Question

2 Consider the following model that relates the percentage of a household's budget spent on alcohol WALC to total expenditure TOTEXP, age of the household head AGE, and the number of children in the household NK Using 1200 observations from a London survey, this equation was estimated with and without the AGE variables included, giving the following results: WALC = 8.149 + 2.884 ln(TOTEXP)-1.217NK-0.5699AGE + 0.0055 15AGE = 62048 (se) (0486) (0.382) (0.1790) (0.002332) WALC--1.206 +2.152In(TOTEXP)-1421NK 6.3196 (se) σ = 0.482) (0.376) Use an F-test and a 5% significance level to test whether AGE and AGE should be included in the equation. Use an F-test and a 5% significance level to test whether NK should be included in the first equation. a. b. c. Use an F-test, a 5% significance level and the first equation to test,.p2-3.5 against the alter- native H : P2 #3.5.

Answer 1

dl. Null hypothesis. Hol : β4 =0 us. Alternative hypothesis. H11 : β4 0 Value of Test statistic-F = (-0.5699/0.1790)2- 10.1366 p - value P(F 10.1366 FF1,1200-5)0.0015< 0.05 Hence we reject H01 at 5% level of significance and conclude that AGE should be included in the equation. Null hypothesis. H02 : β5=0 us. Alternative hypothesis, H12: β5 0 Value of Test statisticF (0.005515/0.002332)25.5929 p - value-P(F> 5.5929F F1,1200-5) 0.0182 0.05 Hence we reject H02 at 5% level of significance and conclude that AGE should be included in the equation. Null hypothesis. H03 : β3 = 0 us. Alternative hypothesis. H13 : β3〆0 Value of Test statistic = F = (- 1.217/0.382 )2-10.1497 p - valuePF 10.1497F F1,1200-5) 0.0015 0.05 Hence we reject H03 at 5% level of significance and conclude that NK should be included in the equation.

Null hypothesis. H04 : β2 = 3.5 us. Alternative hypothesis. H14 : β2メ3.5 Value of Test statisticF ((2.884-3.5)/0.486)2 1.6065 p-value = P(F > 1.6065|F ~ F11200-5) 0.2052 > 0.05 Hence we fail to reject H04 at 5% level of significance