1. Solution 7 W=sum of ranks of group A (best student=rank 1, etc).
(a) (i). Reject H0 if W is too small.
Reject H0 if W<=7, alpha=2/35=.0571; Reject H0 if W<=8, alpha=4/35=.1143.
(ii). Reject H0 if W is too large.
Reject H0 if W>=17, alpha=2/35=.0571; Reject H0 if W>=16, alpha=4/35=.1143.
(iii). Reject H0 if W is too small or too large.
Reject H0 if W<=6 or W>=18, alpha=2/35=.0571;
Reject H0 if W<=7 or W>=17, alpha=4/35=.1143,
Reject H0 if W<=6 or W>=17, alpha=3/35=.0857, (Not equal tailed)
Reject H0 if W<=7 or W>=18, alpha=3/35= (Not equal tailed)
Note: In each case, the nearest test to the nominal size 10%, has actual size 11.43%. My recommendation is not to exceed the nominal size alpha, and go for a test of smaller size.
(b) W Probability Possible combination
/ (1,2,3)
/ (1,2,4)
/ (1,2,5) (1,3,4)
/ (1,2,6) (1,3,5) (2,3,4)
/ (1,2,7) (1,3,6) (1,4,5) (2,3,5)
/ (1,3,7) (1,4,6) (2,3,6) (2,4,5)
/ (1,4,7) (1,5,6) (2,3,7) (2,4,6) (3,4,5)
/ (1,5,7) (2,4,7) (2,3,6) (3,4,6)
/ (1,6,7) (2,5,7) (3,4,7) (3,5,6)
/ (2,6,7) (3,5,7) (4,5,6)
/ (3,6,7) (4,5,7)
/ (4,6,7)
/ (5,6,7)
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SA3202, Solution 7

2. Thus E(W)=12, Var(W)=8
Note: The distribution is symmetric; even for such small sample sizes (3 and 4) it is not very non-normal.
(c ). The point for parts (c ) and (d) is this: for large sample sizes, we need to use normal approximation, for which we need to know the mean and variance.
We can find the mean by noting that the distribution of W is symmetric. So the mean is the “middle” value, which is 12. This method can be applied for large samples also (find the min and max value and take average). But there is an easier way as described below.
When ranking the 7 students, the average rank is (1+2+…+7)/7=7*8/2/7=4. If there is no difference between A and B, each student would get one of the ranks at random. So each student is expected to get a rank 4. So the sum of the ranks for the 3 students in group A is expected to be 4+4+4=12.
The above argument can be used to find the mean of W in general, with sample sizes n1 and n2, and n=n1+n2. In fact, when ranking n students, the average rank is (n+1)/2. So for the n1 students in group A, the sum of the ranks is expected to be E(W)=n1(n+1)/2.
More formally, W=R1+R2+R3 where R1, R2, and R3 are the ranks of the three students in group A. Each R takes one of the values 1,2, …,7 with equal probability:
Pr(Ri=j)=1/7, j=1,2,…,7
From which, we have E(Ri)=4, Var(Ri)=4. Hence E(W)=E(R1+R2+R3)=3*4=12.
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SA3202, Solution 7

3. (d). Note that for different i and j :
Pr(Ri=k, Rj=l)= Pr(Rj=l | Ri=k) Pr(Ri=k)=1/6*1/7=1/42.
Thus for different i and j:
E(RiRj)=46/3, E(Ri)E(Rj)=16, Cov(Ri,Rj)=E(RiRj)-E(Ri)E(Rj)=-2/3.
Thus Var(W)=Var(R1+R2+R3)=Var(R1)+Var(R2)+Var(R3)+2Cov(R1,R2)+2Cov(R1,R3)+2Cov(R2,R3)
=4*3-2*2/3*3=8.
Note that the above argument can be used for the general case with sample sizes n1, n2, and n=n1+n2.
Pr(Ri=j)=1/n, j=1,2,…,n
Pr(Ri=k,Rj=l)=Pr(Ri=k|Rj=l)Pr(Rj=l)=1/(n-1)*1/n=1/[(n-1)n]
E(Ri)=(1+2+….+n)/n=n(n+1)/2/n=(n+1)/2.
E(Ri^2)=(1+2^2+…+n^2)/n=n(n+1)(2n+1)/6/n=(n+1)(2n+1)/6.
Var(Ri)=(n+1)(2n+1)/6-(n+1)^2/4=(n^2-1)/12.
E(RiRj)=[(1+2+…+n)^2-(1+2^2+…+n^2)]/[n(n-1)]=(n+1)(3n+2)/12
Cov(Ri,Rj)=(n+1)(3n+2)/12-(n+1)^2/4=-(n+1)/12.
It follows that
E(W)=E(R1+…+Rn1)=n1(n+1)/2
Var(W)=n1(n^2-1)/12-n1(n1-1)(n+1)/12=n1n2(n+1)/12=n1n2(n1+n2+1)/12.
1/18/2019
SA3202, Solution 7

4. (a) X~U[10,30], so F(x)=(x-10)/20, it follows that xp=10+20p. Thus,
the median is Q2=10+20*.5=20.
The quartiles are Q1=10+20*.25=15, Q2=20,Q3=25
The deciles are 12,14,16,…,28
(b) X~Exp(20). Then F(x)=1-exp(-x/20), x>=0. It follows that xp=-20 log(1-p). Using
this formula, we can compute the median, quartiles and deciles respectively.
(c) X~N(10,4), Z=(X-10)/2~N(0,1). F(x)=Phi((x-10)/2). Thus xp=10+2 zp where zp is the quantiles of Z. Using the normal cdf table, we can compute any quantile of X. For example, the median of X is 10+2 z(.5)=10.
1 , when x>2
3. X~U[0,2], so f(x)=1/2 , 0<=x<=2, and F(x)= x/2, when 0<=x<=2
0, when x<0.
Then, the density function for X(r ) is
for r=1,2,3,4,5, and n=5.
1/18/2019
SA3202, Solution 7