Question

The median wage for economics degree holders is determined by the following equation: log( wage) = Be + B educ + B, exper+ B temure + B.age+ B married + u where educ is the level of education measured in years, exper is the job-market experience in years, tenure is the time spend with the current company in years, age is the age in years and married is a dummy variable indicating if a person is married. 935 reg Iwage educ exper tenure age married Source SS df MS Model 29.5195594 5 5.90391189 Residual 136. 136724 929 .146541145 Total 165.656283 934 .177362188 Number of obs FC 5, 929) Prob > F R-squared Adj R-squared Root MSE 0.1782 0.1738 -38281 Iwage Coef. Std. Err. t P>ltl (95% Conf. Intervall e duc exper tenure age married cons .0726795 .0111184 0121407 .0073614 . 1953893 5.165718 .0066729 0038882 .00259 .0049 0408696 1584856

Answer 1

log (mage) = Bot B E + Pe EX + BaT + By A - t B5 m + w All of lwage) = 1. dw -dws W dm 10 olm as (do 7x100 = (% change ) x dm. marred un marrjed $ 0.1953 Intrapeilatua! 2 the person married then the median wage is 19,53 % more as compared to an unmarried person.

DELTA I Ho Bp = 0 vs two third list Hi: Bog 70 x = 0.05 (ginen) t calculated = B3 - PB S.elB7) 0 = 4.7903 0121407 - .00259 fakutated at x = 0.05 and dol = n-R = 935 - 6 = 9291 - tops, 929 1.96 = 22 (from tables) 4, 7903 clearly I toos, aaa < 4,7903 Hence we ryect the null hypothers Brnu is signifient at 5% lend y significana

DELTA a 95% confidence intiemal PL 8 1 - (to n-R) Selp) = p. 58, + then) Set 6) J = 0.95 7903 P[ .0726795 - (1.96) 1 : 0066729) $ B1 = .0728795 + (1.96) (9.60667) 0.08575 29 Hence 95% C I ů a 20.059 6 new Ho: B = 0.07 vs H : B170.07 cleanly 0.07 € [0.0596, 0.08515] Hence. Ho is accepted at 95% CI |

DELTA PO 1. The Siad 5000 metres Army. Reasoning fon (C) In repeated sampling the confidence intianal will contain the true nalue of Bi = 0.02, 95 out od 100 times (d) Test of geint significance f = R/ R 1 ~ toner 2. The India largest milita surrendered 3. The India means the There is s executed. dif = (R-1, nk) = (6-1, 935-6 ) = (5,929). Falantated (0.1782)/5 = 40.28907 (1 - 0.1782)/929 15,909 = 3.02, Charly 3.02 2 40.28901 Hence we rufect the nullo 4. The C Adolf Hi