Question

hrice-A' + β, ๒(asses) + ln(lotsize) + β3 ln(agrft) + β.bdmns + น where price: house price assess : the assessed housing value (before the house was sold); lotsize : size of the lot, in feet; sgr ft: square footage; and bdrms : number of bedrooms. The econometrician uses Stata 'reg' command, i.e, uses OLS estimation, to get the following results: sumber of abs - F 1, 6)280.94 1 6.13852904 Prob > F 0.0000 R-squared 0.7656 dj R-squared 0.7629 14782 Residual 1.87907448602184970 Total01760352 87 092156362 Root NSE price Coef Std. Err thir/ [95% Conf. Interval] ASSESS1.01 3407 0604609 16.76 0.000 634743 3460739 -0.47 0.642 8932147 1.133599 8494464 5264978 Table A df Number of obs » F 4. 8370.58 prob R-squared 0.7728 Adj R-squared0.7619 Hode 6196074731 54901868 02194613 0.0000 esdus 82352879 Total 0603527 092156362 14814 ersce Coef. Std. Er 151446 1032384 1384305 Pltl [95% Conf. Interval] 7418453 1.344285 1720942 692972 1.396783 t -0430S 6.89 0.000 าร๐ts120| ,0074379-0385615 0.190.848 -0.75 0.458 1.53 0.129 263743 696647 46 0.645 0692593 78571 0101135 084135 bdrms 0338 392 0220983 Table B a) Consider Table B. Test the hypothesis that Ho Bo 0 against one-sided alternative, Ha: B,s table b) Consider Table B. Test the hypothesis that Ho ; β.-1 against two-sided alternative, HA 5. c) Test the hypothesis that β2-A+ β.-0. for 5% significance level. (You can use normal for 10% significance level. d) If the variance of price changes with independent variables, what can you say about t test for Be alternative, HA Do

Answer 1

A)

For β₀, the P value from the table B is 0.0000 which is smaller than 0.05. In this case, the alternate hypothesis is accepted. The coefficient can't take the value of 0 in this regard.

Here, β₀ < 0 is Accepted.

B)

For β₁, the P value from the table B is 0.848 which is larger than 0.05. In this case, the null hypothesis is accepted. The coefficient can take the value of 0 in this regard.

C)

For β₂ = β₃ = β₄, the P value from the table B for each of the coefficient are larger than 0.05 (0.848, 0.458, 0.129 respectively). In this case, the null hypothesis is accepted. The coefficients can take the value of 0 in this regard.

D)

For β₀, the P value from the table B is 0.0000 which is smaller than 0.05. In this case, the alternate hypothesis is accepted. The coefficient can't take the value of 0 in this regard.

Here, β₀ > 0 is Accepted..

End of the Solution...