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Department of Economics Economics 1123

Harvard University Fall 2003

Midterm Exam

Solutions

Question 1 (24 points)

The measure of a “no-school day” used in regression (1) (column (1) in Table 2) is whether the

day is a teacher meeting day. The measure of a “no-school day” in regression (2) is whether the

day is a break day (Thanksgiving break, etc.).

a)  (9 points) For regression (1):

(i)   Provide the estimated effect on the number of incidents of having a “no-school day.”

estimated effect = .76

(ii)

Is this estimated effect large in a real-world sense? Briefly, explain.

This means that there is on average .76 more juvenile property crime incidents on a teacher

meeting day, which seems moderately large – big enough that it shouldn’t be ignored. The mean

number of incidents is 3.35, so an increase of .76 represents an increase of just over 20%,

relative to the mean, and an increase in teen criminal activity of 20% is large enough to be of

concern. (Reaching a different judgment is OK if the reasoning is sound.)

(iii)  Test the hypothesis that this effect is zero, against the alternative that it is nonzero, at

the 5% significance level.

t  = .76/.26 = 2.92, so |t | > 1.96 so the hypothesis is rejected at the 5% significance level.

b)  (9 points) Repeat (i) – (iii) for regression (2)

(i) estimated effect = .173

(ii) This means that there is on average .173 more juvenile property crime incidents on a break

day, which seems pretty small – one crime for every 6 break days; certainly much less than the

estimate in regression (1). (Reaching a different judgment is OK if the reasoning is sound.)

(iii) t  = .173/.114 = 1.52, so |t | < 1.96, so the null hypothesis is not rejected.

c)  (6 points) Suggest a plausible reason, based on the definitions of the variables in

regressions (1) and (2), why the two estimates differ.

The two types of days are different. On a teacher meeting days, one would expect less parental

supervision and the teens to have more free time, because this is a normal work day for the

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parents. On holidays like Thanksgiving or Christmas, teens are more likely to be spending time

with their family rather than hanging. (Other explanations are OK if the logic is sound.)

Question 2 (30 points)

a)

(6 points) Explain what is meant by the SER in regression (3).

The SER is 3.89, which means that a typical prediction error from the regression has absolute

magnitude 3.89 (units of incidents per day)

b)  (6 points) Interpret the coefficient on teacherday in regression (3).

The effect on the number of incidents of having a teacher meeting day is .87 (an increase of .87

incidents/day), holding constant population, whether the day is a break day, and whether the day

falls in the summer. (Here is another way to say the same thing: holding constant population,

the effect of having a teacher meeting day on what would otherwise be a normal school day is to

increase juvenile property crime incidents by .87 [this wording is OK because including breakday and summer  means that the base case for the comparisons is a normal school day]).

c)  (6 points) Suggest a reason why the errors in regression (3) might be heteroskedastic;

explain.

Heteroskedasticity occurs when the variance of the error term depends on one or more of the

regressors.

In this regression, var(u) could plausibly depend on population. Bigger cities will have more

incidents because there are more teens, and it is plausible that the variability of the number of

incidents from day to day is greater in big cities than small cities (the distribution for a small city might range from 0 to 5; for a big city, it might range from 2 to 20). Thus the spread of the

distribution, as well as its mean, depends on the population; that is, var(u), will be a function of

population, one of the regressors.

d)  (6 points) Using regression (3), compute the predicted value of the number of incidents on a

teacher meeting day for a city with a population of 200,000 (so pop = 2).

n# incidents  = 1.39 + .87teacherday + .129breakday + .292 summer  + 1.58 pop

= 1.39 + .87×1 + .129×0 + .292×0 + 1.58×2 = 5.42

e)  (6 points) The school superintendent in a city with population 200,000 is contemplating

changing a normal school day into a teacher meeting day. Use regression (6) (not

regression (3)) to estimate the effect of this decision on the number of juvenile property

crime incidents.

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This can be done using a “before” and “after” calculation. Note that the terms that do not

involve teacherday, such as the term in breakday, are the same in the “before” and “after”

scenarios and thus cancel out when “before” is subtracted from “after.” The operational part of

the calculation (ignoring the terms that drop out) thus is:

“after:” n# incidents  = –.82teacherday + 3.05(teacherday× pop) – 1.21(teacherday× pop

2 )

+ .13(teacherday× pop 3 )

= –.82×1 + 3.05×(1×2) – 1.21×(1×2 2 ) + .13×(1×2

3 ) = 1.48

“before:” n# incidents  = –.82teacherday + 3.05(teacherday× pop) – 1.21(teacherday× pop

2 )

+ .13(teacherday× pop 3 )

= –.82×0 + 3.05×(0×2) – 1.21×(0×22) + .13×(0×23) = 0

The predicted effect of the decision is “after” – “before” = 1.48, that is, according to regression

(6) the predicted effect of the superintendent’s decision is an increase of 1.48 juvenile property

crime incidents on the proposed teacher meeting day.

Question 3 (26 points)

a)  (8 points) One possibility is that pop enters the population regression function nonlinearly.

The hypothesis of linearity implies that the population coefficients on pop 2

and pop 3

are zero. This is tested using the F -statistic, which is 171.5, so with p < .001 the hypothesis is rejected at

the 5% (1%, etc.) significance level. This provides evidence that pop enters the population

regression function for specification (5) nonlinearly. ( Note: it is not a complete answer to

examine only the individual t -statistics on the two individual coefficients on pop 2  and pop

3 ,

unless you do so using the Bonferroni critical values.)

b)  (8 points) Another possibility is that the effect on crime of a “no-school day” is different in

possibility? Briefly explain (be precise).

To examine whether the effect of teacherday depends on population, we need to examine regressions with interactions between teacherday and pop. There are two such regressions, (4)

and (6).

Regression (4) There is a single interaction term, teacherday× pop. The t -statistic is t  = .40/.31

= 1.29 so the hypothesis that this coefficient is zero in population is not rejected at the 5%

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significance level. So, there is no statistical evidence of an interaction effect in this regression.

Regression (6) There are three interaction terms, teacherday× pop, teacherday× pop 2 , and

teacherday× pop 3 . The hypothesis of no interaction effect is equivalent to saying that the

coefficients on all three terms must be zero in the population regression function. This is tested

using the F -statistic for these three coefficients, which is F  = 1.75. Because its p-value of .155

exceeds .05, we cannot reject (at the 5% significance level) the null hypothesis that all three

coefficients are zero. So, there is no statistical evidence of an interaction effect in this

regression.

Overall, there is no statistical evidence that the effect of a teacher meeting day varies with the

population of the city.

c)  (10 points) In words, briefly summarize your conclusions from Table 2 about the effect on

juvenile property crime of having a “no-school day” because of a teacher meeting day.

x  There is no statistical evidence that the effect of a teacher meeting day varies with the

population of the city.

x  There is statistical evidence that population enters the regression function nonlinearly

x  These two observations point to regression (5) as being the preferred specification

x  In regression (5), after controlling for population, the estimated effect of a teacher meeting

day, relative to a normal school day, is to increase the number of incidents of juvenile

property crime by .87, or nearly one. In a real-world sense this seems pretty big: hold a

teacher meeting day, you get a juvenile property crime incident. This effect is statistically

significant at the 5% (and 1%) significance level. [Also, could add: The 95% confidence

interval of (.42, 1.32), which in a practical