Statistics Winter 2007 Final Exam Solutions

Question 1

(a) 1
(b) 2
(c) 1 +/- 2 = (-1,3)
(d) .25
(e) That the data look like normal draws.
Based on the histogram, this looks reasonable.

Question 2

(a) .11
(b) -.54
(c) .90
(d) -.97
(e) 5
(f) .9*sqrt(5)*sqrt(1) =  2.012461 


Question 3

(a)  0.54 
(b) values are 1,2,3 with probs  0.25 0.06 0.69 
(c) p(3,3)/p(F=3) =  0.7826087 
(d) the marginal probs for S are  0.3 0.06 0.64 which are not the same
as those of F, so they are not identically distributed.
(e) P(S=3) is not the same as P(S=3 | F=3) so they are not independent.
(f)  2.34 
(g) conditional probs are  0.1739130 0.04347826 0.7826087 so we get  2.608696 
(h)  S has the bigger variance, F has more prob at one end of the scale,
(i) positive, there is lots of prob at (1,1) and (3,3).
(j) sum row 3 and column 3 then subtract off p(3,3) =  0.79 

Question 4

(a) .1
(b) .1 +/- .2 = (-.1,.3)
(c) P(X < mu - sigma) = .16
(d) P = .012 + .4*R
(e) E(P) = .012 + .4E(R) =  0.052 
(f) normal with mean = 50(1+E(P)) and variance (50^2) * Var(P) which are  52.6  and  4 
(g) N(np,np(1-p)) = N( 84 , 13.44 ) 
(h) N(p,p(1-p)/n) = N( 0.84 , 0.001344 ) 

Question 5

(a) .2*.2 + .8*.15 =  0.16 
(b) .8*.2*.1 =  0.016 
(c) (.2^2)*.04 + (.8^2)*.01 + 2*.2*.8* 0.016  =  0.01312 
(d) .5*E(P) + .5*E(R3) =  0.23 
(e) Q = .1*R1 + .4*R2 + .5*R3, 
so Var(Q) = (.1^2)*.04 + (.4^2)*.01 + (.5^2)*.09 +
2*.1*.4* 0.016  + 2*.1*.5*.048 =  0.03058 

Question 6

(a) iid N(5,4)
(b) random walk, p(up 4)=.5, p(down 2) = .5
(c) iid Bernoulli(.2)
(d) iid Bernoulli(.5)

Question 7

(a) .7
(b) se = sqrt(phat(1-phat)/n) =  0.01449138 
phat +/- 2*se = .7 +/-  0.02898275 
(c) z = (phat-p0)/sqrt(po(1-po)/n =  12.64911 
which is very big, clear reject
(d) 0.

Question 8

(a) line + error looks very reasonable
(b) 21.39 looks right from the plot
(c) 105.165 +/- 2*7.954 = 105.165 +/-  15.908 
(d) from the output the t-statistic is 13.221 
and the corresponding p-value is (of course) tiny, so we reject
(e) t-stat = (105.165 - 100)/7.954 =  0.6493588 so fail to reject 
(f) (d) is 0 from output, (e) I guessed .4, the correct value is  0.5161065 
(g) a+bx +/- 2se =  250.9905  +/-  42.78 
(h) sqrt(R^2) =  0.8004998 
(i) same as (h)

Question 9

(1) F
(2) F
(3) T
(4) T
(5) F
(6) F
(7) F
(8) F
(9) T
(10) T
(11) F
(12) F
(13) T
(14) F
(15) T
(16) F
(17) T
(18) F
(19) T
(20) F

Question 10

I created the diagram, and then from the diagram I created the
following table:

m	v	d	p(m,v,d)
--	--	--	-------
1	1	1	.24
1	1	0	.16
1	0	1	.04
1	0	0	.06
0	1	1	.06
0	1	0	.04
0	0	1	.16
0	0	0	.24

(a) This one I got from my diagram,
P(D=1|M=1) = .8*.6 + .2*.4 =  0.56 
P(D=1|M=0) = .2*.6 + .8*.4 =  0.44 
so, if you randomly pick a trial, a white is more likely to get the death penalty.
(b) This time I just summed over v in the table above.
With M on the rows and D on the columns:
.28  .22
.22  .28
note that .28/(.28+.22) =  0.56  so we could get (a) from here.
(c)  no way, note that from (b) both M and D are Bernoulli(.5) ,
if they were independent, the table in (b) would be all .25.
(d) take the rows in the (m,v,d) table and the divide the probs by the sum.
That is, take
.04  .06
.16  .24
and divide by .5, giving
.08  .12
.32  .48
(e) yes
(f)
.24  .16
.06  .04
divided by .5 is
.48  .32
.12  .08
(g) yes
(h) no, this is immediately apparent from the second two probabilities
given in the question, that is, P(V=1|M=1) is not the same as p(V=1|M=0).
(i) as in (b) I can figure out the joint of (D,V):
.3  .2
.2  .3
Joints are not prods of marginals - so no, they are not independent.
(j)
Just looking at (a) we see that whites are more likely to get the death penalty.
Looking at the last 4 probs given in the question, we see that once you
know V, you don't have to know M to guess D.

Also, from parts (e) and (g) we see that given V, M and D are independent.
What is going on is that a murderer is more likely to get the death
penalty if the victim is white and a white is more likely to kill a white,
so a white is more likely to get the death penalty.

I took real data from a book and simplified the probabilities so this is
not just an imaginary story!

The point is that in real life, if you just think part (a) you might
conclude that white murderes get a harder time.
The full analysis suggests juries care less about non-white victims!!

Question 11

The key idea here is that because of the independence,
you can just plug in the given values, without changing
the distributions of the rest.

(a) E(Y) = 0, Var(Y) = 14
(b) E = 6, V = 5
(c) E=8 V=1