5.30, IN NOTES-8
Modeling faculty salary. The administration of
a midwestern university commissioned a salary
equity study to help establish benchmarks for
faculty salaries. The administration utilized the
following regression model for annual salary,
y:

CASE 1 Abigail Faith
This problem is one which I worked on as a consultant. Nothing is changed except the name of
the CEO and the company.
Abigail Faith is the CEO of AFB Company, a purveyor of computer-aided design software.
Abigail has instructed her di

E(y)=A+B1X+B2X2
Y=A+BIXI+B2X2+E
YC=A+B1X1+B2X2
3.64 Learning the mechanics. Consider the 10 data
points shown in the table.
Model Summary
R
Adjusted R Std. Error of
b
Model
R
Square
Square
the Estimate
1
.998a
.997
.996
1.58690
a. Predictors: VAR00002
b.

1.
Between-Subjects
Factors
N
VAR00002 1.00
8
2.00
8
3.00
8
Tests of Between-Subjects Effects
Dependent Variable: VAR00001
Type III Sum
Mean
Source
of Squares
df
Square
F
Corrected
112.000a
2
56.000
15.474
Model
Intercept
1176.000
1
1176.000 324.947
VAR00

Chapter 4 Model Selection Procedures
Stepwise regression M&S 6.2
Stepwise model selection procedure (Section 6.2):
First determine SLENTRY (the level of significance to enter the model) and SLSTAY (the level
of significance to stay in the model). Suppose

Chapter 3 Multiple Regression Analysis
Reading: M&S Chapter 4
I. Case of two independent variables
Population model
Y = 0 + 1 X 1+ 2 X 2 +
(ERROR)
Prediction equation (lease square equation)
Y^ =b 0 +b1 X 1 +b2 X 2
Residuals (prediction errors)(EPISION)
e

1
ST625 Final Exam
Write question numbers clearly. If the answer is a numerical value or a set of numerical values,
you must provide the formula first, and also enclose the results in a box.
1. You are asked to fit the following data by a simple regressio

ST625 Assignment 4
(Multiple regression model, quadratic model, model with interaction)
Due (extended): Midnight April 5, 2015 (Sunday)
Exercise 1 (4.32, p. 204-205, M&S, 6th edition) Data: MovieGross
The data table gives the worldwide gross avenues for t

ST625 Assignment 3 Solutions
Note: I did both orange juice and fin-tubes data sets.
For two datasets in Assts 1 and 2, you have done the following work:
Calculated intercept and slope estimates; calculated confidence interval for the population slope
beta

ST-625
Topic 6 HW
In notes 6
a)
Let H0 be, there is no regression relation between X and Y
The H1 will be, there exists a regression relation between X and Y
Since the p-value is .000<.05, we reject H0 and conclude that there is a
regression relation betw

Chapter 2 Simple Regression: Inference
(Confidence Interval and Test of Hypothesis for Population Slope)
1. Notations and formulas revisit.
For the purpose of statistical inference, we often write the regression model as
Y = 0 +X +
is called population

Chapter 1 Regression Analysis (Descriptive)
ST625 Asst 1 name
Q1
DueBefore the next class
Subject Title: ST625 Asst1/Question Name
File name: ST625 Asst1 Name
Reading: M&S 7th edition, Chapter 3.1 to 3.3. It is optional to read Chapters 1 and 2, but
do no

Assignment 1
1.
a. command seq
i. How to generate a sequence of numbers
108, 111, 114 ?
like 102, 105,
Command : seq(102,114,3)
ii. How to generate a sequence of numbers
15,.,1000?
like 5, 10,
Command : seq(5,1000,5)
b. command c
i. Create a vector y with

12/10/2012
Chapter 11 Mini Case
Situation
Shrieves Casting Company is considering adding a new line to its product mix, and the capital budgeting
analysis is being conducted by Sidney Johnson, a recently graduated MBA. The production line would be set
up

1.
Between-Subjects Factors
N
Temperatur 1.00
6
e
2.00
6
3.00
6
MountType 1.00
9
2.00
9
Tests of Between-Subjects Effects
Dependent Variable: Performance
Type III Sum
Mean
Source
of Squares
df
Square
F
Corrected
.415a
5
.083
1.487
Model
Intercept
14.045
1

ST 625 HW5
MingluLi,ChakLamCheong
a) Obtaina95%confidenceintervalforthemeansatisfactionforcustomersage
28.Interpretyourinterval.
95%confidenceintervalformeansatisfactionforcustomersageat28is
between3.06148and3.34037.
Therefore,weare95%confidenttotellthatt

Assignment 2
1. Textbook: p.p. 74 #1.96
Hint: If a random variable follows a Normal distribution with mean
and standard deviation . The probability that this random variable is
less than or equal to a constant value, say c, can be found in R by
pnorm(c,m

Assignment 3
1. R functions for t distribution
Using appropriate R functions, compute the following values. Please show
both your R command and output.
The critical value of t for a 90% confidence interval, with df=
(a)
17
Answer: > qt(0.95, df=17)
[1] 1.

Assignment 4
1. Body Fat: simple linear regression
(a) Find the regression of Pct.BF on Weight, i.e. Pct.BF is the
response and Weight is the predictor.
Answer: > attach(bodyfat)
> View(bodyfat)
> lm(Pct.BF~Weight)
Call:
lm(formula = Pct.BF ~ Weight)
Coef

Assignment 5
1. Model fit assessment
A simple linear regression analysis performed on a sample of n = 24
points provides the following information:
b0 = 1, b1 = 2, s(b0) = 0.05, s(b1) = 0.25, SST = 117.2873, SSE =
30.0.
(a)
Based on the above information,

1a)
Source of
Variability
Diet
Error
Total
SSQ
Df
MSQ
Fcalc
27000
57600
84600
3
96
99
9000
600
15
Fcalc= MSBC/MSW. Therefore, 15=MSBc/600 which is equal to 9000.
MSQ= SSQ/Df. Therefore, 9000=SSQ/Df which is equal to 27000.
Df for Error= (R-1)*C. Therefore