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CHAPTER 4
SOLUTIONS TO PROBLEMS
4.2 (i) and (iii) generally cau 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 ca where the correlation is one.
4.3 (i) While the standard error on hrmp has not changed, the magnitude of the coefficient has incread by half. The t statistic on hrmp 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.)
crappy(ii) If we add and subtract 2βlog(employ ) from the right-hand-side and collect terms, we have
eachoflog(scrap ) = 0β + 1βhrmp + [2βlog(sales) – 2βlog(employ )] + [2βlog(employ ) + 3βlog(employ )] + u = 0β + 1βhrmp + 2βlog(sales /employ ) + (2β + 3β)log(employ ) + u ,
where the cond equality follows from the fact that log(sales /employ ) = log(sales ) – log(employ ). Defining 3θ ≡ 2β + 3β gives the result.
(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 u 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.
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(We can u the standard normal critical value, 1.96.) So log(mktval ) is statistically significant. Becau the coefficient is an elasticity, a ceteris paribus 10% increa in market value is predicted to increa salary by 1%. This is not a huge effect, but it is not negligible, either.
(iii) The 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 increas salary by about 1.71%. On the other hand, another year with the company, but not as CEO, lowers salary by about .92%. This cond finding at first ems 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 the people are bid up. More non-CEO years with a company makes it less likely the person was hired as an outside superstar.
4.7 (i) .412 ± 1.96(.094), or about .228 to .596.
(ii) No, becau the value .4 is well inside the 95% CI.
(iii) Yes, becau 1 is well outside the 95% CI.
4.8 (i) With df = 706 – 4 = 702, we u 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/
thelady5.88 ≈ −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.
(iii) Not really. The 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 ud assume homoskedasticity, in addition to the other CLM assumptions. If there is heteroskedasticity in the equation, the tests are no longer valid.
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 increa in sales,
roadmanthis ems like a practically small effect.
(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 H0 in favor of H1 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.
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SOLUTIONS TO COMPUTER EXERCISES
footwear
C4.1 (i) Holding other factors fixed,
111log()(/100)[100log()]
(/100)(%),
voteA expendA expendA expendA βββΔ=Δ=⋅Δ≈Δ
where we u the fact that 100log()expendA ⋅Δ ≈ %expendA Δ. So 1β/100 is the (ceteris paribus) percentage point change in voteA when expendA increas by one percent.
(ii) The null hypothesis is H 0: 2β = –1β, which means a z% increa in expenditure by A and a z% increa in expenditure by B leaves voteA unchanged. We can equivalently write H 0: 1β + 2β = 0.
(iii) The estimated equation (with standard errors in parenthes below estimates) is
odayn 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 increa in spending by candidate A increas 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 increa in spending by B reduces n voteA
by about .66 percentage points. The 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 1ˆβ + 2
ˆβ, 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 ( 1θ)≈ .533. The t statistic for the hypothesis in part (ii) is –.532/.533 ≈ –1. Therefore, we fail to reject H 0: 2β = –1β.
小学课堂纪律二十六个字母21
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.
Therefore, 1ˆθ= 150(.000379) + .0289 = .0858, which means that an additional 150 square foot bedroom increas 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 1ˆθ= .0858; now we also get
(1
ˆθ) = .0268. The 95% confidence interval reported by my software package is .0326 to .1390
(or about 3.3% to 13.9%).
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荆棘鸟英文
= .625.
Now hrunsyr is very statistically significant (t statistic ≈ 4.99), and its coefficient has incread by about two and one-half times.
(ii) The equation with runsyr , fldperc , and sbasyr added is
