The data set busind.dta contains information on Gross National Income
(GNI) per capita and the number of days to open a business and to enforce a contract in a sample of 135 countries. It was extracted from the “Doing Business” dataset, a dataset collected by the World
Bank based on expert opinions in each country. The variable gnipc measures GNI per capita in thousand $. The variable daysopen measures the average number of days needed to open a business in that country, and daysenforce measures the average number of days needed to enforce a given type of contract.
(i) Find the average GNI per capita and the average number of days to open a business, and the average number of days to enforce a contract.

 Stata command: use busind,clear su daysenforce daysopen gnipc
Variable Obs Mean Std. Dev. Min Max
daysenforce 129 352.9612 162.1636 27 909
daysopen 135 50.75556 38.42408 2 203
gnipc 135 6.56983 10.02707 .09 43.35
 (ii) In how many countries does it take on average less than 5 days to open a business? What is the maximum number of days to open a business in the dataset? In which countries does it take more than 200 days to open a business?

 Stata command: su daysopen if daysopen<=5
Variable Obs Mean Std. Dev. Min Max
daysopen 4 3.5 1.290994 2 5 list country if daysopen>=200
country
135. Haiti

 Estimate the following simple regression model: gnipc
 
0
 
1 daysopen
 u
 Give a careful interpretation of estimates b
1 the signs what you expected them to be?
and b
0
. Are

 Stata commands: reg gnipc daysopen
Source SS df MS Number of obs = 135
F( 1, 133) = 20.08
Model 1766.98652 1 1766.98652 Prob > F = 0.0000
Residual 11705.6636 133 88.0125084 R-squared = 0.1312
Adj R-squared = 0.1246
Total 13472.6501 134 100.542165 Root MSE = 9.3815
gnipc Coef. Std. Err. t P>|t| [95% Conf. Interval]
daysopen -.0945063 .0210919 -4.48 0.000 -.1362253 -.0527873
_cons 11.36655 1.340889 8.48 0.000 8.714322 14.01878

 Question: What kind of factors are contained in u? Are these likely to be correlated with the number of days to open a business?
 Answer: Factors contained in u are factors that explain the GNI par capita apart from the number of days to open a business. You might be conscious that there are many other factors, such as economic institutions, education, savings, consumption, R&D… Some factors are likely to be correlated with the number of days to open a business, such as the quality of economic institutions.

.
 Question: What is according to this model the predicted income for a country where it takes 5 days to open a business? And the predicted income for a country where it takes 200 days to open a business? Show how you can calculate the answers by hand (once you have obtained the estimation results). Do the obtained levels of income seem reasonable? Explain.

 You can compute predicted values for the dependent variable in two ways: by “displaying” gn i
ˆ pc
  ˆ
0
  ˆ
1
* when daysopen=5 and daysopen=200 daysopen
Stata commands: display _b[daysopen]*5+_b[_cons]
10.894018
display _b[daysopen]*200+_b[_cons]
-7.5347099

 or by generating the fitted value of the dependent variable : reg gnipc daysopen predict gnipc_hat
. list gnipc_hat if daysopen==5
gnipc_~t
4. 10.89402
. list gnipc_hat if daysopen==200
A problem arises with this second method as there is no observation with daysopen=200, so that it is impossible to get the value of gnipc_hat for daysopen=200.

 To illustrate our fitted values, we can draw the OLS regression line: scatter gnipc daysopen||lfit gnipc daysopen
0 50 100 150 number of days to open a business gross national income ('000 US $) Fitted values
200

Estimate the following simple regression model and give a careful interpretation of

1.
 
0
 
1


 Stata command: reg gnipc daysenforce
Source SS df MS Number of obs = 129
F( 1, 127) = 33.90
Model 2768.61703 1 2768.61703 Prob > F = 0.0000
Residual 10372.442 127 81.6727718 R-squared = 0.2107
Adj R-squared = 0.2045
Total 13141.0591 128 102.664524 Root MSE = 9.0373
gnipc Coef. Std. Err. t P>|t| [95% Conf. Interval]
daysenforce -.0286796 .0049258 -5.82 0.000 -.0384269 -.0189322
_cons 16.66315 1.912056 8.71 0.000 12.87954 20.44676

. reg lngnipc daysopen
Source SS df MS Number of obs = 135
F( 1, 133) = 18.32
Model 42.8638154 1 42.8638154 Prob > F = 0.0000
Residual 311.155695 133 2.3395165 R-squared = 0.1211
Adj R-squared = 0.1145
Total 354.01951 134 2.64193664 Root MSE = 1.5295
lngnipc Coef. Std. Err. t P>|t| [95% Conf. Interval]
daysopen -.0147194 .0034388 -4.28 0.000 -.0215212 -.0079176
_cons 1.4396 .218617 6.59 0.000 1.007184 1.872016



Comparing the estimates of the models in (iii) and (v), which one explains more of the variation in income per capita across countries.
Can you infer whether the duration to open a business or the duration for enforcing contracts is more strongly correlated with income per capita?
Answer: How much of the variation of GNI per capita (y) is explained by an independent variable is given by the R 2 . The greater the R 2 , the more variation of y is explained by x. The R 2 of the regression of GNI per capita on the number of days to open a business is about 13% and the R 2 of the regression of GNI per capita on the number of days to enforce a contract 21%. That means that this variable explains more of the variation of the gni per capita than the former. It means that the duration for enforcing contract is more strongly correlated with income per capita than the number of days to open a business. Here, the correlation between gnipc and daysenforce is equal to -0.46 and the correlation between gnipc and daysopen is equal to -0.36.

 Estimate the following simple regression model and give a careful interpretation of

1.
log(
)
 
0
 
1


 Stata commands: gen lngnipc=ln(gnipc) reg lngnipc daysopen

 Do these results allow you to draw conclusions regarding the desirability of policies aimed at reducing the number of days for opening a business in certain developing countries?
 The dataset contains 135 countries, and hence does not contain information about all the countries in the world. Do you think one should account for that when interpreting the regression results. Why?

 The dataset nepalind.dta contains data from 706 children of
15 years old in Nepal. The data come from the 2003 Nepal
Living Standard Survey (NLSS) Living Standard
Measurement Survey (LSMS). We want to analyze this data to understand the number of years of education. Illiteracy and low levels of education are a major concern in Nepal, so it would be good to know which type of factors could be explaining education of the present generation, to know what type of policies to implement. The dataset has some information on household characteristics and characteristics of the child, and of the household head.
The NLSS is a LSMS-type survey, which are country-wide representative surveys that statistical offices in developing countries conduct with the support of the World Bank to determine poverty levels, determinants of poverty, etc. See www.worldbank.org/lsms for more info.

 Write a paragraph describing the dataset using the standard descriptive statistics (also called summary statistics, or “D-stats”). Add a table with the d-stats.

. su
Variable Obs Mean Std. Dev. Min Max
r2_sex 706 1.473088 .4996292 1 2 r2_healths~t 655 1.309924 .4726228 1 3
nrchild 706 3.478754 1.739072 1 14
nractad 706 3.002833 1.487534 0 13
nrold 706 .3427762 .6291367 0 5
head_age 706 46.56232 10.56877 15 85
head_educ 706 2.827195 4.070464 0 14 value_jewe~y 706 13984.75 26725.61 0 400000
distschool 656 .2925051 .3067064 .0166667 2.5
r2_supown 706 .7404253 1.053909 0 9.811792
educ 706 5.51983 3.565441 0 16

Child characteristics
Male (%)
Health status (%)
Good
Fair
Poor
Years of education
52
69.5
30
0.5
5.5 (3.6)

Household characteristics
Number of household members under 18 years old between 18 and 59
60 or older
Age of the head
Education of the head
Land owned (in ha)
Value of jewelries (in rupees)
Distance to school (in hours)
Number of observations
Standard errors into parenthesis
6.8 (2.73)
3.5 (1.74)
3.0 (1.49)
0.3 (0.63)
46 (10.5)
2.8 (4.1)
0.74 (1.05)
13985 (26726)
0.29 (0.31)
706

Question 2: Show the distribution of the different values of years of education in the dataset. Drop the variables that have values higher than 10. Explain why that might be a smart thing to do, before doing any regression analysis.
. hist educ,discrete
(start=0, width=1)

 Question (3): Specify a model that allows explaining the number of years of education as a function of father’s age, the number of active adults (between 18 and 60 years old) and the number of elderly (60 or older) and all other variables you think are interesting and appropriate.
Make sure only to include variables that are exogenous and discuss why the variables you include can be considered exogenous. Estimate the model and give a careful interpretation of each of the coefficients (sign, size, and significance!). Do you find any of your results counterintuitive?

 Each variable that you add into the model must be related to educ in some way, and should not violate the ZCM assumption=>they must be exogenous=>ask yourself:
 x caused by y? i.e. possibility of reverse causality?
 One third factor determines both x and y? in this case correlation is not causation, and x is not exogenous.
 u and x related for some other reason?
 Gender? Head´s age? Nb of active adults? Number of elderly? Head´s education? Land owned? distance to school? Value jewelry? Nb of children?
Health?

educ
 
0
 
1 headage
 
2 headeduc
 
3 nractad
 
4 nrold
 
5 sup own
 
6 dist
 
7 female
 u
 Expected signs of coefficients? Argue.
. do "C:\Users\Yaya\AppData\Local\Temp\STD03000000.tmp"
. drop if educ>10
(9 observations deleted)
. ge male= r2_sex==1 //we create a dummy, =1 if r2_sex equals 1
. reg educ head_age head_educ nractad nrold r2_supown distschool male
Source SS df MS Number of obs = 649
F( 7, 641) = 20.22
Model 1463.51225 7 209.073179 Prob > F = 0.0000
Residual 6628.93151 641 10.3415468 R-squared = 0.1808
Adj R-squared = 0.1719
Total 8092.44376 648 12.4883391 Root MSE = 3.2158
educ Coef. Std. Err. t P>|t| [95% Conf. Interval]
head_age .033665 .0135985 2.48 0.014 .0069621 .0603679
head_educ .3317472 .0339146 9.78 0.000 .2651501 .3983444
nractad -.093363 .0904375 -1.03 0.302 -.2709525 .0842265
nrold .1106236 .2244013 0.49 0.622 -.3300268 .5512741
r2_supown .3162876 .1255367 2.52 0.012 .0697746 .5628006
distschool -.5371565 .418622 -1.28 0.200 -1.359193 .2848797
male 1.061818 .2538363 4.18 0.000 .5633668 1.560269
_cons 2.509268 .6762122 3.71 0.000 1.181409 3.837127

Question 4:What is the minimum significance level at which one can reject that hypothesis that age of the household head does not affect education levels?
 The p-value gives the smallest significant level at which an hypothesis H0 can be rejected. In other words, a low p-value indicates that the tested hypothesis is unlikely. The minimum significance level at which one can reject the hypothesis that the age of the household head does not affect education levels is given by the p-value of the test β
1
=0. Then, one can directly read on the stata output that this minimum significance level is
1.4%.

 Do your results allow you to conclude that the effects of the number of active adults in the household is different than the effect of elderly? State the null hypothesis and the alternative hypothesis you are testing, and the significance level you are considering. Does your answer differ depending on which significance level you consider?



Need to test null hypothesis: H
0
: β
3
= β
4
You just need command "test".
against H
1
: β
3
≠ β
4
. test nractad = nrold
( 1) nractad - nrold = 0
F( 1, 641) = 0.72
Prob > F = 0.3965

 Test whether the characteristics of the household head are jointly significant. Show how to do this in stata, and calculate the test by hand in 2 different ways. What can you conclude about the role of household head characteristics on education of the children?

. test (head_age=0)(head_educ=0)
( 1) head_age = 0
( 2) head_educ = 0
F( 2, 641) = 47.96
Prob > F = 0.0000

 Run the unrestricted and restricted models, and compute either
SSR or R2 form of the F-statistic.

F

( 1

(
2
R ur
2
R ur
)

/(
R r
2 n
)

/ k q

1 ) reg educ head_age head_educ nractad nrold r2_supown distschool male scalar r2_ur=e(r2) scalar df=e(df_r) reg educ nractad nrold r2_supown distschool male scalar r2_r=e(r2)
. dis ((r2_ur-r2_r)/2)/((1-r2_ur)/df)
47.961109

Child characteristics
Male (%)
Health status (%)
Good
Fair
Poor
Years of education
Non missing Missing
53 48
69.5
30
0.5
5.3
74
26
.
6.8

Household characteristics
Number of household members under 18 years old between 18 and 59
60 or older
Age of the head
Education of the head
Land owned (in ha)
Value of jewelries (in rupees)
Distance to school (in hours)
Number of observations
Non missing
6.9
3.5
3.0
0.3
46.5
2.6
0.77
12212
0.29
600
Missing
6.1
2.8
2.9
0.4
48.2
5.5
0.29
35488
.
46