I am using SAS for producing ROC curves. But the "PROC LOGISTIC" does not give me the confidence-interval for sensitivity and specificity.
Does any one know if there is an option in order to produce the lower and upper band for sensitivity and specificity ?
If it is not the case, does anyone know another method ?
Thk an lot,
when I use basic stats, I use proc freq for associations.
proc freq data=tempds noprint;
tables variable1*std_variable2 / chisq measures;
output out=outds pchi n OR FISHER;
run;
The output dataset "outds" now contains RROR(OR), L_RROR(Lower CI), U_RROR(Upper CI). Is this what you are looking for?
If proc logistic doesn't directly support this, you could try bootstrapping - produce many ROC plots for random samples of your data (e.g. using proc surveyselect) and then calculate the p5 and p95 points for each x and y value in the plot using proc summary. This should give a good approximation provided that you use a large enough number of samples.
Related
I am analyzing a temporal trend(yr) of certain chemicals(a b & c).
I use proc sgplot and series statement to draw a plot and found there was a decreasing trend.
Becuase the data is right-skewed, I used the median concentration of each year to draw the plot.
Now I would like to conduct a statistical test on the trend. My data came from the NHANES and need to use the proc survey** to perform analysis. I know I can do an ANOVA test based on proc surveyreg and use ANOVA option in the MODELstatement.
proc suveyreg data=a;
stratum stra;
cluster clus;
weight wt;
model a=yr/anova;
run;
But since the original data is right-skewed, I think maybe it is better to use Kruskal-Wallis test on the original data. But I don't know how to write a code in SAS and I didn't find information in proc survey**-related document.
My plan B is to use the log-transformed data and ANOVA test. But I am not sure if that is an appropriate approach. Can somebody tell me how to get the normality test of the residual in ANOVA while using proc surveyreg? I would also like to know if I can test a b & c in one procedure or I should write multiple procedures with changes in MODEL statement.
Looking forward to your engagement.Thank you!
I am doing a logistic regression of a binary dependent variable on a four-value multinomial (categorical) independent variable. Somebody suggested to me that it was better to put the independent variable in as multinomial rather than as three binary variables, even though SAS seems to treat the multinomial as if it is three binaries. THeir reason was that, if given a multinomial, SAS would report std errors and confidence intervals for the three binary variables 'relative to the omitted variable', whereas if given three binaries it would report them 'relative to all cases where the variable was zero'.
When I do the regression both ways and compare, I see that nearly all results are the same, including fit statistics, Odds Ratio estimates and confidence intervals for odds ratios. But the coefficient estimates and conf intervals for those differ between the two.
From my reading of the underlying theory,as presented in Hosmer and Lemeshow's 'Applied Logistic Regression', the estimates and conf intervals reported by SAS for the coefficients are consistent with the theory for the regression using three binary independent variables, but not for the one using a 4-value multinomial.
I think the difference may have something to do with SAS's choice of 'design variables', as for the binary regression the values are 0 and 1, whereas for the multinomial they are -1 and 1. But I don't really understand what SAS is doing there.
Does anybody know how SAS's approach differs between the two regressions, and/or can explain the differences in the outputs?
Here is a link to the SAS output:
SAS output
And here is the SAS code:
proc logistic data=tab descending;
class binB binC binD / descending;
model y = binD binC binB ;
run;
proc logistic data=tab descending;
class multi / descending;
model y = multi;
run;
My goal is to fit a data to any distribution which has positive support. (weibull(2p), gamma(2p), pareto(2p), lognormal (2p),exponential(1P)). First attempt,i used proc univariate.This is my code
proc univariate data=fit plot outtable=table;
var week1;
histogram / exp gamma lognormal weibull pareto;
inset n mean(5.3) std='Standar Deviasi'(5.3)
/ pos = ne header = 'Summary Statistics';
axis1 label=(a=90 r=0);
run;
The first thing i noticed, there's no kolmogorov statistic shown for weibull distribution.Then i used proc severity instead.
proc severity data=fit print=all plots(histogram kernel)=all;
loss week1;
dist exp pareto gamma logn weibull;
run;
Now, i got the KS statistic for weibull distribution.
Then i compared KS statistic produced by proc severity and proc univariate. They're different. Why? Which one should i use?
I do not have access to SAS/ETS so cannot confirm this with proc severity, but I imagine that the difference you are seeing come down to the way the distribution parameters are fitted.
With your proc univriate code you are not requesting estimation for several of the parameters (some are in some cases set to 1 or 0 by default, see sigma and theta in the user guide). For example:
data have;
do i = 1 to 1000;
x = rand("weibull", 5, 5);
output;
end;
run;
ods graphics on;
proc univariate data = have;
var x;
/* Request maximum liklihood estimate of scale and threshold parameters */
histogram / weibull(theta = EST sigma = EST);
/* Request maximum liklihood estimate of scale parameter and 0 as threshold */
histogram / weibull;
run;
You will note that when an estimate of theta is requested SAS also produces the KS statistic, this is due to the way that SAS estimates the fit statistic requiring know distribution parameters (full explanation here).
My guess is that you are seeing different fit statistics between the two procedures because either they are returning slightly different fits, or they use different calculations for the estimation of fit statistics. If you are interested you can investigate how they perform their parameter estimation in the user guide (proc severity and proc univariate). If you wanted to investigate further you could force the distribution parameters to match in both procedures and then compare the fit statistics to see how far they differ.
I would recommend that if possible you use only one of the procedure, and that you select the one that best fits your needs in terms of output.
I'm not so familiar with SAS proc glm. All I have done using proc glm so far is to output parameter estimates and predicted values on training datasets. But I also need to use the fitted model to make prediction on testing dataset. (both point estimates and interval estimates)
Here is my code.
ods output ParameterEstimates=Pi_Parameters FitStatistics=Pi_Summary PredictedValues=Pi_Fitted;
proc glm data=Train_Pi;
class Area Fo5 Tye M0 M1 M2 M3;
model Pi = Dow Area Fo5 Tye M0|HC M1|HC M2|HC M3|HC/solution p ss3 /*tolerance*/;
run;
But how to proceed to next step? something like predict(Model_from_Train_Pi,Test_Pi)
If you're on SAS 9.4 see Jake's answer from this question:
How to predict probability in logistic regression in SAS?
If not on 9.4, my answer applies for adding the data in to the original data set.
A third option is PROC SCORE - documentation has an example for proc reg that's almost identical to your question:
http://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_score_sect018.htm
is there a way to detect an outlier from proc means while calculating min max Q1 and Q3?
the box plot procedure is not working on my SAS and I am trying to perform a boxplt in excel with the values from SAS.
Assuming you have a specific definition for what an outlier is, PROC UNIVARIATE can calculate the value that appears at that percentile using the PCTLPTS keyword on the OUTPUT statement. It also will identify extreme observations individually, so you can see the top few observations (if you have few enough observations that the number of extremes is likely to be <= 5).
The paper A SAS Application to Identify and Evaluate Outliers goes over a few of the ways you can look at outliers, including box plots and PROC UNIVARIATE, and includes some regression-based approaches as well.
If you want a 'standard boxplot' use the outbox= option in SAS to create the standard data set used for a box plot.
proc boxplot data=sashelp.class;
plot age*sex / outbox = xyz;
run;