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\begin{document}
\title{Econometrics Homework 2}
\author[1]{Felicia Cowley}%
\affil[1]{George Mason University}%
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\date{\today}
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\textbf{Questions 2, 3, and 9 from Chapter 4 on pp. 141-143}
\par\null
\textbf{Question 2. Consider an equation to explain the salaries of CEOs
in terms of annual firms sales, return on equity (roe, in percentage
form), and return on the firm's stock (ros, in percentage
form):~}\(log(salary) = \beta_0 + \beta_1 log(sales) +\beta_2 roe +\beta_3 ros +u\)
\textbf{(i) In terms of the model parameters, state the null hypothesis
that, after controlling for~}\emph{\textbf{sales}} \textbf{and}
\emph{\textbf{roe}}\textbf{,} \emph{\textbf{ros}} \textbf{has no effect
on CEO salary. State the alternative that better stock market
performance increases a CEO's salary.}
\(\)\(H_0 : \beta_3 =0\)
\(H_1 : \beta_3 >0\)
\textbf{(ii) Using the data in CEOSAL1, the following equation was
obtained by OLS:}
\textbf{~}\(\hat{log(salary)}=4.32 +.280 log(sales) +.0174 roe +.00024 ros\)
\textbf{~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (.32)~ ~ ~ ~ ~ ~ ~
~(.035)~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (.0041)~ ~ ~ ~ ~ ~ ~ ~ ~ (.00054)}
\textbf{n=209,~}\(R^2\)\textbf{=.283}
\textbf{By what percentage is \emph{salary} predicted to increase if
\emph{ros} increases by 50 points? Does \emph{ros} have a
practically~large effect on \emph{salary}?}
\(.00024 (50)=0.012\)~~~~~~~~~~~~~~~~~\(0.012(100)=1.2%\)~ ~ This 1.2\% is
practically a small effect on \emph{salary}.
\textbf{(iii) Test the null hypothesis that} \emph{\textbf{ros}}
\textbf{has no effect on~}\emph{\textbf{salary}} \textbf{against the
alternative that} \emph{\textbf{ros}} \textbf{has a positive effect.
Carry out the test at the 10\% significance level.}
~~\(t \hat{\beta}=\frac{.00024}{.00054}=0.4\bar{4}\)~ ~ ~ ~ ~ ~ ~ ~ and the~\(c=1.282\) because
of the degrees of freedom at a 10\% significance level.That
is,~\(n-k-1=209-3-1=205\) which is greater than the largest numerical
significance level on the critical values table (ie, use infinity). This
means that we fail to reject the null at a 10\% significance level. For
all intensive purposes, the relationship between ros and salary is
practically indistinguishable from zero.
\textbf{(iv) Would you include \emph{ros} in a final model explaining
CEO compensation in terms of firm performance? Explain.}
Including \emph{ros} might not be a negative generally as its
relationship is almost zero, however, this depends on its correlation
with other variables.
\par\null
\textbf{Question 3. The variable \emph{rdintens} is expenditures on
research and development (R\&D) as a percentage of sales. Sales are
measured in millions of dollars. The variable \emph{profmarg} is profits
as a percentage of sales. Using the data in RDCHEM for 32 firms in the
chemical industry, the following equation is estimated:~}
\(\hat{rdintens} =.472 + .321 log(sales) +.050 profmarg\)
\textbf{~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (1.369)~ ~(.216)~ ~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~ ~ ~ ~ (.046)}
\textbf{n=32,~}\(R^2=.099\)
\textbf{(i) Interpret the coefficient on log(\emph{sales}). In
particular, if sales increases by 10\%, what is the estimated percentage
point change in \emph{rdintens}? Is this an economically large effect?}
If sales goes upward by 10\%, rdintens will increased by approximately
.0321 percentage points (\(\frac{.321}{100}\)) which shows that the sales
change is not practically large.
\textbf{(ii) Test the hypothesis that R\&D intensity does not change
with sales against the alternative that does increase with sales. Do the
test at the 5\% and 10\% levels.}
\(t=\frac{.321}{.216} \approx1.486\)~ ~ ~ ~ ~The degrees of freedom
are~\(n-k-1=32-2-1=29\) which means using the critical values table,
c=1.699 at a 5\% significance level and c=1.311 at a 10\% significance
level. This means at 5\% significance level, we can reject
since~\(1.699>1.486\) and we cannot reject at a 10\% significance
level as~\(1.311<1.486\).
\textbf{(iii) Interpret the coefficient on}
\emph{\textbf{profmarg}}\textbf{. Is it economically large?}
The coefficient is not economically large. For instance, a 1\% increase
in \emph{profmarg} only lends to a 0.05\% increase in \emph{rdintens}.
\textbf{(iv) Does profmarg have a statistically significant effect on
rdintens?}
\(t=\frac{\hat{\beta_1} -\hat{\beta_2}}{SE(\hat{\beta_1} -\hat{\beta_2})}=\frac{0.05-0}{.046}\approx1.087\)~ ~That answer is no since the t statistic is below
the 10\% critical value for the one-tailed test.
\textbf{Question 9. In Problem 3 in Chapter 3, we estimated the
equation}
\textbf{~}\(\hat{sleep}=3638.25-.148 totwrk -11.13 educ +2.20 age\) \textbf{~ ~ ~ ~ ~}
\textbf{~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (112.28)~ ~ ~ ~ ~ ~ ~ (.017)~ ~ ~ ~ ~
~ ~ ~ ~ (5.88)~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (1.45)}
\textbf{n=706,~}\(R^2=.113\)
\textbf{where we now report standard errors along the estimates.}
\textbf{(i) Is either \emph{educ} or \emph{age} individually significant
at the 5\% level against a two-sided alternative? Show your work}.
~ ~\(\)\(t_{educ}=\frac{\hat{\beta_1} -\hat{\beta_2}}{SE(\hat{\beta_1} -\hat{\beta_2})}=\frac{-11.13+0}{5.88}\approx -1.893\) and the critical value is 1.96
found by the critical values table when~\(n-k-1=706-3-1=702\). This means
that since the absolute value of 1.893 \textless{} 1.96, we fail to
reject the null which is~\(H_0 : \beta_{educ} =0\) at the 5\% significance
level.~
\(t_{age}=\frac{\hat{\beta_1} -\hat{\beta_2}}{SE(\hat{\beta_1} -\hat{\beta_2})}=\)\(\frac{2.20-0}{1.45}\approx 1.517\)~ and the critical value is 1.96
found by the critical values table with 702 degrees of freedom.
Accordingly, 1.517\textless{}1.96 meaning that we fail to reject the
null~\(H_0 : \beta_{age}=0\). Both variables are not statistically
significant at a 5\% significance level.
\textbf{(ii) Dropping \emph{educ} and \emph{age} from the equation
gives~}
\(\hat{sleep}=3586.38-.151 totwrk\)
\textbf{~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (38.91)~ ~ ~ ~ ~ ~ ~(.017)}
\textbf{n=706,~}\(R^2=.103\)
\textbf{Are \emph{educ} and \emph{age} jointly significant in the
original equation at the 5\% level? Justify your answer.}
\(F=\frac{(R_{ur} ^2 -R_r ^2)/q}{(1-R_{ur} ^2)/df_{ur}}\)\(=\frac{(.113-.103)/2}{(1-.113)/(702)}=\frac{0.01/(2)}{0.887(1/702)}=0.005/0.00126353=3.96\)
The 5\% critical value is c=3.00 with infinite degrees of freedom. This
means that \emph{educ} and \emph{age} are jointly significant in the
original equation.
\textbf{(iii) Does including} \emph{\textbf{educ}} \textbf{and}
\emph{\textbf{age}} \textbf{in the model greatly affect the estimated
tradeoff between sleeping and working?}
No it doesn't. While the variables are jointly significant, including
the variables only change the coefficient on \emph{totwrk} from -.151 to
-1.48 which is practically small.
\textbf{(iv) Suppose that the sleep equation contains
heteroskedasticity. What does this mean about the tests computed in
parts (i) and (ii)?}
If heteroskedasticity existed, then the test computed in those parts are
not valid and the results are biased.
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