Harvard University Economics1123
Department of Economics Spring2006
Problem Set6–Suggested Answers
Thank you for smoking
Due:Thursday6April.
Question1.
(a)Estimate the probability of smoking for
(i)all workers in the sample
(ii)workers a?ected by smoking bans
(iii)workers not a?ected by smoking bans
Using the summarize command we?nd that24.2%of the sample are smokers.From the linear probability model estimates(next question)I have that29.0%of those without a smoking ban are smokers and29.0-7.8=21.2%are smokers in places where there are bans.
(b)What is the di?erence in the probability of smoking between workers a?ected by a workplace smoking ban and workers not a?ected by a workplace smoking ban?Use a linear probability model to test whether or not this di?erence is statistically signi?cant.
The results of the linear probability model are
\
smo ker=0.290?0.078?smkban
(0.007)(0.009)
The hypothesis we want to test is H0:β1=0versus H a:β1=0.The t statistic is-0.078/0.009=-8.66.This is far outside the acceptance region of being within±1.96and so we reject the hypothesis that banning smoking in the workplace has no e?ect on whether or not the person is a smoker.
(c)Estimate a linear probability model with smoker as the dependent variable and the following regressors:smkban,female,age,age2,hsdrop,colsome,colgrad,black and https://www.doczj.com/doc/1d12104466.html,e the robuster-rors https://www.doczj.com/doc/1d12104466.html,pare the estimated e?ect on smoking of the ban in this regression to that of1(b). What is the change in the e?ect,and suggest a reason for the di?erence.
The Stata output for this regression is
(d)In the regression in1(c),test(at the5%level)that smkban has no e?ect on smoking. The hypothesis we want to test is still H0:β1=0versus H a:β1=0.The t statistic is -0.047/0.009=-5.27.This is still statistically signi?cantly di?erent from zero.Hence we still reject, after controlling for other variations in the propensity for people to smoke,that banning smoking has no e?ect on the probability that a person smokes.
(e)Test the hypothesis that the level of education has no e?ect on smoking in the regression in 1(c).Use a test with size5%.
There are four education variables,we can use an F test to test the coe?cients for zero.The null hypothesis is that the coe?cients on hsdrop,hsgrad,colsome,colgrad are all jointly zero against the alternative that one or more is non zero.The reason is that if any are nonzero then education has an e?ect on the probability that a person smokes.The value for the F statistic is140.09,which is easily larger than the critical value of2.37and hence we https://www.doczj.com/doc/1d12104466.html,cation does have an e?ect on the probability that a person smokes.
The Stata output is
Question2.Estimate a probit model using the same regressors as in1(c).Use the robust errors option.
The Stata output is
(a)Test the same hypothesis as in part1(d)but now for this probit regression.
The hypothesis we want to test is still H0:β1=0versus H a:β1=0.The t statistic is 0.-0.159/0.029=-5.45.Hence we still reject in the probit regression that the ban has no e?ect on the probability that a person smokes.
(b)Consider a hypothetical individual,Mr A.,who is white,non-hispanic,30years old and a high school dropout.Suppose this person is subject to the wordplace smoking ban,and caclulate the predicted probability that this person https://www.doczj.com/doc/1d12104466.html,pute the probability this person smokes if they are not subject to a work ban.
(c)Consider another hypothetical individual,Ms B,who is the same demographically to Mr A except she completed high https://www.doczj.com/doc/1d12104466.html,pute the probabilities that she smokes when there is a workplace ban and when there is no ban.
For the answers in both(b)and(c),we can place them in a table.Note that I had Stata compute these,see the Stata?le at the end of the problem set.
Mr A Ms B
Ban0.4450.306
No Ban0.5080.363
Di?erence-0.063-0.057
(d)Are the results on the e?ect of the ban the same for each of these hypothetical people?Why or why not?
The results are basically the same.They need not have been,given that the estimated proba-bility depends on all of the factors.
(e)Compare the results you obtain from2(b)and2(c)and compare them to the e?ect in the linear probability model in1(c).What is going on?
The linear probability model gives an e?ect of a decrease of4.7%for all possible demographic groups.This is less a problem with the linear probability model than the fact that the speci?cation of the LPM does not allow for demographic e?ects to impact the e?ect of the smoking ban.We could of course have interacted the smkban dummy variable with some of the demographic variables, and then we would have had di?ering e?ects just as the probit has di?ering e?ects.Of course,the
reason we use a probit in the?rst place is that it is sensible to impose nonlinearities on the model to ensure that the predicted probabilities remain inside the bounds for all possible values for the covariate,which would be hard to do for the LPM.
Stata do?le to compute the results
clear
#delimit;
capture log close;
/*Ps6—smoking example*/
log using c:\1123\ps6_01.log,replace;
use C:\1123\smoking.dta;
desc;
su;
gen age2=age*age;
summarize;
regress smoker smkban,robust;
regress smoker smkban female age age2hsdrop hsgrad colsome colgrad black hispanic,robust;
test hsdrop hsgrad colsome colgrad;
probit smoker smkban female age age2hsdrop hsgrad colsome colgrad black hispanic,robust;
sca mra1=_b[_cons]+_b[smkban]*1+_b[female]*0+_b[age]*30+_b[age2]*900
+_b[hsdrop]*1+_b[hsgrad]*0+_b[colsome]*0+_b[colgrad]*0+_b[black]*0
+_b[hispanic]*0;
dis"mr A,ban"mra1;
gen pmra1=normprob(mra1);
dis pmra1;
sca mra0=_b[_cons]+_b[smkban]*0+_b[female]*0+_b[age]*30+_b[age2]*900
+_b[hsdrop]*1+_b[hsgrad]*0+_b[colsome]*0+_b[colgrad]*0+_b[black]*0
+_b[hispanic]*0;
dis"mr A,no ban"mra0;
gen pmra0=normprob(mra0);
dis pmra0;
sca msa1=_b[_cons]+_b[smkban]*1+_b[female]*1+_b[age]*30+_b[age2]*900
+_b[hsdrop]*0+_b[hsgrad]*1+_b[colsome]*0+_b[colgrad]*0+_b[black]*0
+_b[hispanic]*0;
dis"ms A,ban"msa1;
gen pmsa1=normprob(msa1);
dis pmsa1;
sca msa0=_b[_cons]+_b[smkban]*0+_b[female]*1+_b[age]*30+_b[age2]*900
+_b[hsdrop]*0+_b[hsgrad]*1+_b[colsome]*0+_b[colgrad]*0+_b[black]*0
+_b[hispanic]*0;
dis"ms A,no ban"msa0;
gen pmsa0=normprob(msa0);
dis pmsa0;
log close;
clear; exit;