Solution to HW #5 36-220/247
1. (7.2, 10 pts) Y1 , . . . , Yn are Poisson()
(a) To nd the unbiased estimators, need to nd their expected value and check if it is equal
to what they are estimating, .
1
Solution to HW #2 36-220/247/747
1. (3.33, 20 pts) (a) Dene the following events:
A = cfw_ Alarm A goes o , B = cfw_ Alarm B goes o , I = cfw_ There is an intruder .
P (A|I) = 0.9
P (B|I) = 0.95
P (A|
Part I: For each the following, simply give the answer. It is not required that any derivation
0r justication be provided, and there is no partial credit. Each of these is worth three points.
1. Supp
Solution to HW #6 36-220/247
1. (8.10) (2 points)
Let stand for the egg-hatching rate.
H0 : = 0.5
Ha : > 0.5
2. (8.16) (2 points each part)
(a): Let = mean surface roughness.
H0 : = 2
Ha : = 2
(b): Si
Solution to HW #4 36-220/247/747
1. (5.30, 15 pts)
Let Y be the injected quantity, so Y is normal with mean = 10, sd = 0.2.
(a) Want P(underlled) = P (Y < 10)
= P ( Y 10 < 1010 ) = P (Z < 0) = 1
.2
.2
Solution to HW #9 36-220/247/747
1. (11.23)
(a.) (2pts) The regression output is as follows:
Regression Analysis: Voltage versus Disperse Phase Volume, Salinity, .
The regression equation is
Voltage =
Solution to HW #7 36-220/247/747
1. Let X denote the sample mean. (5 points each part)
(a): P(Type II error) = P(fail to reject | = 10.5)
= 0.05 corresponds to a z-value of 1.65. So the rejection reg
Solution to HW #3 36-220/247
1. (4.5, 10 pts) (a) Y = Number of free links.
There are two total links, so the possible values of Y are 0, 1, 2.
(b) We must nd P (Y = 0), P (Y = 1), P (Y = 2). (Actuall
Solution to HW #1 36-220/247
1. (a) The population of interest in this case is the thion level in air as a result of contamination
due to extensive application of pesticides to crops, at any location
36-220/247/747 Exam Three April 11, 2014
VERSION A
Name
Email
Group
Part Possible Points
I
16
II - 1
8
II - 2
5
II - 3
5
Free
1
1
Score
36-220/247 Exam Three April 16, 2014
VERSION B
Name
Email
Group