Question

Question 2 (6 points) The following table gives the joint probability distribution between employ- employed or looking tor work ment status and college graduation among those either (unemployed) in the working-age U.S. population for 2012. Joint Distribution of Employment Status and College Graduation in the U.S.Population Aged 25 and Older, 2012 Non-college grads (X- 0) College grads (x-1) Total Unemployed (Y = O) 0.053 0.015 0.068 Employed (Y= 1) 0.586 0.346 0.932 Total 0.639 0.361 1.000 (a) Compute E[Y] (b) Calculate EY|X = 1] and EY|X = 이 (c) Are educational achievement and employment status independent? Explain

Answer 1

X Y X=0 X =1 P(Y) Y=0 0.053 0.015 0.068 Y = 0.586 0.346 0.932 P(X) 0.639 0.361 1 Calculate E (Y) E(Y)= XY-P(Y) =[(y=0)xP(Y = 0)]+[(Y =1)*P(Y =1)] =(0x0.068) +(1+0.932) = 0.932 Calculate E (Y|X =1) and E (Y| X = 0) E(Y|X =1)= y.P(Y|X = 1) P(Y = 0 | X =1)=- P(Y = On X =1)_0.015 - 0 0416 P( X =1) 0.361 PY=1 X-1) - P(Y=1n X =1)_0.346 P(X =1) 0.361 0.9584 E(Y|X =1)=[(Y = 0)-P(Y =0 | X =1)]+[(y =1).P(Y = 1| X =1)] = (0x0.0416)+(1+0.9584) = 0.9584 E(Y|X = 0) = {y-P(Y|X=0) P(Y=0 X-0)- P(Y=0 X = 0) 0.053 P(X=1) =0.639 = 0.0829 P(Y=1nX = 0) 0.586 P(Y =1| X = 0) = P ( X =0) 620 = 0.9171 E(Y|X = 0) = [(y = 0)- P(Y = 0 X =0)]+[(y =1). P(Y =1| X =0)] = (0x0.0829)+(1x0.9171) = 0.9171 Check the independent condition. P(Y = 0 X =0)=P(X=0)xP(Y=0) P(Y=0 X =0) = 0.053 P( X = 0)*P(Y =0)=0.639x0.068 = 0.043 P(Y = 0 X =0)*P( X = 0)P(Y =0) The independent condition not holds. Therefore, X and Y are not independent.