Question

4) You have obtained a sub-sample of 1744 individuals from the Current Population Survey (CPS) and are interested in the relationship between weekly earnings and age. The regression, using heteroskedasticity-robust standard errors, yielded the following results:

EARN= 239.16 + 5.20×Age , R2 = 0.05, SER = 287.21., (20.24) (0.57) where Earn and Age are measured in dollars and years respectively.

(a) Interpret the slope and intercept coefficients and the measure of fit.

(b) The average age in this sample is 37.5 years. What is annual income in the sample?

(c) Construct a 95% confidence interval for both the slope and the intercept.

(d) Is the relationship between Age and Earn statistically significant?

(e) Do the results suggest that there is a guarantee for earnings to rise for everyone as they become older? Can we interpret the relationship between Age and Earn as a causal effect? Please discuss why or why not it is causal in detail

Answer 1

a).

Here the regression equation is, => Earn = 239.16 + 5.2*Age. So, the slope coefficient is “5.2”, => if the “Age” of an individual increases by 1 year, the average weekly earning of the individual will also increases by “$5.2”. Now, the intercept coefficient is “239.16”, => if the “Age” of an individual is zero, the average weekly earning of the individual is “$239.16” which is positive, which is not economically justified.

b).

The average age is “37.5 years”, then the average weekly income is “239.16 + 5.2*37.5 = 434.16”. In a particular year there are 52 weeks, => the average annual income is “52*434.16 = $22,576.32”.

c).

Here the sample size is “n=1744”, => the sample size is large enough. So, here the confidence interval of the slope coefficient is given below.

=> b2^ – z(0.025)*SE(b2) < b2 < b2^ + z(0.025)*SE(b2), where “SE(b2) = 0.57”, “z(0.025) = 1.96” and “b2^ = 5.2”.

=> b2^ – z(0.025)*SE(b2) < b2 < b2^ + z(0.025)*SE(b2), => 5.2 – 1.96*0.57 < b2 < 5.2 + 1.96*0.57,

=> 4.0828 < b2 < 6.3172, the confidence interval of the slope coefficient.

Here the confidence interval of the intercept coefficient is given below.

=> b1^ – z(0.025)*SE(b1) < b1 < b1^ + z(0.025)*SE(b1), where “SE(b1) = 20.24”, “z(0.025) = 1.96” and “b1^ = 239.16”.

=> 239.16 – 1.96*20.24 < b2 < 239.16 + 1.96*20.24,

=> 199.4896 < b2 < 278.8304, the confidence interval of the intercept coefficient.

d).

The “Coefficient Determination(R^2)” is the goodness of fit measure of a regression equation. Here the “R^2” is “0.05 = 5%”, which is quite low. So, here the regression equation explains the 5% of the variation in “Earning”. So, the relationship is not statistically significant.