Economics 421/521

Winter 2012

Solution to Homework #1

**Part I. Hypothesis Testing**

1. Suppose that you estimate a model of house prices to determine the impact of having beach frontage on the value of a house. You do some research, and you decide to use the size of the lot instead of the size of the house for a number of theoretical and data availability reasons. Your results (standard errors in parentheses) are:

PRICE_{i} = 40 + 35.0 LOTi – 2.0 AGEi + 10.0 BEDi – 4.0FIREi + 100 BEACHi

(29) (5.0) (1.1) (10.0) (3.0) (9.0)

n = 30, R^{2} = .63

where,

PRICEi = the price of the *i*th house (in thousands of dollars)

LOTi = the size of the lot of the *i*th house (in thousands of square feet)

AGEi = the age of the *i*th house in years

BEDi = the number of bedrooms in the *i*th house

FIREi = a dummy variable for a fireplace (1 = yes for the *i*th house)

BEACHi = a dummy for having beach frontage (1 = yes for the *i*th house)

a) You expect the variables LOT, BED, and BEACH to have positive coefficients. Create and test the appropriate hypotheses to evaluate these expectations at the 5 percent level.

For LOT;

Ho: βLOT = 0

Ha: βLOT > 0

*t* – score: (35.0) / (5.0) = 7.0

*t-*critical: 1.711 because d.f. is 24 and 5% level of significance.

Since 7.0 > 1.711, we can reject the null hypothesis that the true coefficient of LOT is not positive.

For BED;

Ho: βBED = 0

Ha: βBED > 0

*t* – score: (10.0) / (10.0) = 1.0

*t-*critical: 1.711 because d.f. is 24 and 5% level of significance.

Since 1.0 < 1.711, we cannot reject the null hypothesis that the true coefficient of BED is not positive.

For BEACH;

Ho: βBEACH = 0

Ha: βBEACH > 0

*t* – score: (100) / (0.9) = 11.1

*t-*critical: 1.711 because d.f. is 24 and 5% level of significance.

Since 10.0 > 1.711, we can reject the null hypothesis that the true coefficient of BEACH is not positive.

b) You expect AGE to have a negative coefficient. Create and test the appropriate hypothesis to evaluate these expectations at the 10 percent level.

For AGE;

Ho: βAGE = 0

Ha: βAGE < 0

*t* – score: ( - 2.0) / (1.1) = - 1.81

*t-*critical: 1.318 because d.f. is 24 and 10% level of significance.

Since | - 1.81 | > |1.318|, we can reject the null hypothesis that the true coefficient of AGE is not negative.

c) At first you expect FIRE to have a positive coefficient, but one of your friends says that fireplaces are messy and are a pain to keep clean, so you are not sure. Run a two-sided *t*-test around zero to test these expectations at the 5 percent level.

For FIRE;

Ho: βFIRE = 0

Ha: βFIRE ≠ 0

*t* – score: ( - 4.0 ) / (3.0) = - 1.3

*t-*critical:: 2.064 because d.f. is 24 and 5% level of significance.

Since | - 1.3 | < 2.064, we cannot reject the null hypothesis that the true coefficient of FIRE is not different from zero.

2. Consider the following regression:

log(Qci) = 921.6 – 1.3log(Pci) + 0.7log(Pai) + 11.4log(Inci)

(121) (0.3) (0.05) (2.8)

n = 30, R2 = 0.82

where,

Qci = the total sales of CAMRY in the ith city, the year of 2003

Pci = the price of a CAMRY in the ith city, the year of 2003 (in thousands)

Pai = the price of a ACCORD in the ith city, the year of 2003 (in thousands)

Inci = the average income in the ith city, the year of 2003 (in thousands)

Numbers in the parentheses are standard errors.

a) What does the constant term (= 921.6) mean?

The constant term means is the value of the dependent variable when all the independent variables are zero.

b) How would you interpret the coefficient on log(Pci). Be explicit and explain, in terms of economic theory, the importance of its magnitude.

Totally differentiate the estimated equation w.r.t. Qci and Pci to obtain the elasticity:

Elasticity. |-1.3| > 1, so elastic. If the price of CAMRY increases by 1%, then the total sales of CAMRY decreases by 1.3%. (Thus, the coefficient gives the elasticity.)

c) Get t-values of each of all coefficients in the regression. Are all of our coefficients statistically significant at the 5% level of significance? How about at the 1% level of significance?

Pci Pai Inci

t-score -4.3 14. 4.08

Two-tailed test at the 5% level of significance, t-critical = 2.056 because d.f. is 26.

Two-tailed test at the 1% level of significance, t-critical = 2.779 because d.f. is 26.

Therefore, all of our coefficients statistically significant at the both 1% and 5% level of significances.

d) Interpret R2. What does it mean?

R2 tells how well the sample regression line fits the data.

(TSS = ESS + RSS)

Thus, it is the ratio of the explained variation in the dependent variable divided by the total variation and hence measures the percentage of the total variation that is explained by the variable sin the model.

** **

**Part II. Short Answer**

1. State the Gauss-Markov Theorem and explain the term BLUE.

Given the classical assumptions (model is linear and correctly specified, X's are exogenous, no perfect multicollinearity, error has zero mean, homoskedasticity, errors are independent, errors and X's are uncorrelated, errors normally distributed), the OLS estimator is the minimum variance estimator from among the set of all linear unbiased estimators. OLS is BLUE. This means it is the Best (minimum variance) Linear Unbiased Estimator.

**Part III. Estimation**

1. Given data on M2, real GDP, and the T-bill rate, estimate the following regression and test whether the coefficients differ from zero. Do the coefficients have the expected signs?:

M_{t} = β_{0} + β_{1}RGDP_{t} + β_{2}Tbill_{t} + e_{t}

To do this problem, first create a new workfile by following these steps (some of the figures are pop-ups that bring up larger, clearer versions):

Next, read in the data set (save it to your computer first by right clicking on the link in the homework set):

Finally, run the regression of M2 on a constant, RGDP and the TBillRate:

Running this regression gives:

The critical value at the 1% level is 2.576, thus the coefficients are significant. They also have the expected sign: People tend to demand more money as income (RGDP) increases. They also tend to demand less money as the T-bill rate increases – this is because as the T-bill rate goes up, the opportunity cost to holding money also goes up.

Economics 421/521

Winter 2012

Solution to homework #2

1. Using the EAEF data set, regress LGEARN on S, EXP, and ASVABC. Use F-tests to determine whether the coefficients on S and EXP are (a) jointly significant, and (b) equal. [Parts (a) and (b) are two separate tests.]

First, read in the data:

Then, take the log of earnings:

Next, estimate the unrestricted model This is needed for both parts (a) and (b). To estimate the UR model, regress lnearnings on a constant, exper, and asvabc:

The results are:

(a) The restricted model for the null hypothesis that the coefficients on s and exper are zero is:

F= [(149.31-130.35)/2]/[130.35/(540-4) =38.98

The critical value for this test is F(2,536)= 3.05 (approx. for 536), so reject that both coefficients are zero.

(b) For this part, the null hypothesis is that the coefficients on s and exper are equal. To impose this on the model, start with:

*lnsalary = b*_{1} + b_{2}*s + b_{3}*exper + b_{4}*asvabc + u

Then impose that *b*_{2}=b_{3}:

*lnsalary = b*_{1} + b_{2}*s + b_{2}*exper + b_{4}*asvabc + u

Group terms:

*lnsalary = b*_{1} + b_{2}*(s + *exper) +** b*_{4}*asvabc + u

This is the model we need to estimate. Thus, the first step is to obtain data on the sum of s and exper:

Run the restricted regression:

Here are the results:

Finally, use these to calculate the F-statistic:

F= [(138.29-130.35)/1]/[130.35/(540-4) = 32.65

The critical value for this test is F(1,536)= 3.90 (approx.), so reject that the coefficients are equal.

2. Problem 7.1 in the text.

In this case the F statistic is simply the ratio of the residual sum of squares (more generally it is the rss divided by n-k, i.e. F = [rss_{2}/(n_{2}-k)]/ [rss_{1}/(n_{1}-k)], where n_{1} and n_{2} are the number of observations in each sample, but when n_{1}=n_{2} the terms cancel):

F=28,101/321 = 87.54

The critical value (5%) for and F(8,8) = 3.44, so the null of homoskedasticity is rejected.

3. Moved to the next homework.