Problem Set 1
Statistics and Econometrics
Due: 12pm, 4 November 2019
Marios Zoulias 01766825
Question 1
The data set ceosal2.RData contains information on chief executive officers for U.S. corporations. Two variables of interest are the annual compensation (\(salary\)) and the prior number of years as company CEO (\(ceoten\)).
First i want to load the data and check them a little, in order to have a first view of them. Then save them in a format that i can analyse easier in the next questions (taking only the columns i need).
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionload("ceosal2.RData")
ceo_data = data
head(ceo_data)summary(ceo_data)## salary age college grad
## Min. : 100.0 Min. :33.00 Min. :0.0000 Min. :0.0000
## 1st Qu.: 471.0 1st Qu.:52.00 1st Qu.:1.0000 1st Qu.:0.0000
## Median : 707.0 Median :57.00 Median :1.0000 Median :1.0000
## Mean : 865.9 Mean :56.43 Mean :0.9718 Mean :0.5311
## 3rd Qu.:1119.0 3rd Qu.:62.00 3rd Qu.:1.0000 3rd Qu.:1.0000
## Max. :5299.0 Max. :86.00 Max. :1.0000 Max. :1.0000
## comten ceoten sales profits
## Min. : 2.0 Min. : 0.000 Min. : 29 Min. :-463.0
## 1st Qu.:12.0 1st Qu.: 3.000 1st Qu.: 561 1st Qu.: 34.0
## Median :23.0 Median : 6.000 Median : 1400 Median : 63.0
## Mean :22.5 Mean : 7.955 Mean : 3529 Mean : 207.8
## 3rd Qu.:33.0 3rd Qu.:11.000 3rd Qu.: 3500 3rd Qu.: 208.0
## Max. :58.0 Max. :37.000 Max. :51300 Max. :2700.0
## mktval lsalary lsales lmktval
## Min. : 387 Min. :4.605 Min. : 3.367 Min. : 5.958
## 1st Qu.: 644 1st Qu.:6.155 1st Qu.: 6.330 1st Qu.: 6.468
## Median : 1200 Median :6.561 Median : 7.244 Median : 7.090
## Mean : 3600 Mean :6.583 Mean : 7.231 Mean : 7.399
## 3rd Qu.: 3500 3rd Qu.:7.020 3rd Qu.: 8.161 3rd Qu.: 8.161
## Max. :45400 Max. :8.575 Max. :10.845 Max. :10.723
## comtensq ceotensq profmarg
## Min. : 4.0 Min. : 0.0 Min. :-203.077
## 1st Qu.: 144.0 1st Qu.: 9.0 1st Qu.: 4.231
## Median : 529.0 Median : 36.0 Median : 6.834
## Mean : 656.7 Mean : 114.1 Mean : 6.420
## 3rd Qu.:1089.0 3rd Qu.: 121.0 3rd Qu.: 10.947
## Max. :3364.0 Max. :1369.0 Max. : 47.458- Find the average salary and the average tenure in the sample.
Average salary is 865.86 and average tenure is 7.95
mean(ceo_data$salary)## [1] 865.8644mean(ceo_data$ceoten)## [1] 7.954802- How many CEOs are in their first year as CEO (that is, \(ceoten=0\))? What is the longest tenure as a CEO?
We can see that 5 CEOs are in their first year (ceoten = 0), and the max ceo tenure is 37 years.
ceo_data %>%
count(ceoten == 0)max(ceo_data$ceoten)## [1] 37- What is the average salary for CEOs with tenure longer than or equal to the average tenure? What is the average salary for CEOs with tenure shorter than the average tenure?
We can see that the average salary for ceos with tenure longer or equal to average tenure is 1003.5 and for ceos with less than average tenure is 766.98.
above_av = ceo_data %>%
filter(ceoten >= 7.95) %>%
select(salary)
mean(above_av$salary)## [1] 1003.5bellow_av = ceo_data %>%
filter(ceoten < 7.95) %>%
select(salary)
mean(bellow_av$salary)## [1] 766.9806- Create a graph to examine the relationship between \(salary\) and \(ceoten\) for all CEOs. Comment.
To examine this relationship we obviously need a scatterplot of salary and ceoten columns. We can see that the salary of a CEO does not strongly affected by the number of previous years as CEO. That might change from company to company, however we can see CEOs with 0-7 years of CEO experience receive the same income as CEOs with 7+ years of experience (to be sure, i would say that the relationship between salary and ceoten is slightly positive).
library(ggplot2)
ggplot(data = ceo_data, aes(x = ceoten, y = salary)) + geom_point() + stat_smooth(method = "lm")
- Estimate the simple regression model \[\log(salary)=\beta_0+\beta_1 ceoten+u,\] and report your results. What is the (approximate) predicted percentage increase in salary given one more year as a CEO?
Given E(u) = 0, an increase of 1 year of ceo experience, gives on average an increase of 0.97% in salary.
model = lm(log(salary) ~ ceoten, data=ceo_data)
summary(model)##
## Call:
## lm(formula = log(salary) ~ ceoten, data = ceo_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.15314 -0.38319 -0.02251 0.44439 1.94337
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.505498 0.067991 95.682 <2e-16 ***
## ceoten 0.009724 0.006364 1.528 0.128
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.6038 on 175 degrees of freedom
## Multiple R-squared: 0.01316, Adjusted R-squared: 0.007523
## F-statistic: 2.334 on 1 and 175 DF, p-value: 0.1284Question 2
The data set bwght.RData contains data on births to women in the United States. Two variables of interest are the infant birth weight in ounces (\(bwght\)), and the average number of cigarettes the mother smoked per day during pregnancy (\(cigs\)).
I will continue with the same strategy and i will try first to explore a little bit the data.
load("bwght.RData")
birth_data = data
head(birth_data)summary(birth_data)## faminc cigtax cigprice bwght
## Min. : 0.50 Min. : 2.00 Min. :103.8 Min. : 23.0
## 1st Qu.:14.50 1st Qu.:15.00 1st Qu.:122.8 1st Qu.:107.0
## Median :27.50 Median :20.00 Median :130.8 Median :120.0
## Mean :29.03 Mean :19.55 Mean :130.6 Mean :118.7
## 3rd Qu.:37.50 3rd Qu.:26.00 3rd Qu.:137.0 3rd Qu.:132.0
## Max. :65.00 Max. :38.00 Max. :152.5 Max. :271.0
##
## fatheduc motheduc parity male
## Min. : 1.00 Min. : 2.00 Min. :1.000 Min. :0.0000
## 1st Qu.:12.00 1st Qu.:12.00 1st Qu.:1.000 1st Qu.:0.0000
## Median :12.00 Median :12.00 Median :1.000 Median :1.0000
## Mean :13.19 Mean :12.94 Mean :1.633 Mean :0.5209
## 3rd Qu.:16.00 3rd Qu.:14.00 3rd Qu.:2.000 3rd Qu.:1.0000
## Max. :18.00 Max. :18.00 Max. :6.000 Max. :1.0000
## NA's :196 NA's :1
## white cigs lbwght bwghtlbs
## Min. :0.0000 Min. : 0.000 Min. :3.135 Min. : 1.438
## 1st Qu.:1.0000 1st Qu.: 0.000 1st Qu.:4.673 1st Qu.: 6.688
## Median :1.0000 Median : 0.000 Median :4.787 Median : 7.500
## Mean :0.7846 Mean : 2.087 Mean :4.760 Mean : 7.419
## 3rd Qu.:1.0000 3rd Qu.: 0.000 3rd Qu.:4.883 3rd Qu.: 8.250
## Max. :1.0000 Max. :50.000 Max. :5.602 Max. :16.938
##
## packs lfaminc
## Min. :0.0000 Min. :-0.6931
## 1st Qu.:0.0000 1st Qu.: 2.6741
## Median :0.0000 Median : 3.3142
## Mean :0.1044 Mean : 3.0713
## 3rd Qu.:0.0000 3rd Qu.: 3.6243
## Max. :2.5000 Max. : 4.1744
## - Estimate the simple regression model \[bwght=\beta_0+\beta_1 cigs+u,\] and report your results.
So we see that the equation is \[bwght=119.77 -0.51* cigs+u,\]
model = lm(bwght ~ cigs, data=birth_data)
summary(model)##
## Call:
## lm(formula = bwght ~ cigs, data = birth_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -96.772 -11.772 0.297 13.228 151.228
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 119.77190 0.57234 209.267 < 2e-16 ***
## cigs -0.51377 0.09049 -5.678 1.66e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 20.13 on 1386 degrees of freedom
## Multiple R-squared: 0.02273, Adjusted R-squared: 0.02202
## F-statistic: 32.24 on 1 and 1386 DF, p-value: 1.662e-08- What is the predicted birth weight when \(cigs = 0\)? What about when \(cigs = 20\) (one pack per day)? Comment on the difference.
With E(u) = 0 and :
cigs = 0 we have \[bwght=119.77 -0.51 * 0 = 119.77\]
cigs = 20 we have \[bwght=119.77 -0.51 * 20 = 109.57\]
The difference between those two values is 20 X β1. The negative relationship between cigs and bwgt means that the more cigaretes the mother smokes during her pregnancy the less the infant’s weight on birth. - Does this simple regression necessarily capture a causal relationship between the child’s birth weight and the mother’s smoking habits? Explain.
In this regression we see a negative relationship between the two variables, as stated above. However this relationship does not necessarily means that more smoking habbits reduces infant’s weight. So we dont necessarily have a causal effect from smoking habbits on infant weight, especially having an R-squared equal to 0.02, meaning that cigs variable explains only 0.02 of bwght (also meaning that we ignore many other factors). - To predict a birth weight of \(125\) ounces, what would \(cigs\) have to be? Comment.
To predict a weight of 125 ounces we solve the equation \[125 = 119.77 - 0.51 * cigs\] In this way we find that cigs is equal to -10.25. A negative value is impossible for our case , as a woman can’t smoke -10 cigs per day. Probably given the data we have, we are not able to make good predictions - The proportion of women in the sample who do not smoke while pregnant is about \(85\%\). Does this reconcile your finding from part 4?
Obviously, we made a mistake. We made a model using the whole population, while our predictions can be made sufficiently only for the 15% of this. Our model would be better if we had only data of women who smoked during pregnancy. As a result, what we found in question 4 is absolutely logically false. We need better data and assumptions.