09-19-2024
Kai123
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Latest posts by Kai123
Subject Views Posted 2324 09-11-2024 04:33 PM 2357 09-11-2024 01:48 PM 2408 09-11-2024 10:04 AM 2439 09-11-2024 09:16 AM 2510 09-10-2024 04:33 PM 2586 09-09-2024 04:41 PM 2634 09-09-2024 02:22 PM -
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- Posted Re: What CV should be entered for lognormal data in Power Procedure, PairedMeans? on Statistical Procedures. 09-11-2024 04:33 PM
- Posted Re: What CV should be entered for lognormal data in Power Procedure, PairedMeans? on Statistical Procedures. 09-11-2024 01:48 PM
- Posted Re: What CV should be entered for lognormal data in Power Procedure, PairedMeans? on Statistical Procedures. 09-11-2024 10:04 AM
- Posted Re: What CV should be entered for lognormal data in Power Procedure, PairedMeans? on Statistical Procedures. 09-11-2024 09:16 AM
- Posted Re: What CV should be entered for lognormal data in Power Procedure, PairedMeans? on Statistical Procedures. 09-10-2024 04:33 PM
- Posted Re: What CV should be entered for lognormal data in Power Procedure, PairedMeans? on Statistical Procedures. 09-09-2024 04:41 PM
- Posted What CV should be entered for lognormal data in Power Procedure, PairedMeans? on Statistical Procedures. 09-09-2024 02:22 PM
- Tagged What CV should be entered for lognormal data in Power Procedure, PairedMeans? on Statistical Procedures. 09-09-2024 02:22 PM
- Tagged What CV should be entered for lognormal data in Power Procedure, PairedMeans? on Statistical Procedures. 09-09-2024 02:22 PM
09-11-2024
04:33 PM
This is not due to random error, the two CVs are two different things. Look at the dataset below, %CVs are 80 and 113, very different. There is no proof that the two values should be similar. I don't think you understand my question. If you are a staff in SAS, could you please refer me to the right team? Thanks. Original Scale Ln scale Original Scale (Assuming Normal) Original Scale (LogNormal) Ln scale 0.182934103 -1.698629283 n 47 47 0.335884261 -1.090988639 1.768862198 0.570336514 GM 0.921088052 0.322330226 -1.132178713 Mean 1.300955496 -0.0822 1.869962942 0.625918614 GSD 2.482316779 4.092048456 1.40904569 SD 1.039364637 0.909192 1.615234314 0.479480032 %CV 80 113 0.586349193 -0.533839775 0.854526206 -0.157208108 0.368046246 -0.999546681 1.92127702 0.65299008 0.694299929 -0.364851237 0.062064248 -2.779585165 0.522980221 -0.648211634 1.603548144 0.472218764 0.495035732 -0.703125334 3.263487488 1.182796405 0.991658216 -0.008376771 0.807604862 -0.213682372 2.166518473 0.773121489 0.367439171 -1.001197496 0.796960379 -0.226950314 0.437185347 -0.827398039 1.649246645 0.500318605 0.28163224 -1.267153174 3.533515975 1.262293402 1.850595694 0.615507584 0.50108275 -0.690984021 0.578666907 -0.547028257 2.031517886 0.708783241 1.004733644 0.004722476 1.748168892 0.558568893 0.727650154 -0.317934903 1.325797122 0.28201388 0.242380432 -1.417246756 0.843038824 -0.170742268 2.131842779 0.75698676 1.011822918 0.011753574 4.396706363 1.480855707 2.611930193 0.960089486 0.961948844 -0.038794006 2.065023208 0.725141465 0.251275204 -1.38120651 0.792867525 -0.232099127 0.426753947 -0.851547668 2.196685175 0.786949486 1.853787599 0.617230897
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09-11-2024
01:48 PM
Thanks for taking the time to create an example. 1. x1+x2+...+xn = exp(log (x1+x2+...+xn)) , which are not equal to exp(log(x1)*log(x2)*....*log(xn)), as log (x1+x2+...+xn) is not equal to log(x1)*log(x2)*....*log(xn) 2. As you mentioned, in your example, the two CVs are "almost the same value", meaning they are different, no matter how similar they are in this particular example. For a statistical software like SAS, I do believe we should aim at an accurate result, instead of some results similar. Even a small difference could mean a lot in different industries/context. 3. The key is to understand how SAS uses this CV value at the back end to calculate the sample size/power. I'm hoping the SAS team that wrote the SAS documentation can answer my question.
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09-11-2024
10:04 AM
I don't understand what you meant by "valid". CVs calculated from these two formulas are different, both can be entered in Power Procedure, and get different results. As a SAS user, I want to know which is the right one.
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09-11-2024
09:16 AM
Yes. That's why I think CV=sqrt(exp(SDlog2)-1) should be entered in the CV option, instead of the ratio of SD and Mean of original scale data, as in SAS documentation.
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09-10-2024
04:33 PM
CV=sqrt(exp(SDlog2)-1) CV2=exp(SDlog2)-1 CV2+1= exp(SDlog2) log(CV2+1)= SDlog2 SDlog=( log(CV2+1))1/2 Note "2" is referring to square and "1/2" represents square root. This window does not allow me to type it. which is the same as in SAS documentation
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09-09-2024
04:41 PM
Thanks Jill for your reply. This reference is for correlation conversion, I don't have it now, will appreciate if you can share the part for derivation. Regarding my original question about CV, it makes more sense if CV=sqrt(exp(SDlog2)-1) where SDlog is the SD from natural Log scale data. I suspect it is a mistake in SAS documentation to indicate it as the ratio of the standard deviation to the mean on the original data scale, but I want to get the confirmation from the experts in SAS.
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09-09-2024
02:22 PM
Hi, I have a question about what CV should be entered in CV or PairCVs option in Pairmeans of Power Procedure for lognormal data. See the SAS documentation for CV below. CV=number-list specifies the coefficient of variation that is assumed to be common to both members of a pair. The coefficient of variation is defined as the ratio of the standard deviation to the mean on the original data scale. You can use this option only with DIST=LOGNORMAL. For information about specifying the number-list, see the section Specifying Value Lists in Analysis Statements. This sounds like, when data are assumed lognormally distributed, CVs entered should be STD/mean, and STD and Mean are calculated from data in the original scale. However, in SAS/STAT® 13.1 User’s Guide The POWER Procedure, Page 6445-6446 If CV=SD/Mean, where SD and Mean are calculated from data in original scale, how the σ* (SD from log-transformed data) can be derived from sqrt(log(CV2+1))? Instead, if CV is calculated from CV=sqrt(exp(SDlog2)-1) where SDlog is the SD from natural Log scale data, then it makes sense that σ* (SDlog) can be derived by back-calculating. However, the documentation clearly indicates the inputted CV is from data in original scale. This confuses me. Also have a similar question about the correlation, as the documentation states to enter "correlation of the original untransformed pairs (Y1; Y2)". Thanks,
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