I conducted a BOX-COX analysis using the following code: ods rtf file='/folders/myfolders/regr4BCoxNresuls/CLnrGFR1.rtf' style=journal; ods TRACE ON;   ods output FitStatistics=FitStat  ParameterEstimates=param; proc transreg data=boot2 plots=all; /*plots(only)=FITPLOT(stats=none); */ model boxcox(CLnr/convenient lambda=-3 to 3 by 0.125)=identity(GFR1)/cl; /*SPEC;*/ output out=pred   ; proc print data=fitstat;run; proc print data=param;run; proc print data=pred;run; run; quit;   ods rtf close;   The best lamda was =0 which means that the log transformation for CLnr was best.  The data is posted below (Nstudyr4.csv)set which if I use log transformation for CLnr would give negative values.  On line it was suggested that I can do log transformation  if I add a constant to each value.  My question is how do I interpret the value for the intercept if I do a regression when I take the antilog with the added constant? I have attached the data set and the BOX-Cox output. Subject CL CLr CLnr GFR1  GFR2 1 2.02 1.73 0.29 125.5 124.1 2 2.18 1.63 0.55 114.3 113.9 3 1.95 1.28 0.67 102.2 102 4 1.83 1.16 0.67 101.6 100.7 5 1.64 1.47 0.17 96.6 95.6 6 1.94 1.72 0.22 90.7 89.2 7 1.98 1.47 0.51 87.8 86.2 8 1.67 1.07 0.6 84.5 83.6 9 1.25 1.19 0.06 84.2 83.4 10 0.9 0.74 0.16 77.7 76.5 11 1.53 1.06 0.47 75.6 75.2 12 1.52 0.82 0.7 68 67.5 13 1.58 1.23 0.35 66.5 65.3 14 0.93 0.75 0.18 64.4 64.2 15 0.93 0.95 . 60 59.6 17 0.61 0.48 1.2 52.3 52.5 18 0.76 0.46 0.15 52 52.3 19 0.57 0.44 0.32 51.7 52.3 20 0.66 0.4 0.17 49.3 49.7 21 0.51 0.15 0.51 37.5 37.7 22 0.41 0.32 0.19 35.5 35.7 23 0.65 0.57 . 34.3 34.4 24 0.9 0.58 0.07 31 31.4 25 0.5 0.29 0.61 28.1 29.3 26 0.31 0.22 0.28 26.6 26.8 27 0.3 0.21 0.1 24.8 24.7 28 0.48 0.38 . 18 18.6 29 0.42 0.19 0.29 16.3 16.6 30 0.31 0.15 0.27 16 16.6 31 0.2 0.12 0.19 12.7 12.7 32 0.18 0.2 . 12.7 12.7 33 0.25 0.18 . 12.7 12.7 34 0.35 0.13 0.12 10.3 12.7 35 0.2 0.12 0.23 9.8 12.7 36 0.15 0.08 0.12 8 11.5 37 0.15 0.21 . 7.7 10.2 38 0.11 0.092 0.058 6.5 10.2 39 0.046 0.092 0.018 4.7 8.9 40 0.076 0.061 . 4.1 6.4 41 0.14 0.034 0.042 4.1 3.8 42 0.16 0.018 0.122 4.1 3.8 43 0.18 0.015 0.145 1.8 1.5 44 0.15 0.018 0.162 1.5 1.5 ... View more

I have data for which I did a regression and the White test for constant variance had a p value =0.0016 indicating heterosdedasticity of the variance.  See attached graph for residuals vs predicted value.  The data was normally distributed see atached distribution graph which had a p=0.79 for the Shapiro Wilk test.  These results indicate that I need to do a weighted regression.   proc reg;              /* weighted linear regression */    model y = x;    weight w;   In the literature I read, "  If however we know the noise variance σ 2 i at each measurement i, and set wi = 1/σ2 i , we get the heteroskedastic MLE, and recover efficiency."   My question is how do we know this weight value and based upon my data what would be an appropriate weight i.e., 1/y or something else? ... View more

I have the following code which runs fine:     ods rtf file='/folders/myfolders/regr1resuls/CLCLr.rtf' style=journal; ods TRACE ON; ods output FitStatistics=FitStat ParameterEstimates=param; proc reg data=boot1 plots=all; /*plots(only)=FITPLOT(stats=none);*/ model CL=CLr /clb SPEC; output out=pred predicted=p lcl=lcl lclm=lclm ucl=ucl uclm=uclm residual=r; proc print data=fitstat; run; proc print data=param; run; proc print data=pred; run; run; quit; ods rtf close;   The problem is that the outputted graphs are truncated (i.e., the X axis is not visible).  I have attached an output for your perusal. Can someone tell me why this occurs and how to correct it?   ... View more

  I am using the following code within a macro to export files in SAS University and the resulting file is a word document. proc export data=normal&i outfile="folders/mufolders/regr/nnormal&i" dbms=csv replace;run;   How can I edit this code to have the output be a *.csv file?     ... View more