 ≈ .132; this is very small, and we fail to reject whether we specify a one- or two-sided alternative.

4.4 (i) In columns (2) and (3), the coefficient on profmarg is actually negative, although its t statistic is only about –1. It appears that, once firm sales and market value have been controlled for, profit margin has no effect on CEO salary.

(ii) We use column (3), which controls for the most factors affecting salary. The t statistic on log(mktval ) is about 2.05, which is just significant at the 5% level against a two-sided alternative. (We can use the standard normal critical value, 1.96.) So log(mktval ) is statistically significant. Because the coefficient is an elasticity, a ceteris paribus 10% increase in market value is predicted to increase salary by 1%. This is not a huge effect, but it is not negligible, either.

(iii) These variables are individually significant at low significance levels, with t ceoten ≈ 3.11 and t comten ≈ –2.79. Other factors fixed, another year as CEO with the company increases salary by about 1.71%. On the other hand, another year with the company, but not as CEO, lowers salary by about .92%. This second finding at first seems surprising, but could be related to the “superstar” effect: firms that hire CEOs from outside the company often go after a small pool of highly regarded candidates, and salaries of these people are bid up. More non-CEO years with a company makes it less likely the person was hired as an outside superstar.

4.5 (i) With df = n – 2 = 86, we obtain the 5% critical value from Table G.2 with df = 90. Because each test is two-tailed, the critical value is 1.987. The t statistic for H 0:0β = 0 is about -

.89, which is much less than 1.987 in absolute value. Therefore, we fail to reject 0β = 0. The t

statistic for H 0: 1β = 1 is (.976 – 1)/.049 ≈ -.49, which is even less significant. (Remember, we

reject H 0 in favor of H 1 in this case only if |t | > 1.987.)

(ii) We use the SSR form of the F statistic. We are testing q = 2 restrictions and the df in the unrestricted model is 86. We are given SSR r = 209,448.99 and SSR ur = 165,644.51. Therefore,

(209,448.99165,644.51)8611.37,165,644.512F ???=?≈????

which is a strong rejection of H 0: from Table G.3c, the 1% critical value with 2 and 90 df is

4.8

5.

课后答案网 w w w .k h d a w .c o m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 31 (iii) We use the R -squared form of the F statistic. We are testing q = 3 restrictions and there are 88 – 5 = 83 df in the unrestricted model. The F statistic is [(.829 – .820)/(1 – .829)](83/3) ≈

1.46. The 10% critical value (again using 90 denominator df in Table G.3a) is

2.15, so we fail to reject H 0 at even the 10% level. In fact, the p -value is about .2

3.

(iv) If heteroskedasticity were present, Assumption MLR.5 would be violated, and the F statistic would not have an F distribution under the null hypothesis. Therefore, comparing the F statistic against the usual critical values, or obtaining the p -value from the F distribution, would not be especially meaningful.

4.6 (i) We need to compute the F statistic for the overall significance of the regression with n = 142 and k = 4: F = [.0395/(1 – .0395)](137/4) ≈ 1.41. The 5% critical value with 4 numerator df and using 120 for the numerator df , is 2.45, which is well above the value of F . Therefore, we fail to reject H 0: 1β = 2β = 3β = 4β = 0 at the 10% level. No explanatory variable is

individually significant at the 5% level. The largest absolute t statistic is on dkr , t dkr ≈ 1.60, which is not significant at the 5% level against a two-sided alternative.

(ii) The F statistic (with the same df ) is now [.0330/(1 – .0330)](137/4) ≈ 1.17, which is even lower than in part (i). None of the t statistics is significant at a reasonable level.

(iii) We probably should not use the logs, as the logarithm is not defined for firms that have zero for dkr or eps . Therefore, we would lose some firms in the regression.

(iv) It seems very weak. There are no significant t statistics at the 5% level (against a two-sided alternative), and the F statistics are insignificant in both cases. Plus, less than 4% of the variation in return is explained by the independent variables.

4.7 (i) .412 ± 1.96(.094), or about .228 to .596.

(ii) No, because the value .4 is well inside the 95% CI.

(iii) Yes, because 1 is well outside the 95% CI.

4.8 (i) With df = 706 – 4 = 702, we use the standard normal critical value (df = ∞ in Table G.2), which is 1.96 for a two-tailed test at the 5% level. Now t educ = ?11.13/

5.88 ≈ ?1.89, so |t educ | =

1.89 < 1.96, and we fail to reject H 0: educ β = 0 at the 5% level. Also, t age ≈ 1.52, so age is also statistically insignificant at the 5% level.

(ii) We need to compute the R -squared form of the F statistic for joint significance. But F =

[(.113 ? .103)/(1 ? .113)](702/2) ≈ 3.96. The 5% critical value in the F 2,702 distribution can be obtained from Table G.3b with denominator df = ∞: cv = 3.00. Therefore, educ and age are jointly significant at the 5% level (3.96 > 3.00). In fact, the p -value is about .019, and so educ and age are jointly significant at the 2% level.

课后答案网 w w w .k h d a w .c o m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied,

or distributed without the prior consent of the publisher.

32

(iii) Not really. These variables are jointly significant, but including them only changes the coefficient on totwrk from –.151 to –.148.

(iv) The standard t and F statistics that we used assume homoskedasticity, in addition to the other CLM assumptions. If there is heteroskedasticity in the equation, the tests are no longer valid.

4.9 (i) H 0:3β = 0. H 1:3β ≠ 0.

(ii) Other things equal, a larger population increases the demand for rental housing, which should increase rents. The demand for overall housing is higher when average income is higher, pushing up the cost of housing, including rental rates.

(iii) The coefficient on log(pop ) is an elasticity. A correct statement is that “a 10% increase in population increases rent by .066(10) = .66%.”

(iv) With df = 64 – 4 = 60, the 1% critical value for a two-tailed test is 2.660. The t statistic is about 3.29, which is well above the critical value. So 3β is statistically different from zero at the 1% level.

4.10 (i) We use Property VAR.3 from Appendix B: Var(1?β ? 32?β) = Var (1

?β) + 9 Var (2

?β

) – 6 Cov (1

?β,2

?β).

(ii) t = (1?β? 32?β ? 1)/se(1?β? 32?β), so we need the standard error of 1?β ? 32

?β.

(iii) Because

1θ = 1β – 3β2, we can write 1β = 1θ + 3β2. Plugging this into the population model gives y = 0β + (1θ + 3β2)x 1 + 2βx 2 + 3βx 3 + u

= 0β + 1θx 1 + 2β(3x 1 + x 2) + 3βx 3 + u .

This last equation is what we would estimate by regressing y on x 1, 3x 1 + x 2, and x 3. The coefficient and standard error on x 1 are what we want.

4.11 (i) Holding profmarg fixed, n rdintens

Δ = .321 Δlog(sales ) = (.321/100)[100log()sales ?Δ] ≈ .00321(%Δsales ). Therefore, if %Δsales = 10, n rdintens Δ ≈ .032, or only about 3/100 of a percentage point. For such a large percentage increase in sales,

this seems like a practically small effect.

课

后

答案

网

w

w w

.k

h d a

w .

c o

m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 33 (ii) H 0:1β = 0 versus H 1:1β > 0, where 1β is the population slope on log(sales ). The t

statistic is .321/.216 ≈ 1.486. The 5% critical value for a one-tailed test, with df = 32 – 3 = 29, is obtained from Table G.2 as 1.699; so we cannot reject H 0 at the 5% level. But the 10% critical value is 1.311; since the t statistic is above this value, we reject H 0 in favor of H 1 at the 10% level.

(iii) Not really. Its t statistic is only 1.087, which is well below even the 10% critical value for a one-tailed test.

课后答案网 w w w .k h d a w .c o m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied,

or distributed without the prior consent of the publisher.

34

SOLUTIONS TO COMPUTER EXERCISES

C4.1 (i) Holding other factors fixed,

111log()(/100)[100log()]

(/100)(%),

voteA expendA expendA expendA βββΔ=Δ=?Δ≈Δ

where we use the fact that 100log()expendA ?Δ ≈ %expendA Δ. So 1β/100 is the (ceteris paribus) percentage point change in voteA when expendA increases by one percent.

(ii) The null hypothesis is H 0: 2β = –1β, which means a z% increase in expenditure by A and a z% increase in expenditure by B leaves voteA unchanged. We can equivalently write H 0: 1β + 2β = 0.

(iii) The estimated equation (with standard errors in parentheses below estimates) is

n voteA = 45.08 + 6.083 log(expendA ) – 6.615 log(expendB ) + .152 prtystrA (3.93) (0.382) (0.379) (.062)

n = 173, R 2 = .793.

The coefficient on log(expendA ) is very significant (t statistic ≈ 15.92), as is the coefficient on log(expendB ) (t statistic ≈ –17.45). The estimates imply that a 10% ceteris paribus increase in spending by candidate A increases the predicted share of the vote going to A by about .61

percentage points. [Recall that, holding other factors fixed, n voteA

Δ≈(6.083/100)%ΔexpendA ).] Similarly, a 10% ceteris paribus increase in spending by B reduces n voteA

by about .66 percentage points. These effects certainly cannot be ignored.

While the coefficients on log(expendA ) and log(expendB ) are of similar magnitudes (and

opposite in sign, as we expect), we do not have the standard error of 1?β + 2

?β, which is what we would need to test the hypothesis from part (ii).

(iv) Write 1θ = 1β +2β, or 1β = 1θ– 2β. Plugging this into the original equation, and rearranging, gives

n voteA = 0β + 1θlog(expendA ) + 2β[log(expendB ) – log(expendA )] +3βprtystrA + u ,

When we estimate this equation we obtain 1θ

≈ –.532 and se( 1θ)≈ .533. The t statistic for the hypothesis in part (ii) is –.532/.533 ≈ –1. Therefore, we fail to reject H 0: 2β = –1β.

课

后

答案

网

w

w w

.k

h d a

w .

c o

m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied,

or distributed without the prior consent of the publisher.

35

C4.2 (i) In the model

log(salary ) = 0β+1βLSAT +2βGPA + 3βlog(libvol ) +4βlog(cost)+5βrank + u ,

the hypothesis that rank has no effect on log(salary ) is H 0:5β = 0. The estimated equation (now with standard errors) is

n log()salary = 8.34 + .0047 LSAT + .248 GPA + .095 log(libvol ) (0.53) (.0040) (.090) (.033)

+ .038 log(cost ) – .0033 rank (.032) (.0003)

n = 136, R 2 = .842.

The t statistic on rank is –11, which is very significant. If rank decreases by 10 (which is a move up for a law school), median starting salary is predicted to increase by about 3.3%.

(ii) LSAT is not statistically significant (t statistic ≈ 1.18) but GPA is very significance (t statistic ≈ 2.76). The test for joint significance is moot given that GPA is so significant, but for completeness the F statistic is about 9.95 (with 2 and 130 df ) and p -value ≈ .0001.

(iii) When we add clsize and faculty to the regression we lose five observations. The test of their joint significant (with 2 and 131 – 8 = 123 df ) gives F ≈ .95 and p -value ≈ .39. So these two variables are not jointly significant unless we use a very large significance level.

(iv) If we want to just determine the effect of numerical ranking on starting law school salaries, we should control for other factors that affect salaries and rankings. The idea is that there is some randomness in rankings, or the rankings might depend partly on frivolous factors that do not affect quality of the students. LSAT scores and GPA are perhaps good controls for student quality. However, if there are differences in gender and racial composition across

schools, and systematic gender and race differences in salaries, we could also control for these. However, it is unclear why these would be correlated with rank . Faculty quality, as perhaps measured by publication records, could be included. Such things do enter rankings of law schools.

C4.3 (i) The estimated model is

n log()price = 11.67 + .000379 sqrft + .0289 bdrms (0.10) (.000043) (.0296)

n = 88, R 2 = .588.

课

后

答

案

网

w

w w

.k

h d a

w .

c o

m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 36Therefore, 1?θ= 150(.000379) + .0289 = .0858, which means that an additional 150 square foot

bedroom increases the predicted price by about 8.6%.

(ii) 2β= 1θ – 1501β, and so

log(price ) = 0β+ 1βsqrft + (1θ – 1501β)bdrms + u = 0β+ 1β(sqrft – 150 bdrms ) + 1θbdrms + u .

(iii) From part (ii), we run the regression

log(price ) on (sqrft – 150 bdrms ), bdrms ,

and obtain the standard error on bdrms . We already know that 1?θ= .0858; now we also get se(1

?θ) = .0268. The 95% confidence interval reported by my software package is .0326 to .1390 (or about 3.3% to 13.9%). C4.4 The R -squared from the regression bwght on cigs , parity , and faminc , using all 1,388 observations, is about .0348. This means that, if we mistakenly use this in place of .0364, which is the R -squared using the same 1,191 observations available in the unrestricted regression, we would obtain F = [(.0387 ? .0348)/(1 ? .0387)](1,185/2) ≈ 2.40, which yields p -value ≈ .091 in an F distribution with 2 and 1,1185 df . This is significant at the 10% level, but it is incorrect. The correct F statistic was computed as 1.42 in Example 4.9, with p -value ≈ .242. C4.5 (i) If we drop rbisyr the estimated equation becomes n log()salary = 11.02 + .0677 years + .0158 gamesyr (0.27) (.0121) (.0016) + .0014 bavg + .0359 hrunsyr (.0011) (.0072) n = 353, R 2 = .625.

Now hrunsyr is very statistically significant (t statistic ≈ 4.99), and its coefficient has increased by about two and one-half times.

(ii) The equation with runsyr , fldperc , and sbasesyr added is 课后答案网 w w w .k h d a w .c o m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 37 n log()salary =

10.41 + .0700 years + .0079 gamesyr (2.00) (.0120) (.0027)

+ .00053 bavg + .0232 hrunsyr

(.00110) (.0086)

+ .0174 runsyr + .0010 fldperc – .0064 sbasesyr

(.0051) (.0020) (.0052) n = 353, R 2 = .639.

Of the three additional independent variables, only runsyr is statistically significant (t statistic = .0174/.0051 ≈ 3.41). The estimate implies that one more run per year, other factors fixed,

increases predicted salary by about 1.74%, a substantial increase. The stolen bases variable even has the “wrong” sign with a t statistic of about –1.23, while fldperc has a t statistic of only .5. Most major league baseball players are pretty good fielders; in fact, the smallest fldperc is 800 (which means .800). With relatively little variation in fldperc , it is perhaps not surprising that its effect is hard to estimate.

(iii) From their t statistics, bavg , fldperc , and sbasesyr are individually insignificant. The F statistic for their joint significance (with 3 and 345 df ) is about .69 with p -value ≈ .56. Therefore, these variables are jointly very insignificant.

C4.6 (i) In the model

log(wage ) = 0β + 1βeduc + 2βexper + 3βtenure + u

the null hypothesis of interest is H 0: 2β = 3β.

(ii) Let 2θ = 2β – 3β. Then we can estimate the equation

log(wage ) = 0β + 1βeduc + 2θexper + 3β(exper + tenure ) + u

to obtain the 95% CI for 2θ. This turns out to be about .0020 ± 1.96(.0047), or about -.0072 to

.0112. Because zero is in this CI, 2θ is not statistically different from zero at the 5% level, and

we fail to reject H 0: 2β = 3β at the 5% level.

C4.7 (i) The minimum value is 0, the maximum is 99, and the average is about 56.16.

(ii) When phsrank is added to (4.26), we get the following:

n log(

) wage = 1.459 ? .0093 jc + .0755 totcoll + .0049 exper + .00030 phsrank (0.024) (.0070) (.0026) (.0002) (.00024) 课后答案网 w w w .k h d a w .c o m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 38 n = 6,763, R 2 = .223

So phsrank has a t statistic equal to only 1.25; it is not statistically significant. If we increase phsrank by 10, log(wage ) is predicted to increase by (.0003)10 = .003. This implies a .3% increase in wage , which seems a modest increase given a 10 percentage point increase in phsrank . (However, the sample standard deviation of phsrank is about 24.)

(iii) Adding phsrank makes the t statistic on jc even smaller in absolute value, about 1.33, but the coefficient magnitude is similar to (4.26). Therefore, the base point remains unchanged: the return to a junior college is estimated to be somewhat smaller, but the difference is not significant and standard significant levels.

(iv) The variable id is just a worker identification number, which should be randomly assigned (at least roughly). Therefore, id should not be correlated with any variable in the regression equation. It should be insignificant when added to (4.17) or (4.26). In fact, its t statistic is about .54.

C4.8 (i) There are 2,017 single people in the sample of 9,275.

(ii) The estimated equation is

n nettfa = ?43.04 + .799 inc + .843 age ( 4.08) (.060) (.092)

n = 2,017, R 2 = .119.

The coefficient on inc indicates that one more dollar in income (holding age fixed) is reflected in about 80 more cents in predicted nettfa ; no surprise there. The coefficient on age means that, holding income fixed, if a person gets another year older, his/her nettfa is predicted to increase by about $843. (Remember, nettfa is in thousands of dollars.) Again, this is not surprising.

(iii) The intercept is not very interesting as it gives the predicted nettfa for inc = 0 and age = 0. Clearly, there is no one with even close to these values in the relevant population.

(iv) The t statistic is (.843 ? 1)/.092 ≈ ?1.71. Against the one-sided alternative H 1: β2 < 1, the p-value is about .044. Therefore, we can reject H 0: β2 = 1 at the 5% significance level (against the one-sided alternative).

(v) The slope coefficient on inc in the simple regression is about .821, which is not very different from the .799 obtained in part (ii). As it turns out, the correlation between inc and age in the sample of single people is only about .039, which helps explain why the simple and multiple regression estimates are not very different; refer back to page 84 of the text.

课后答案网 w w w .k h d a w .c o m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 39C4.9 (i) The results from the OLS regression, with standard errors in parentheses, are

n log(

) psoda =?1.46 + .073 prpblck + .137 log(income ) + .380 prppov (0.29) (.031) (.027) (.133)

n = 401, R 2 = .087

The p -value for testing H 0: 10β= against the two-sided alternative is about .018, so that we

reject H 0 at the 5% level but not at the 1% level.

(ii) The correlation is about ?.84, indicating a strong degree of multicollinearity. Yet each

coefficient is very statistically significant: the t statistic for log()?income β is about 5.1 and that for

?prppov

β is about 2.86 (two-sided p -value = .004).

(iii) The OLS regression results when log(hseval ) is added are

n log(

) psoda =?.84 + .098 prpblck ? .053 log(income ) (.29) (.029) (.038)

+ .052 prppov + .121 log(hseval )

(.134) (.018)

n = 401, R 2 = .184

The coefficient on log(hseval ) is an elasticity: a one percent increase in housing value, holding the other variables fixed, increases the predicted price by about .12 percent. The two-sided p -value is zero to three decimal places.

(iv) Adding log(hseval ) makes log(income ) and prppov individually insignificant (at even the 15% significance level against a two-sided alternative for log(income ), and prppov is does not have a t statistic even close to one in absolute value). Nevertheless, they are jointly significant at the 5% level because the outcome of the F 2,396 statistic is about 3.52 with p -value = .030. All of the control variables – log(income ), prppov , and log(hseval ) – are highly correlated, so it is not surprising that some are individually insignificant.

(v) Because the regression in (iii) contains the most controls, log(hseval ) is individually significant, and log(income ) and prppov are jointly significant, (iii) seems the most reliable. It holds fixed three measure of income and affluence. Therefore, a reasonable estimate is that if the proportion of blacks increases by .10, psoda is estimated to increase by 1%, other factors held fixed.

课后答案网 w w w .k h d a w .c o m

This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 40C4.10 (i) Using the 1,848 observations, the simple regression estimate of bs β is about .795?. The 95% confidence interval runs from 1.088 to .502??, which includes ?1. Therefore, at the 5% level, we cannot reject that 0H :1bs β=? against the two-sided alternative.

(ii) When lenrol and lstaff are added to the regression, the coefficient on bs becomes about ?.605; it is now statistically different from one, as the 95% CI is from about ?.818 to ?.392. The situation is very similar to that in Table 4.1, where the simple regression estimate is ?.825 and the multiple regression estimate (with the logs of enrollment and staff included) is ?.605. (It is a coincidence that the two multiple regression estimates are the same, as the data set in Table 4.1 is for an earlier year at the high school level.)

(iii) The standard error of the simple regression estimate is about .150, and that for the multiple regression estimate is about .109. When we add extra explanatory variables, two factors work in opposite directions on the standard errors. Multicollinearity – in this case, correlation between bs and the two variables lenrol and lstaff works to increase the multiple regression standard error. Working to reduce the standard error of ?bs βis the smaller error variance when lenrol and lstaff are included in the regression; in effect, they are taken out of the simple regression error term. In this particular example, the multicollinearity is modest compared with the reduction in the error variance. In fact, the standard error of the regression goes from .231 for simple regression to .168 in the multiple regression. (Another way to summarize the drop in the error variance is to note that the R -squared goes from a very small .0151 for the simple regression to .4882 for multiple regression.) Of course, ahead of time we cannot know which effect will dominate, but we can certainly compare the standard errors after running both regressions.

(iv) The variable lstaff is the log of the number of staff per 1,000 students. As lstaff increases, there are more teachers per student. We can associate this with smaller class sizes, which are generally desirable from a teacher’s perspective. It appears that, all else equal, teachers are willing to take less in salary to have smaller class sizes. The elasticity of salary with respect to staff is about ?.714, which seems quite large: a ten percent increase in staff size (holding enrollment fixed) is associated with a 7.14 percent lower salary.

(v) When lunch is added to the regression, its coefficient is about ?.00076, with t = ?4.69. Therefore, other factors fixed (bs , lenrol , and lstaff ), a hire poverty rate is associated with lower teacher salaries. In this data set, the average value of lunch is about 36.3 with standard deviation of 25.4. Therefore, a one standard deviation increase in lunch is associated with a change in lsalary of about ?.00076(25.4) ≈ ?.019, or almost two percent lower. Certainly there is no evidence that teachers are compensated for teaching disadvantaged children.

(vi) Yes, the pattern obtained using ELEM94_95.RAW is very similar to that in Table 4.1, and the magnitudes are reasonably close, too. The largest estimate (in absolute value) is the simple regression estimate, and the absolute value declines as more explanatory variables are added. The final regressions in the two cases are not the same, because we do not control for lunch in Table 4.1, and graduation and dropout rates are not relevant for elementary school children.

课后答案网 w w w .k h d a w .c o m

本文来源：https://www.bwwdw.com/article/k9ce.html