Suppose that you estimate a model of house prices to determine the impact of having beach frontage

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Question:

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:

PRICEi = 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, R2 = .63

where,

PRICEi = the price of the ith house (in thousands of dollars)
LOTi = the size of the lot of the ith house (in thousands of square feet)
AGEi = the age of the ith house in years
BEDi = the number of bedrooms in the ith house
FIREi = a dummy variable for a fireplace (1 = yes for the ith house)
BEACHi = a dummy for having beach frontage (1 = yes for the ith 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.