Question

I need to know the code to do this question in R

10. (8 marks) (using dataset: "meap93", in R: data(meap93, package-'wooldridge)) We want to explore the relationship between the math pass rate (scil/) and average teachers' compensation (salary + benefits) in the school (totcomp) In the population model: scill,-β0 + βι log(totcompi-u, prove that β/ 10 is the percentage point change in scill given a 10% increase in totcom. ii Use the data in MEAP93 to estimate the model from part (i). Report the estimated equation in the usual way, including the sample size and R-squared 111 How big is the estimated compensation effect? Namely, if compensation increases by 10%, what is the estimated percentage point increase in scill? iv One might worry that regression analysis can produce fitted values for sci11 that are greater than 100. Why is this not much of a worry in this data set?

thanks

Answer 1

i) the population model value for sci11, when totcomp =tc  is

$\text{sci11}(tc)=\beta_0+\beta_1\log(\text{tc})+u$

The value of sci11 when totcomp is increased by 10%, that is when totcomp=tc*1.1 is

= ß0 + β1 (log(tc) log( 1.1)) + u using the fact that log(ab) = log(a) o Bi log(tc)1 log(1.1) -scil1 (tcB1 x 0.095 log(b)

That means for a 10% increase in totcomp, the percentage point change in sci11 is

$\begin{align*} \beta_1\times 0.095\approx \beta_1/10 \end{align*}$

ii) The R code to estimate the model is below (all statements starting with # are comments and can be removed)

#install the package for the first time
install.packages('wooldridge')

#Load the data meap93
data(meap93,package='wooldridge')
#print some records
head(meap93)

#estimate the regression equation
fit<-lm(sci11~log(totcomp),data=meap93)
summary(fit)

# get this output

Call: lm (formula = sci II ~ log (totcomp), data = meap93) Residuals: 10 Median 3Q Max Min 42.991 -7.810 -0.2507.783 39.534 Coefficients: Estimate Std. Error t value Pr (>Itl 42.911 -1.626 0.10474 (Intercept)-69.771 log (totcomp)11.286 4.071 2.772 0.00582 Signif. codes: 0.001' 0.01 0.05 '.' 0.1'' 1 Residual standard error: 12.42 on 406 degrees of freedom Multiple R-squared: 0.01858, Adjusted R-squared: 0.01616 F-statistic: 7.686 on 1 and 406 DF, p-value: 0.005821

the estimated equation is

-69.771 + 11.286 log(totcomp) scil 1

R20.01858

Sample size n=408

iii) from part i) we can say that if the total compensation increases by 10%, then the estimated percentage point increase in scill is

R code

inc<-log(1.1)*fit$coef[2]
sprintf('The increase in scill is %.2f',inc)

##output is

> sprint f( 'The increase in scill is %.2f', inc) [11 "The increase in scill is 1.08" 13 I

Or using the approximation from part i) the increase is

percentage points

11,286/10 = 1.13

ans: If compensation increases by 10% the estimated percentage point increase in sci11 is 1.1

iv) Solving for totcomp using the fitted model we get

sci-69.771 11.286 log(totcomp) log(totcomp)- SCi11 +69.771 11.286 scil1+69.771 totcomp e 11.286

For the fitted values of sci11 to be greater than 100, we need the

-3, 411,343 totcomp 112863

That is the total compensation for the teachers to be more than $3,411,343, which is very unlikely for a school teacher in the US to aspire to. The maximum total compensation in the dataset is $63,518.

max (meap93stotcomp) 1 63518

Hence it is unlikely that in this dataset corresponding to the salary of school teachers, the fitted values of sci11 can be greater than 100.