Solution to HW9:
518. They are not independent, since the joint pdf can not expressed as the product of two function related to x and y , respectively.
c
0
3 x+2
( x + y)dydx = xy +
3
y2 2
x+2
dx
x
= x( x + 2) +
0 3
3
[
x
0
( x+ 2)2 2
 x2 
x2 2
]dx
=
Section 62 614. NOTE: The data set contains one additional measurement of 91.2 that should be included in the book. Solutions use this additional measurement. Stemandleaf display of octane rating N = 82
Leaf Unit = 0.10 834 represents 83.4
1 834 3 8
729. f ( x) =
e x!

x
L ( ) =
i =1
n
e e = xi !
xi
n

 n n
n xi i =1
x !
i i =1
ln L( ) = n ln e + x i ln  ln x i !
i =1 i =1
n
d ln L( ) 1 = n + d
n
x
i =1 i
n
i
0
= n +
x
i =1
=0
x
i =1
n
i
= n
^ =
x
i =1
n
i
n
735 a)
E ( X 2 ) = 2 =
^ =
1 n
Solution to HW7: Problem 1: a) h( x) =
x , so h' ( x) =
1 3 / 2 1 1 / 2 and h' ' ( x) =  x x . 4 2
1 1 1 h' ' ( x)( y  x) 2 = h( x) + x 1 / 2 ( y  x)  x 3 / 2 ( y  x) 2 8 2 2
h( y ) h( x) + h' ( x)( y  x) +
Take y = s
2
x = 2 , so 1 1 s + 1 ( s
Solution to HW10:
^ 1118 0
^ ^ ^ + 1 x = ( y  1 x ) + 1 x = y
1119 a) The slopes of both regression models will be the same, but the intercept will be shifted. ^ b) y = 2132.41  36.9618 x
^ 0 = 2625.39 ^ 1 = 36.9618
Problem 3.
a) x
vs.
^ 0 = 2132.41
Solution to HW8:
Problem 1: H 0 : 1 = 2 H a : 1 2
X = X i n and s 2 = ( X i  X ) 2 (n  1)
i =1 i =1
n
n
Male: n1 = 7, X 1 = 69.57143 and s12 = 7.285714
2 Female: n 2 = 8, X 2 = 64.5 and s 2 = 13.14286 Pooled ttest statistic:
s2 = p
2 (n1  1) s12 + (n
Chapter 5
Important Distributions and Densities
5.1 Important Distributions
In this chapter, we describe the discrete probability distributions and the continuous probability densities that occur most often in the analysis of experiments. We will a
Statistics 511
Homework 2
Fall 2006
1. In a study of a new drug to reduce the number of epilepsy episodes, an inbred line of
epileptic mice were used. It is well known from previous experience that the mean
weekly number of episodes for these mice is 9,