Hello all, I am trying to install my SAS 9.4 on windows 8.1, and I have encountered the most weird error, never seen anything like this. I ran the setup file, and choose "install SAS software", and then this message came: "This machine is currently using windows 7 pro 64 bit. However, you need windows 7 pro 64 SP1 or later, please refer to the document....." Ha ? I have windows 8.1....legal, activated...what makes SAS think I have windows 7 pro ? When I press OK, I go backwards, meaning, I can't install.... I went to my PC's system, it said windows 8.1. I went to the Microsoft website, where you can check your version online, it said windows 8.1 Only SAS thinks I have windows 7...how do I get out of this situation and install the software ? Will appreciate a solution ! Thanks. ... View more

Hello all, I am interested in analyzing data with a time to event response variable. My response variable measures time until a treatment succeeds to do what it is meant to be doing. My response variable (Y) gets the values: 0,1,2,3,4,5 and 10 min (times when the status is being checked). After 10 minutes, if the treatment did not work, it is a failure, some sort of censoring. Naturally, lower values of Y are better. In this dataset there are two treatments, a new interventional and a control, which is the standard of care. The main question is comparing the two treatments, to find superiority of the new one over the existing one. Each patient enrolled for this trial, received the above procedure once or more. It is most frequent to find patients with either 1 procedure or 2. It is rare, however not impossible, to see even 3 procedures. All procedures within a patient are treated with the same treatment, either the new one or the control. There are two types of these procedures, I'll call them A and B. Every patient is having one of these two, other possible procedure exist, but were not chosen for this trial. In other words, every patient have 1, 2 and rarely 3 procedures, all from the same type, either A or B, and the treatment is either the new or the control. In each procedure, Y is measured like mentioned above. The correlation within a patient is assumed to be high. The trial is also multi-center. Summary: Y - time to event X1 - treatment - fixed factor Z1 - procedure type - random factor Z2 - center - random factor Subject - repeated measures within a patient How would you analyze this kind of data using SAS 9.3/9.4 ? I was thinking about GLMM and PROC GLIMMIX, but not sure how to setup the code and more importantly, the rationale. Is this a nested design, blocked ? And one more question, perhaps harder. If you had to plan something like this, which approach would you use for the power and sample size calculations ? Thank you in advance ... View more

Hello all, I have set up an GLMM model, and I am not 100% sure I have set the right model, while on the other hand struggle to make good interpretation of some of the results. I will try to simplify the problem, and in order to do so, I attach a diagram of the design. I am comparing 2 treatments used in some sort of surgery, a standard treatment, and a new proposed treatment. The outcome of the surgery is success or failure (0/1), i.e., binary. Two departments were chosen randomly. In each department, doctors were chosen randomly to perform the surgeries. In the 1st department there were 16 doctors chosen, while in the 2nd only 10 (these are big departments). Each doctor performed between 1 to 4 surgeries. Each surgery was randomly assigned to a treatments (red - standard, blue - new). Each surgery gave an outcome of success or failure. The diagram is how I see things. Now, I have set the following model: proc glimmix data = ... method = quad;      class department doctor_id treatment;      model outcome = treatment / dist = binary oddsratio solution;      random int / solution subject = doctor_id(department); run; This syntax ran well, with no major errors. The treatment variable was statistically significant, and the model showed no over dispersion (Pearson Chi-Square / DF = 0.86). The covariance parameter estimate for the intercept of doctor(department) was 0.326. When I tried to add another random statement:  "random int / solution subject = doctor_id;", I got an error saying that the G matrix is not positive definite. I still got 0.326 for doctor_id(department), and got 4.41E-13 for doctor_id. It didn't look good, so I removed it and went back to the syntax I wrote above (didn't like the error message). I understand that 0.326 is the estimated variance of the random doctor in department intercept on the logit scale. My main questions thus are: 1. How do I know if the source of the variance is the doctors or departments ? From descriptive view, I am quite confident that it's coming from departments and not doctors, but theoretically, how can I know, if I add another random statement and get an error...by the way, trying to write the syntax different, i.e., adding a random statement for department only and not doctors, doesn't even run. 2. How do I make an interpretation of this results. Let's say that I need to explain this result to someone who doesn't know what logit scale is. How do I do it ? I read somewhere that I can calculate the inter-class correlation (ICC) using this formula: ICC = 0.326 / (0.326+3.29) = 0.0901. It makes sense, the inter-class correlation is not high, meaning that most variance is due to the different treatments and not the random effects (right ?). I am not sure if this formula is correct or not, but with or without a relation to it, how would you report the result of 0.326 so that someone with very basic statistical knowledge will understand you ? Thank you in advance for any help. ... View more

Hello all, I need your help to understand something weird in the SAS output of the logistic regression. I have an ordinal dependent variable with 3 levels: 0 - no outcome  1 - normal outcome  2 - double outcome I am trying to fit a model, I have several independent variables, some discrete and some continuous, however, my main variable of interest is discrete with not less than 5 levels! This variable called "Type" represents 5 treatments, from which the last 2 are combinations of two of the first three treatments. Before controlling for other variables, I wish to fit a simple model with this independent variable only. I ran a logistic regression with both SAS and JMP. The results were identical. My SAS code was simple: proc logistic data=A descending;   class Type_ (ref='5') ;   model Y = Type_; run; My main problem is, that the full model was not significant, however one of the dummy variables created by SAS is. And I don't understand why...I am attaching you some of my output. I attach the frequency histogram of both Y and Type(X). I attach a Mosaic plot JMP gave me, showing the two-way table (blue=no outcome, gray = normal outcome, red=double outcome - X axis is the categories of the IV), and I am attaching most of SAS's output of the logistic regression. The P value of Type=1 is significant, however the model isn't. I understand why it can be significant, you can see that Type=1 has most red area, while Type=5, which is reference, doesn't have red at all. But why isn't the model significant then ? A chi square test and Fisher exact both were NOT significant. My second question regarding this analysis is for when I added variable. Two continuous variables I added were very significant, making the entire model to be (and destroying the significant P value of Type=1). However, when I tried drawing a box-plots (reverse analysis) of the two variables vs. Y, I couldn't see any difference, I mean, the boxes were parallel, and perhaps the median was slightly different, nothing that catches the eye. How can I tell if the significant result is due to a large sample size, due to type I error, or real ? For t-test we have effect size, what shall I do here ? Thank you in advance ! ... View more