# Econometrics Homework Help

The econometrics homework help solution is based on the concept of linear regression with one regressor. STATA was the software used for the statistical analysis. Data sources used are commonly available in public websites and analysis was carried out over slope-intercept values, scatter plots and regression outputs. Such an example for econometrics assignment help can serve as a good reference material for students. From the given data, the equation of linear regression line was derived and then the values were predicted by the model on the basis of previous training input data. Variances in the data and the errors in prediction were also closely observed and analysed while preparing the below report.

E4.1

a.

The Stata command and its output are given below:

regress ahe age, cformat(%9.3f) pformat(%5.3f) sformat(%8.3f)

Source |       SS       df       MS              Number of obs =    7711

————-+——————————           F(  1,  7709) =  230.43

Model |  23005.7375     1  23005.7375           Prob > F      =  0.0000

Residual |  769645.718  7709  99.8372964           R-squared     =  0.0290

Total |  792651.456  7710   102.80823           Root MSE      =  9.9919

——————————————————————————

ahe |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]

————-+—————————————————————-

age |      0.605      0.040   15.180   0.000        0.527       0.683

_cons |      1.082      1.184    0.914   0.361       -1.239       3.404

——————————————————————————

The estimated intercept, _cons, is 1.082. In this data, since no workers would ever be 0 years old, the estimate has no empirical relevance but simply indicates econometrics homework help where the regression line intersects the vertical axis. The t statistic for the intercept, 0.914, has an associated p value of 0.361. The p value implies econometrics project help that if the null hypothesis that the population intercept is zero is true, then the probability of obtaining an intercept 1.082 or more away from 0 is 0.361. Given the relatively high probability of observing an intercept this extreme or more when the null hypothesis is true, the null hypothesis is econometrics assignment  solutionnot rejected, and the conclusion is that there is no evidence in this sample that the intercept is different than 0.

The slope estimate for age indicates the increase of the average of average hourly earnings with each one year increase in age. The estimated coefficient, 0.605, indicates that the average hourly wage increases by 60.5 cents with each 1 year increase in econometrics assignment helpage. The t statistic, 15.180, has an associated p value of less than 0.001, which implies that if the slope on age were 0 in the population, there would be less than a 0.001 probability of obtaining a slope estimate this far from 0 as the result of help with econometrics assignment chance. Given the low probability of the estimated slope from a population with a zero slope, the conclusion is that average hourly wages increase as age increases.

b.

Stata can find prediction intervals by using the command lincom. For Bob, we have

lincom _b[_cons] + 26*_b[age]

( 1)  26*age + _cons = 0

——————————————————————————

ahe |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]

————-+—————————————————————-

(1) |   16.81192   .1824085    92.17   0.000     16.45435    17.16949

——————————————————————————

The point estimate of Bob’s average hourly earnings is \$16.81 per hour, and the 95% confidence interval estimate is [\$16.45, 17.17].

For Alexis, we have

lincom _b[_cons] + 30*_b[age]

( 1)  30*age + _cons = 0

——————————————————————————

ahe |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]

————-+—————————————————————-

(1) |   19.23186   .1150273   167.19   0.000     19.00638    19.45735

——————————————————————————

The point estimate of Alexis’s average hourly earnings is \$19.23 per hour, and the 95% confidence interval estimate is [\$19.01, \$19.46].

The R2 for the regression indicates the proportion of the variationhelp with econometrics homework in the dependent variable that is explained by the regression. In part a, it is shown that the R2 for the regression is 0.0290, so that 2.9% of the variation in average hourly earnings is explained by differences in econometrics homework help age of the workers. Although useful, this regression leaves 97% of the variation in average hourly wages unexplained.

E4.3

a.

The Stata command and output are given below:

regress eddist, cformat(%9.3f) pformat(%5.3f) sformat(%8.3f)

Source |       SS       df       MS              Number of obs =    3796

————-+——————————           F(  1,  3794) =   28.48

Model |  93.0256754     1  93.0256754           Prob > F      =  0.0000

Residual |  12394.3568  3794    3.266831           R-squared     =  0.0074

Total |  12487.3825  3795  3.29048287           Root MSE      =  1.8074

——————————————————————————

ed |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]

————-+—————————————————————-

dist |     -0.073      0.014   -5.336   0.000       -0.100      -0.046

_cons |     13.956      0.038  369.945   0.000       13.882      14.030

——————————————————————————

The estimated intercept is 13.956, which is the average number of help with econometrics assignment years of school completed for a person whosehigh school is 0 miles from the college (or less than 5 miles, so that distance rounds to 0 miles rather than 10 miles). The point estimate of econometrics assignment helpthe interval is statistically significantly different than 0 at the level of significance given by the p value, which is less than 0.001.

The estimated slope on distance of the college from high school in 10s of miles is -0.073, which implies that each additional 10 miles of econometrics problems with answers distance lowers completed years of education by 0.073 years, so that an additional 136.7 miles (=1/0.073) would lower completed years by about 1 year. The estimated coefficient is statistically significantly different than 0 at less than the0.001 level of significance.

The results imply that colleges don’t necessarily need to be next door to high schools, but that large distances, like 100 miles, should be avoided.

The predicted values if Bob lives 10 and 20 miles from school can be found using the lincom command:

For living 10 miles from school, the following prediction is obtained:

lincom _b[_cons] + 1*_b[dist]

( 1)  dist + _cons = 0

——————————————————————————

ed |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]

————-+—————————————————————-

(1) |   13.88248   .0309831   448.07   0.000     13.82174    13.94323

——————————————————————————

The prediction is that if Bob lives 10 miles from school, he will complete 13.9 years of schooling.

For living 20 miles from school, the following prediction is obtained:

lincom _b[_cons] + 2*_b[dist]

( 1)  2*dist + _cons = 0

——————————————————————————

ed |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]

————-+—————————————————————-

(1) |   13.80911   .0295788   466.86   0.000     13.75112     13.8671

——————————————————————————

The prediction is that if Bob lives 20 miles from school, he will complete 13.8 years of school.  The effect of increasing the distance from 10 to 20 miles is the estimated coefficient, which is not very large.

The R2 for the regression indicates the proportion of the total variation in econometrics homework help the dependent variable, ed, years of school completed that is explained by the regression model. From part a, the value of R2 here is 0.0074, very small, which implies make my econometrics homework that less than 1%, 0.74%,  of the variation in years of school completed is explained by distance from school. Although the estimated coefficient for distance is statistically significantly less than zero in the regression, by itself distance leaves more than 99% of the variation in years of college completed unexplained.

The standard error ofthe regression is the same as the Root MSE in the listing in part a, which is the root mean squared error, which is given as 1.8074. This RMSE or standard error is the econometrics questions with answersestimated standard deviation about the regression line, and its units are the same as the units of the standard deviation for the dependent variable, ed, which are years.

E4.4

1. Provide a scatterplot between growth arte and trade share. b.

Malta is the outlier in the upper right hand corner, as the graph below shows. Malta has a much higher trade share than other countries and is an outlier in econometrics assignment helptrade share. However, it looks like the regression line if we removed Malta may not be much different from the regression line when we include Malta.
c.
The regression command and the output are the following:
regress growth trade share, cformat(%9.3f) pformat(%5.3f) sformat(%8.3f)
Source | SS df MS Number of obs = 65
————-+—————————— F( 1, 63) = 8.89
Model | 28.4885066 1 28.4885066 Prob > F = 0.0041
Residual | 201.851551 63 3.20399287 R-squared = 0.1237
Total | 230.340057 64 3.59906339 Root MSE = 1.79
——————————————————————————
growth | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
tradeshare | 2.306 0.773 2.982 0.004 0.761 3.852
_cons | 0.640 0.490 1.307 0.196 -0.339 1.619
——————————————————————————
The estimated intercept is 0.640, which indicates that theeconometrics assignment for money average growth rate of a country with a zero trade share would be 0.640.This estimated coefficient has a p value of 0.196, which implies econometrics homework solution that we cannot reject the null hypothesis that the population intercept is zero.
The estimated slope oneconometrics project help trade share is 2.306, which implies that with each 1 unit increase in trade share, the average growth rate increases by about 2.3%. The p value indicates that the estimated coefficient is statistically significant at the 0.004 level of significance. The point estimate of 2.3% suggests a large effect of trade share do my econometrics homework on growth, but note that the confidence interval estimate is quite wide around this point estimate, implying that we can be 95% sure that the increase in the average growth rate for a 1 unit increase in trade share pay for econometrics homework would be between 0.76 and 3.852.
For a country with a trade share of 0.5, predicted growth is given by the following:
( 1) .5*tradeshare + _cons = 0
——————————————————————————
growth | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
(1) | 1.793482 .2275893 7.88 0.000 1.338681 2.248283
——————————————————————————
Thus estimated growth foreconometrics homework help a country with a 0.5 trade share is 1.79, with a 95% confidence interval of [1.34, 2.25].
For a country with a tradeshare of 1.0, predicted growth is given by the following:
( 1) tradeshare + _cons = 0
——————————————————————————
growth | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
(1) | 2.946699 .4033066 7.31 0.000 2.140755 3.752643
——————————————————————————
Thus estimated growth for a country with a 1.0 trade share is 2.95, with a 95% confidence interval of [2.14, 3.75].
d.
The command that estimates the regression and the estimates follow:
regress growth tradeshare if country_name != “Malta”, cformat(%9.3f) pformat(%5.3f) sformat(%8.3f)
Source | SS df MS Number of obs = 64
————-+—————————— F( 1, 62) = 2.90
Model | 9.28031557 1 9.28031557 Prob > F = 0.0937
Residual | 198.527844 62 3.20206201 R-squared = 0.0447
Total | 207.80816 63 3.29854222 Root MSE = 1.7894
——————————————————————————
growth | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
tradeshare | 1.681 0.987 1.702 0.094 -0.293 3.655
_cons | 0.957 0.580 1.650 0.104 -0.203 2.118
——————————————————————————
The estimated intercept is 0.957, which means that the average growth rate of help with econometrics homework a country with a zero trade share would be 0.957.This estimated coefficient has a p value of 0.104, which implies that we cannot reject the null hypothesis that the population intercept is zero a the 0.10 level of significance.
The estimated slope on trade share is 1.681, which implies that with each 1 unit increase in trade share, the average growth rate increases by about 1.7%. The p value indicates that the estimated coefficient is statistically significant at the 0.094 level of significance.
For a country with a trade share of 0.5, predicted growth is given by the following:
( 1) .5*tradeshare + _cons = 0
——————————————————————————
growth | Coef. Std. Err. t P>|t| [95% Conf. Interval]
————-+—————————————————————-
(1) | 1.797863 .2275613 7.90 0.000 1.342974 2.252752
——————————————————————————
Thus the point estimate is that a country with a trade share of 0.5 would have a growth rate of 1.80%, with a 95% confidence interval of [1.34, 2.25]
For a country with a trade share of 1.0, predicted growth is given by the following: