I don't know about PROC IML, but PROC GLMSELECT can use either the LASSO or the Least Angle Regression method to select variables. ... View more

You state that the value of an unstratified odds ratio should be between the values of its component stratified odds ratios.  This is not true, as the following set of 2-by-2 tables shows:                        10      10                                2          1                            12    11                          1         2                             10        10                            11    12   odds ratio         2.00                                      2.00                                  1.19 ... View more

The usual unstandardized regression coefficients estimated in PROC MIXED are not the standardized regression coefficients that you describe above, which are calculated by multiplying these usual regression coefficients by the ratio of the standard deviation of that independent variable and the standard deviation of the dependent variable.  These standardized regression coefficients are the same as the usual regression coefficients when the dependent variable and all the independent variables are standardized and used in a linear regression.  The largest reduction in R-squared does NOT necessarily occur with the removal of the independent variable with the largest standardized regression coefficient.  Standardized regression coefficients have also been criticized for other reasons (see the reference, Bring J.  How to standardize regression coefficients.  The American Statistician 1994 Aug;48(3):209-213). PROC MIXED does not include an R-squared measure in its output because this procedure is usually used with random-effect models or with mixed models rather than fixed-effect models where PROC GLM and PROC REG would be more appropriate.  However, since PROC MIXED can write raw residuals to an output data set using the OUTP option of the model statement, which can be squared and summarized to form one part of the R-squared statistic.  The sum of the squared difference between the observed dependent variable for each observation and the mean dependent variable across all observations forms the second part of the R-squared statistic.  Thus, the R-squared statistic can be calculated from the output of PROC MIXED. The reference cited above describes how the t-statistic for a given independent variable (obtained by specifying the SOLUTION option in the PROC MIXED MODEL statement) is related to the increment in the R-squared statistic obtained by including that variable as the last independent variable in a model.  Thus, comparing the t-statistics among independent variables in the same model is the same as comparing the decreases in the R-squared statisitc by excluding each of these variables from the model.  Unfortunately, this comparison of independent variables yields an incremental R-squared, not exactly what you're interested in; but this comparison does emphasize  that an independent variable's impact on the variability of the dependent variable is NOT unique but does depend on the other independent variables in the model. ... View more

In your previous e-mails, you described proportions like 0.9 and 0.7 for the treatment and the control groups respectively as the proportions of SUCCESSES over a specific follow-up time, not the proportion of FAILURES over that time.  I would thus characterize the hazard rate of (new) FAILURES for the control group as 0.30/sum of follow-up time [=1.00-0.70] and the hazard rate of (new) FAILURES for the treatment group as 0.10/sum of follow-up time [=1.00-0.90].  The hazard ratio is the ratio of the hazard rate for the treatment group relative to the hazard rate for the control group:       hazard ratio = (0.10/sum of follow-up time) / (0.30/sum of follow-up time)  = 0.333333.... With respect to median survival time for a group, you can get that information from the published literature on previous clinical trials using the treatments of interest, or you can estimate it as the time when half of the group has "survived" [=not yet suffered a new failure] or its equivalent, the time when half of the group has "failed".  Since the form of your survival curves is exponential, you can estimate these median survival times from the proportion surviving over the follow-up time.  Using your example:         How long an interval would it be when the probability of survival = 0.5, given that probability of survival for treated group = 0.9           at 4 weeks/months/years?                         0.5 = (0.9)^x  -->   ln(0.5) = x ln(0.9)   --->   x = ln(0.5)/ln(0.9)  --> 6.58.         Thus, the estimated median survival time equals 6.58 times the duration it took to reach the probability of survival of 0.9 [=4 weeks/months/years] --> 26.3 weeks/months/years. You can estimate the median survival time for the control group in a similar way. ... View more

The number of time subintervals is used in the calculations of sample size when the pattern of follow-up times is complex.  This is not the situation in your example because both survival curves have exponential distributions (constant hazards).  If you change the number of these subintervals in your example using the option, NSUBINTERVALS, the estimated sample size or the statististical power does not markedly change.  The default number of subintervals does not refer to a specific unit of time, whether weeks, months, or years. Although you have specified a one-sided statistical test assuming that the new treatment is better than the standard comparison treatment, the FDA may require you to calculate the statistical power or the sample size based on a two-sided test because a clinical trial wouldn't be needed if you knew that the new treatment was always better than the standard treatment.  The new treatment may have unexpected side effects that make it hard for the patient to continue taking it and may in fact be worse than the standard treatment, In this reply, I'd also like to correct an error I made in a previous reply in defining what a hazard ratio is.  A hazard ratio is the ratio of the rate of new failures in a new treatment group to the rate of new failures in a standard comparison group.  The ratio is NOT of proportions but of rates that have a denominator of person-time (of follow-up):   Rate = (number of new failures in a group) / (sum of person-time follow-up across all members of the group). Rates and follow-up time are relevant to follow-up studies like clinical trials because the expected number of new failures used in calculating sample sizes would increase with increasing duration of follow-up.  The hazard ratio also refers to NEW (incident) failures and excludes already existing (prevalent) failures.  In your specification of characteristics for calculating statistical power in PROC POWER, these distinctions don't matter because these specifications imply similar follow-up in both the new treatment group and the standard treatment group:   Both groups are followed for the same duration, no one from either group is lost to follow-up or withdraws from follow-up, and all failures are ascertained equally in both groups. ... View more

The hazard ratio is the ratio of the proportion of failures in the treatment group to the proportion of failures in the comparison group:     0.15/0.25  = 0.60. The accrual time is the time needed to recruit study subjects for the trial if all those needed are not available at the start of the trial.  The follow-up time is the time that the trial study subjects are followed after the start of the trial either to the end of the trial, to the time that they fail, to the time they are lost-to-follow-up, or to the time they withdraw from the trial.  The total time is the sum of the accrual time and the follow-up time. One SAS code specification to calculate the sample size in each group for the two-group six-month trial you propose above with a two-sided significance level, alpha [=0.05], and a statistical power of 0.80 is the following:      proc power;          twosamplesurvival              test=logrank              alpha=0.05              sides=2              curve("Standard")=6 : 0.75              curve("Proposed")=6 : 0.85               groupsurvival="Standard" | "Proposed"               followuptime=6               totaltime=6               power=0.80               npergroup=.; run; This calculation assumes 100% follow-up and ascertainment of all relevant outcomes. In a multi-center study where the study subjects do not differ across the centers, the follow-up time and the total time are the same, and the outcome results will be aggregated across the centers, the sample size calculations would be the same. ... View more

The SAS method for the two-sample survival analysis using the log-rank method is described in the PROC POWER documentation under the computational methods for the TWOSAMPLESURVIVAL statement.  This documentation cites a 1988 reference by Lakatos, et al. and a 1997 SAS book by Cantor. Does the other tool you use cite a reference that is the source of its formula?  Unless it does, the other tool may be problematic. From what I've been able to read, the discrepancy between the sample sizes you've calculated is much larger that what would be expected.  Until you can compare these formulas and the assumptions underlying these formulas, I'd be careful in choosing one over another. ... View more

PROC POWER in SAS version 9.2 or later has an option, TWOSAMPLESURVIVAL, that should be able to calculate sample size for you.  The documentation for PROC POWER includes an example for comparing two survival curves. I split the observation time into six single monthly intervals for the standard treatment group and six single monthly intervals for the new treatment group.  Then I calculated the estimated failure rates at each of these monthly intervals based on overall values of p1=0.10 for the standard treatment group (decrements from 1.00 to 0.90 over 6 intervals = 0.10/6 = 0.01667 per month) and of p2=0.02 for the new treatment group (decrements form 1.00 to 0.98 over 6 intervals = 0.02/6 = 0.00333 per month).  The followup time = 6, and the total time = 6 (assuming that all the patients would be available in both groups at the beginning of the study).  I used the LOGRANK test that assumes that failures occur relatively evenly across follow-up (TEST=LOGRANK); a one-sided test (SIDES=1); a statistical significance level, alpha, of 0.05 [the default]; and a statistical power of at least 0.80.  The sample size for each group was estimated as 110, which is smaller than 124, but still not as small as the required 100 per group. Note that the monthly decrements are probably unrealistic, especially for the new treatment group, because failures/adverse effects will probably not occur at the monthly follow-up examinations.  If some of the patients are not available at the start of the study, you may have to use an accrual time instead of just a follow-up time as above.  Changes in either of these options probably will affect the sample size estimates.  Finally, multiple interim follow-up examinations may add more to the study cost than adding more patients above 100 per group.  That's why you should provide the study investigators and the funding group multiple scenarios for them to choose from. ... View more

I don't think you can get what you want. Even with the R-squared statistic in linear regression, the proportion of the variance in the dependent variable accounted for by a specific independent variable depends on the sample being used, on other independent variables in the model, and on how the model is specified.  Several attempts have been made to develop measures of variable importance to compare variables.  You should probably Google and read references to this concept of variable importance to see the subtleties and the potential pitfalls in trying to obtain what you are seeking. ... View more

You did not specify your MODEL statement for PROC LOGISTIC. If you specify one effect as nested within another, this implies that the first effect does NOT have an independent ("main") effect on the dependent variable:      model y = Type + Speed(Type); or, equivalently ,      model y = Type + Speed*Type; since SAS parameterizes nested effects as crossed or interaction effects:    Speed(Type)  -->  Speed*Type. Because Speed is nested within Type, Speed is assumed to have no independent effect on Y separate from that through its association with Type. However, if you specify one effect as interacting with another effect, this implies that both effects have independent ("main") effects on the dependent variable separate from their interaction effect on the dependent variable:      model y = Type + Speed + Speed*Type; If the interaction effect, Speed*Type, is statistically significant, you should keep the main effects of Type and Speed even if any one or both of these main effects is not statistically significant (the so-called "hierarchy principle").   I agree with Steve Denham's advice about stepwise variable selection in regression. ... View more

I don't know the FDA's rules, but FDA probably won't allow you to increase your statistical significance level, alpha, to exceed 0.05 because that's the usually accepted standard.  If the financing body will pay for only 100 study subjects per group with your original study design, keep alpha [=0.05, two-sided] and the number of study subjects per group [=100] constant and vary either the proportion expected to fail in the treated group [=p2, as you did above], the proportion expected to fail in the standard care group [=p1], the statistical power [=1.00 - beta], or all three to provide different scenarios to the inventor of the new treatment and the financing body at a prespecified alpha and number of study subjects per group.  Let them decide which scenario based on these criteria is most appropriate. ... View more