Question

This is concemed with the value of houses in towns surounding Boston. It uses the data ofHarison, D., and D. L. Bubinfeld (1973), "Hedonic Prices and the Demand for Clean Air," Joumal of Environmental Economics and Management, 5, 81-102 The output appears in the table below. The variables are defined as follows: VALUE = median value of owner-occupied homes in thousands of dollars CRIME per capita crime rate NITOX = nitric oxide concentration (parts per million) ROOMS = AGE = proportion of owner-occupied units built prior to 1940 DIST-weighted distances to five Boston employment centers in miles ACCESS = index of accessibility to radial highways TAX-full-value property-tax rate per $10,000 PTRATIO-pupil-teacher ratio by town average number ofrooms per dwelling Dependent Variable: VALUE Included observation: 506 Variable Coefficient 28.4067 0.1834 Std. Error 5.3659 0.0365 CRIME NITOX ROOMS 22.8109 6.3715 0478 n.3353 4.1607 0.3924 0.0141 0.2001 0.0723 ACCESS TAX PTRATIO 0.2723 0.0126 168 0.0038 0.1394

160 Table 7. 8 EViews Output for Exercls Dependent Variable: VALUE Sample: 1 506 Included Observations: 506 Std. Error 5.365948 t-Statistic Prob Coefficient Variable 5.293875 CRIME 0.183449 0.036489-5.027548 NITOX 22.81088 4.160741 -5.482407 3.386085 0.0000 0.0000 0.0000 0.0000 0.0008 28.40666 0.392387 16.23784 0.014102 ROOMS AGE DIST ACCESS TAX 6.371512 -0.047750 -1.335269 0.272282 -0.012592 -6.671448 0.0000 0.200147 0.072276 0.003770 -3.339939 020002 0.0009 0.0000 PTRATIO-1.176787 0.139415 8.440868 The variables are defined as follows: VALUE= median value of owner-occupied homes in $1000's CRIME per capita crime rate, NTOX-nitric oxides concentration (parts per million), ROOMS - average number of rooms per dwelling. AGE proportion of owner-occupied units built prior to 1940, DIST weighted distances to five Boston employment centers, ACCESS index of accessibility to radial highways, TAX - full-value property-tax rate per $10,000, PTRATIO pupil-teacher ratio by town. (a) Report briefly on how each of the variables influences the value home (b) Find95% confidence intervals for the coefficients o ACCESS. (c) Test the hypothesis that increasing the number of rooms Dy the value of a house by $7,000. (d) Test as an alternative hypothesis H1 that reducing the pup0 000 y more th Pupil-teacher ratio e of lamb 7.9 Data on per capita consumption of heef th the price of pork ou

Answer 1

a) For one unit increase in per capita crime rate , median value of owner occupied homes fall by $183.449

  For one unit increase in nitric oxide concentration i.e. pollution , median value of owner occupied homes fall by $22810.88 .

  For one unit increase in average number of rooms per dwelling , median value of owner occupied homes increase by $6371.512 .

For one unit increase in owner occupied units built prior to 1940 , median value of owner occupied homes fall by $47.750

For one unit increase in distance to work , median value of owner occupied homes fall by $1335.269 .

For one unit increase in accessibility to radial highways , median value of owner occupied homes increase by $272.282

For $10,000 increase in full value property tax , median value of owner occupied homes fall by $12.592

For one unit increase in pupil-teacher ratio , median value of owner occupied homes fall by $1176.787 .

b) CRIME

alpha = 1 - CI/100 = 1- (95/100) = 0.05

df = n - 9 ( 8 variables and one intercept)

= 506- 9 = 497

Critical t value from t table at 95% CI is for 2 tailed test is -1.648

Margin of error = critical value*SE = -1.648*0.036489 = -0.060

Thus 95 %CI is -0.183449 + -0.060

-0.243449 AND -0.123449

ACCESS

alpha = 1 - CI/100 = 1- (95/100) = 0.05

df = n - 9 ( 8 variables and one intercept)

= 506- 9 = 497

Critical t value from t table at 95% CI is for 2 tailed test is -1.648

Margin of error = critical value*SE = -1.648*0.072276 = -0.119

Thus 95 %CI is 0.272282 + -0.119.

0.153282 AND 0.391282

c) Coefficient is 7000/1000 = 7 .

alpha = 1 - 95/100 = 0.05

Critical probability = 1-alpha/2 = 1-0.05/2 = 0.975.

0.6371 is the coefficient given.

Critical t value from t table at 95% CI is for 2 tailed test is -1.648

Margin of error = critical value*SE = -1.648*392387 = -0.6466

Thus 95 %CI is 6.371 + -0.6466.

7.01776 and 5.7244

7 falls under this CI. So, this is statistically significant.

d) PT RATIO

alpha = 1 - CI/100 = 1- (95/100) = 0.05

df = n - 9 ( 8 variables and one intercept)

= 506- 9 = 497

Critical t value from t table at 95% CI is for 2 tailed test is -1.648

Margin of error = critical value*SE = -1.648*0.139415 = -0.2297

Thus 95 %CI is -1.176787 + -0.2297.

-0.947 AND -1.406

Coefficient is -10,000/1000*10 [ because we are scaling the dollar to 1000 and also looking at 10 unit change and not 1]

= -100.

Also scaling the CI to -94.7 and -140.6 .

Thus THIS IS ALSO STATISTICALLY SIGNIFICANT.