620.202 Statistics - Semester 2, 2004
p
Answers to Problem set 3
1.
(a)
since E( X ) = 2 and var( X ) = 2 (1 - 2 )/n 0 as n . n n X p Therefore since (u) = u is continuous. n , 2 ,
X = E( X ) - E n n X Note that E 2 - n X n 19 30 X n) 2
X n
(b) var(
= 2 -

620.202 Statistics Semester 2, 2003
Answers to Problem set 5
1.
(a) [ = 50]
X50 d = S/ 33
t32 .
53.2250 For this sample, n = 33, x = 53.22 and s = 11.12; so t = 11.12/33 = 1.663. Thus we accept [ = 50] since |t| < 2.037; c0.975 (t32 ) = 2.037. P = 2 Pr(

1
620.202 Statistics Semester 2, 2004
Answers to Problem set 2
1.
d (a) X = N(50,
X 50 d = N(0, 1); so a 90% interval for X is given by 50 1.6449 10/ 100 i.e., 48.36 < X < 51.64.
102 ) 100
or
102 , 100
(b)
99S 2 d 2 99S 2 = 99 Pr(77.05 < < 123.23) = 0.9

620.202 Statistics - Semester 2, 2004
Answers to Problem set 8
1. For the first table: A A B X=6 2 8 B Y =2 4 6 8 6 14
d d
P = Pr(X 6) = Pr(Y 2) = 0.1562, using X = Hg(8, 8, 14), or Y = Hg(6, 8, 14). These probabilities can be obtained using minitab. And

620.202 Statistics - Semester 2, 2003
Answers to Problem set 10
1. All that really needs to be shown here is that if M1 = P1 , P2 , Q1 and Q2 are all 22 matrices, then M1 M2 =
P1 O and M2 = O Q1 P1 P2 O . O Q1 Q2
P2 O
O Q2
, where
A generalisation of this

620.202 Statistics - Semester 2, 2004
Answers to Problem set 7
1.
(a) s2 = 1 (s2 + s2 + s2 ) = 16.34, so W = 27s2 = 441.07; s2 = 31.48, so B = 20s2 = 628.95. 1 2 3 W W B B 3 These sums of squares enable us to complete the ANOVA table: between within total

620.202 Statistics - Semester 2, 2004
Answers to Problem set 1
1. Here is a copy of the word page based on my simulation:
MTB > MTB > SUBC> MTB > Base 641387. random 100 c1; normal 50 10. describe c1
Descriptive Statistics: C1
Variable C1 Variable C1 N 10