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08-26-2016 06:39 AM

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 |

Accepted Solutions

Solution

08-30-2016
03:28 PM

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Posted in reply to jacksonan123

08-30-2016 02:15 PM

So what if ln(CLnr) is negative? There is no requirement that a dependent variable be positive. If you are truly worried, rescale the dependent variable--say by multiplying by 1000. Then the ln values will also be positive. Since I suppose that CLnr is clearance of some metabolite, it is like shifting the measurement from millimolar to micromolar.

Steve Denham

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Solution

08-30-2016
03:28 PM

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Posted in reply to jacksonan123

08-30-2016 02:15 PM

So what if ln(CLnr) is negative? There is no requirement that a dependent variable be positive. If you are truly worried, rescale the dependent variable--say by multiplying by 1000. Then the ln values will also be positive. Since I suppose that CLnr is clearance of some metabolite, it is like shifting the measurement from millimolar to micromolar.

Steve Denham

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Posted in reply to SteveDenham

08-30-2016 03:31 PM

Thanks for the response and it addresses my issue.

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Posted in reply to jacksonan123

08-30-2016 02:29 PM

Because your response data are positive, the log transformed response model is

log(Y) = a + b*x

or

Y = exp(a + b*x) where a is the estimate for the intercept and b is the estimate for the explanatory coefficient.

If you define A=exp(a), you get Y = A*exp(b*x)

So the intercept term multiplies the model: a unit change in the intercept estimate results in a mulitplicative change (a factor of e) in the predicted response.

You can do the same computation for other models. If you add c to the response before you fit the model, then

Y = -c + A*exp(b*x)

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Posted in reply to Rick_SAS

08-30-2016 03:30 PM

Thanks for the response.