ignoring the interaction
15:19 Wednesday, May 7, 2008 The GLM Procedure Number of Observations Read Number of Observations Used
33
12 12 34
ignoring the interaction
15:19 Wednesday, May 7, 2008 The GLM Procedure
Dependent Variable: carbonation

STT 422 Sample Exam 2 Summer 2008 Part 1: Contingency Tables and Chi-Square Tests: 1. Given the following contingency table: High GPA Middle GPA Lower Classman 20 30 Upper Classman 15 40
Low GPA 10 15
a) What is the test statistic for a test for as

EX 1:
data ex1; /*name the data*/ input x y; /*what order the variables will be in*/ datalines; /*the data starts after this line*/ 1 2 2 4 3 5 4 5 ; /*for some reason the semicolon gets its own line*/ proc glm data=ex1; /*do a regression on this dat

We are trying to predict whether a woman has ever used estrogen therapy (coded 1 for yes) based on uterine cancer subtype. There are three categories to the subtype predictor. Adenocarcinoma (coded 0), Adenosquamoous (coded 1), and Other (coded 2).

Ex 4: proc glm data=gpa; class gender; /*tells SAS that gender is categorical*/ model gpa = gender concept iq /solution; /*creates a model with different intercepts for each gender*/ run;
The GLM Procedure Class Level Information Class gender Levels

The ANOVA Procedure Class Level Information Class week group Levels 2 6 Values 4 8 ECM1 ECM2 ECM3 MAT1 MAT2 MAT3
Number of Observations Read Number of Observations Used The SAS System The ANOVA Procedure Dependent Variable: gpi Sum of Squares 27057.

A look at using logistic regression to control for Simpson's paradox and aggregated tables. Death Penalty Example revisited.
The SAS System
14:13 Thursday, June 12, 2008
2
The LOGISTIC Procedure Model Fit Statistics Intercept and Covariates 230.2

STT 422 HW #2 Due: Monday, June 2 Text Questions: 13.1 13.2 13.7 13.16 Additional Questions: 1. Complete the following ANOVA table: Source Sum of Degrees of Squares Freedom Factor A 150 c) Factor B a) 2 Interaction 60 6 Error 40 b) Total 450 21
Mean

The Goal of this multiple logistic regression is to predict whether a patient has coronary heart disease based on smoking status (1 if they smoke), age in years, and cholesterol (continuous).
The SAS System 11:35 Tuesday, June 10, 2008 2
The LOGISTI

HW 1: DUE: Friday, May 23 Text Questions: 11.2 11.3 11.4 11.8 11.10
1. The following is SAS output analyzing the length of time a saw stayed useable. There were two manufacturers tested, and the saws were run at different speeds (RPM). The response

STT 422 HW 3 Due Friday, May 13 BOOK QUESTIONS 9.15 9.18 9.20 a, b 9.22 a 9.38 9.46 ADDITIONAL QUESTIONS 1. Suppose we are trying to see whether there is agreement between how husbands and wives view their satisfaction with their workplace environmen

Sample Exam 1 Part 1: Multiple Linear Regression: 1. A regression line is fit from a sample of size 30. The least squares line is: ^ y 2.2 4.7 * x1 13.2 * x2 , and the standard errors of b1 and b2 are 3.1 and 2.8, respectively. a) Give a 95% confiden

Ex 2:
The SAS System 15:08 Monday, May 5, 2008 80
The GLM Procedure Dependent Variable: gpa Sum of Squares 159.9022855 179.5246058 339.4268914
Source Model Error Corrected Total
DF 2 75 77
Mean Square 79.9511428 2.3936614
F Value 33.40
Pr > F <

Stt 422 HW 4 Due Monday, June 23
1. Consider the following two way contingency table: Lung Cancer Smoker 19 Non Smoker 3 a) b) c) d) e)
No Lung Cancer 86 107
Find the odds of having lung cancer for a smoker. Find the odds ratio of having lung canc

Residual Graph Diagnostics Ideal:
r esi d 2
1
0
-1
-2
-3 0 1 2 3 4 pr ed 5 6 7 8 9
There is no structure here. Ideally we will just get a box with equal amounts of residuals above and below the x axis.
Things that can go wrong:
r esi d 3
2

Model Fit Statistics Intercept and Covariates 427.460 449.519 417.460 4
Criterion AIC SC -2 Log L The SAS System
Intercept Only 440.558 444.970 438.558
05:23 Friday, June 13, 2008
The LOGISTIC Procedure Testing Global Null Hypothesis: BETA=0 Test