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STATA APPLICATIONS Task 1 last year- Computer assignment 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. Answer to question (i) Stata command: use busind,clear su daysenforce daysopen gnipc Variable Obs Mean daysenforce daysopen gnipc 129 135 135 352.9612 50.75556 6.56983 Std. Dev. Min Max 162.1636 38.42408 10.02707 27 2 .09 909 203 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? Answer to Question (ii) 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 Question (iii) Estimate the following simple regression model: gnipc 0 1daysopen u Give a careful interpretation of estimates b1 and b0. Are the signs what you expected them to be? Answer to Question (iii) Stata commands: reg gnipc daysopen Source SS df MS Model Residual 1766.98652 11705.6636 1 1766.98652 133 88.0125084 Total 13472.6501 134 100.542165 gnipc Coef. daysopen _cons -.0945063 11.36655 Std. Err. .0210919 1.340889 t -4.48 8.48 Number of obs F( 1, 133) Prob > F R-squared Adj R-squared Root MSE P>|t| 0.000 0.000 = = = = = = 135 20.08 0.0000 0.1312 0.1246 9.3815 [95% Conf. Interval] -.1362253 8.714322 -.0527873 14.01878 Question (iv) 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 (v) 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. Answer to Question (v) You can compute predicted values for the dependent variable in two ways: by “displaying” gniˆpc ˆ0 ˆ1* daysopen when daysopen=5 and daysopen=200 Stata commands: display _b[daysopen]*5+_b[_cons] 10.894018 display _b[daysopen]*200+_b[_cons] -7.5347099 Answer to question (v) 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 regressio -10 0 10 20 30 40 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 Question (vi) Estimate the following simple regression model and give a careful interpretation of 1. gnipc 0 1daysenforce u Answer to Question (vi) Stata command: reg gnipc daysenforce Source SS df MS Model Residual 2768.61703 10372.442 1 2768.61703 127 81.6727718 Total 13141.0591 128 102.664524 gnipc Coef. daysenforce _cons -.0286796 16.66315 Std. Err. .0049258 1.912056 t -5.82 8.71 Number of obs F( 1, 127) Prob > F R-squared Adj R-squared Root MSE P>|t| 0.000 0.000 = = = = = = 129 33.90 0.0000 0.2107 0.2045 9.0373 [95% Conf. Interval] -.0384269 12.87954 -.0189322 20.44676 Question (viii) . reg lngnipc daysopen Source SS df MS Model Residual 42.8638154 311.155695 1 42.8638154 133 2.3395165 Total 354.01951 134 2.64193664 lngnipc Coef. daysopen _cons -.0147194 1.4396 Std. Err. .0034388 .218617 t -4.28 6.59 Number of obs F( 1, 133) Prob > F R-squared Adj R-squared Root MSE P>|t| 0.000 0.000 = = = = = = 135 18.32 0.0000 0.1211 0.1145 1.5295 [95% Conf. Interval] -.0215212 1.007184 -.0079176 1.872016 Question (vii) 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 R2. The greater the R2, the more variation of y is explained by x. The R2 of the regression of GNI per capita on the number of days to open a business is about 13% and the R2 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. Question (viii) Estimate the following simple regression model and give a careful interpretation of 1. log(gnipc) 0 1daysopen u Answer to Question (viii) 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? Task 2 last year- Computer exercise 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. Question 1 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 r2_sex r2_healths~t nrchild nractad nrold 706 655 706 706 706 1.473088 1.309924 3.478754 3.002833 .3427762 head_age head_educ value_jewe~y distschool r2_supown 706 706 706 656 706 educ 706 Std. Dev. Min Max .4996292 .4726228 1.739072 1.487534 .6291367 1 1 1 0 0 2 3 14 13 5 46.56232 2.827195 13984.75 .2925051 .7404253 10.56877 4.070464 26725.61 .3067064 1.053909 15 0 0 .0166667 0 85 14 400000 2.5 9.811792 5.51983 3.565441 0 16 Question (1) Child characteristics Male (%) 52 Health status (%) Good 69.5 Fair 30 Poor 0.5 Years of education 5.5 (3.6) Question (1) Household characteristics Number of household members 6.8 (2.73) under 18 years old 3.5 (1.74) between 18 and 59 3.0 (1.49) 60 or older 0.3 (0.63) Age of the head 46 (10.5) Education of the head 2.8 (4.1) Land owned (in ha) 0.74 (1.05) Value of jewelries (in rupees) 13985 (26726) Distance to school (in hours) 0.29 (0.31) Number of observations Standard errors into parenthesis 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? Tips to answer question (3) 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? A reasonable model to estimate: educ 0 1headage 2headeduc 3nractad 4nrold 5 sup own 6dist 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 . reg //we create a dummy, =1 if r2_sex equals 1 educ head_age head_educ nractad nrold r2_supown distschool male Source SS df MS Model Residual 1463.51225 6628.93151 7 641 209.073179 10.3415468 Total 8092.44376 648 12.4883391 educ Coef. head_age head_educ nractad nrold r2_supown distschool male _cons .033665 .3317472 -.093363 .1106236 .3162876 -.5371565 1.061818 2.509268 Std. Err. .0135985 .0339146 .0904375 .2244013 .1255367 .418622 .2538363 .6762122 t 2.48 9.78 -1.03 0.49 2.52 -1.28 4.18 3.71 Number of obs F( 7, 641) Prob > F R-squared Adj R-squared Root MSE P>|t| 0.014 0.000 0.302 0.622 0.012 0.200 0.000 0.000 = = = = = = 649 20.22 0.0000 0.1808 0.1719 3.2158 [95% Conf. Interval] .0069621 .2651501 -.2709525 -.3300268 .0697746 -1.359193 .5633668 1.181409 .0603679 .3983444 .0842265 .5512741 .5628006 .2848797 1.560269 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 β1 =0. Then, one can directly read on the stata output that this minimum significance level is 1.4%. Question (5) 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? Answer to Question (5) Need to test null hypothesis: H0: β3=β4 against H1:β3≠β4 You just need command "test". . test nractad = nrold ( 1) nractad - nrold = 0 F( 1, 641) = Prob > F = 0.72 0.3965 Question (6) 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? Answer to Question (6) . test (head_age=0)(head_educ=0) ( 1) ( 2) head_age = 0 head_educ = 0 F( 2, 641) = Prob > F = 47.96 0.0000 Question 6: compute F-test Run the unrestricted and restricted models, and compute either SSR or R2 form of the F-statistic. ( Rur2 Rr2 ) / q F (1 Rur2 ) /(n k 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 Question (9) Non missing Missing 53 48 Good 69.5 74 Fair 30 26 Poor 0.5 . Years of education 5.3 6.8 Child characteristics Male (%) Health status (%) Question (9) Non missing Missing 6.9 6.1 under 18 years old 3.5 2.8 between 18 and 59 3.0 2.9 60 or older 0.3 0.4 Age of the head 46.5 48.2 Education of the head 2.6 5.5 Land owned (in ha) 0.77 0.29 12212 35488 Distance to school (in hours) 0.29 . Number of observations 600 46 Household characteristics Number of household members Value of jewelries (in rupees)