Statistics 3515Q-01/5515-01
Spring 2015
Exam 2 Solutions
_
Name
Instructions: Please answer all 4 questions. Present your statements briefly and clearly.
Good Luck!
(30) 1. In studying the surface finish of steel, five factors were varied each at two leve
5.16. The shear strength of an adhesive is thought to be affected by the application pressure and
temperature. A factorial experiment is performed in which both factors are assumed to be fixed. Analyze
the data and draw conclusions. Perform a test for non
5.3. The yield of a chemical process is being studied. The two most important variables are thought to be
the pressure and the temperature. Three levels of each factor are selected, and a factorial experiment with
two replicates is performed. The yield da
Homework 5
Ex. 5.16
From the interaction plot, it
appears the effects of pressure
are not different at different
temperature levels as the lines
are all perfectly parallel, so it is
safe to conclude there is no
interaction between
temperature and pressure
4.23. An industrial engineer is investigating the effect of four assembly methods (A, B, C, D) on the
assembly time for a color television component. Four operators are selected for the study. Furthermore,
the engineer knows that each assembly method prod
Statistics 3515Q/5515
Handout 8
The Random Effects Model - Components of Variance
In some situations the factor levels (treatments) are not of
intrinsic interest in themselves, and the factor (treatment) has a
large number of possible levels. If the exper
6.15. A nickel-titanium alloy is used to make components for jet turbine aircraft engines. Cracking is a
potentially serious problem in the final part, as it can lead to non-recoverable failure. A test is run at the
parts producer to determine the effects
Statistics 3515Q/5515
Handout 5
One way ANOVA Diagnostics - Analysis of the Residuals.
Most of the diagnostics are based on the analysis of various types of
residuals. We proceed to dene these residuals and discuss their properties
and applications.
The r
Statistics 3515Q/5515
Handout 9
Point Estimates for Variance Components in One Way Random ANOVA
We use the methods of moments approach to derive point estimates for the
variance components:
1. To estimate
2
"
we use the fact that
2
"
E(M SE ) =
and theref
Statistics 3515Q/5515
Handout 10
Randomized Complete Block Design
A randomized complete block design (rcbd) is a restricted
randomization design in which experimental units are first
selected into homogeneous groups, called blocks, and the
treatments are
Statistics 3515Q/5515
Handout 13
Statistical Analysis of the Fixed Eects Model
The following notation will be used here: Let yi: denote the total of all
observations for the i th level of factor A
b
n
P P
yi: =
yijk ;
j=1 k=1
y:j: denote the total of all
Statistics 3515Q/5515
Handout 14
Factorial designs with two factors and one observation per cell
When observations are extremely time-consuming or expensive to
collect, an experiment may be designed to have one observation per cell.
If one can assume that
Statistics 3515Q/5515
Lecture 6
Transformations on the Response Variable when
the Assumptions of Homogeneity of Variances is Violated.
One common cause for heterogeneity of variances between
levels of the treatment factor is a non linear relationship
betw
14.16. A structural engineer is studying the strength of aluminum alloy purchased from three vendors.
Each vendor submits the alloy in standard-sized bars of 1.0, 1.5, or 2.0 inches. The processing of different
sizes of bar stock from a common in'got invo
Chapter 7
Blocking and Confounding in the 2* Factorial Design
Solutions
7.1
Consider the experiment described in Problem 6.1. Analyze this experiment assuming that each
replicate represents a block of a single production shift.
Source of
Sum of
Degrees of
Chapter 6
k
The 2 Factorial Design
Solutions
6.1. An engineer is interested in the effects of cutting speed (A), tool geometry (B), and cutting angle on
the life (in hours) of a machine tool. Two levels of each factor are chosen, and three replicates of a
ID
"
7.26. Suppose that in Problem 6.7 ABCD was confounded in replicate I and ABC was confounded in
replicate II. Perform the statistical analysis of variance.
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square
657.03
13.78
1
1
657.03
13.78
14.3. A manufacturing engineer is studying the dimensional variability of a particular component that is
produced on three machines. Each machine has two spindles, and four components are randomly selected
from each spindle. The results follow. Analyze th
Stat 3515Q/5515
Handout 4
Model Adequacy - Checking the Underlying Assumptions
The model for the one way anova is given by:
Yij = + i + ij
where i=1,.,a and j=1,.,ni.
Fixed Effects
Random Effects
1. i is a fixed parameter.
1. i are iid N(0, 2)
2. ij are i
Statistics 3515Q/5515
Handout 3
Tests on Means Set Prior to Experimentation
in One Way ANOVA
a. Orthogonal Contrasts
A contrast in treatment means is a linear combinations of
the form
a
= 3cii,
i=1
where
a
3ci = 0.
i=1
We say that a linear combination of
Homework 4
Ex. 4.23
H0: MethodA = MethodB =
HA: Not all s are =
MethodC
=
MethodD
The Q-Q plot of residuals and
histogram show that the residuals
follow a normal distribution, so we
can proceed with a Latin Square
Design ANOVA procedure.
The p-value is 0
Homework 1
Ex. 2.30
a) Because the p-value for the two-sample t-test is less than the significance level of .05, we
can conclude there is enough evidence to reject the null hypothesis that there is no
difference between the two groups.
b) The p-value is <
Homework 7
Ex. 13.1
a) Random effects model, treating operator and part as random variables
Yijk = + i + j + ( )ij + ijk
Assume: i~N(0, 2a), j~N(0, 2b), ijk~N(0, 2)
b) H0: 2ab = 0
vs.
HA: 2ab > 0
The overall ANOVA table
shows that at least one variable is
Homework 6
Ex. 7.1
With each replicate treated as a block, the ANOVA table indicates that the significant
variables in the model are A*C, A C, and B. The three-way interaction term and the two-way
interaction terms A*B and B*C are not significantly import
Homework 3
Ex 4.3
Using the complete randomized block design:
H0: 1 = 2 = 3= 4 = 0
HA: 0 for at least one i
As we can see from the overall ANOVA table above, the analysis produces a very small pvalue of <.001. However, when we separate out the treatment c
Homework 2
Ex. 3.28
a)
Because the p-value=0.0021 <.05, there is significant evidence to reject the null hypothesis
that all materials have the same failure time and conclude that material does have some
effect on failure time.
b)
The residual vs. predict
Homework 8
Xinyi Liu
6.15 (a-b, d-f), 6.17, 6.22, 6.26 (a-b, d-g)
6.15
(a)
Model
Model
Model
Model
Model
Model
Model
Error
Error
Error
Error
Error
Error
Error
Error
Term
A
B
C
D
AB
AC
ABC
AD
BC
BD
CD
ABD
ACD
BCD
ABCD
Effect
3.01888
3.97558
-3.59625
1.9577
Homework 7
Xinyi Liu
5.19 5.28 6.1 6.8
5.19
Source
DF
Sum of Squares
Mean Square
F Value
Pr > F
Model
17
1239.333333
72.901961
22.24
<.0001
Error
36
118.000000
3.277778
Corrected Total
53
1357.333333
Source
DF
Type I SS
Mean Square
F Value
Pr > F
time
2
4
Liu 1
Homework 6
Xinyi Liu
5.10, 5.16, 5.18
5.10
a)
Source
DF
Sum of
Squares
Mean
Square
F Value
Pr > F
Model
8
2411750.741
301468.843
824.77
<.000
1
Error
18
6579.333
365.519
Corrected Total
26
2418330.074
Source
D
F
Type I SS
Mean Square
F Value
Pr > F
Homework 9
Xinyi Liu
7.1
Source
D
F
Sum of
Squares
Mean
Square
F Value
Pr >
F
Model
10
1616.895833
161.689583
4.39
0.007
5
Error
13
478.437500
36.802885
Corrected Total
23
2095.333333
Source
DF
Type I SS Mean Square
F Value Pr > F
REPLICAT
2
0.5833333
0.2