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#1
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Weibull function in Excel
I have a survival dataset (see below). I want to fit to Weibull function
S(t)=EXP(-alpha*Time^gamma). If I use Excel Solver, alpha=0.00367 and gamma=2.32. If I use a method tranforming probability to Ln(Ln(1/S(t)) which will be a linear function of Ln (t). But I got a different alpha and gamma. When I compared two alphas adn gammas, the Solver results had a better goodness of fit. What is the problem? THanks Month Probability of survival 0 1 1 1 2 0.92 3 0.90 4 0.88 5 0.83 6 0.76 7 0.74 8 0.69 9 0.57 10 0.48 11 0.39 12 0.27 13 0.27 14 0.13 |
#3
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Weibull function in Excel
Hi Mike, thanks for the reply. So if I want to find a function with best fit,
adding a trendline is not an proper method. I got the weibull function estimates from someone else and then try to replicate the results in Excel. When I used Solver, most of the time I can get the parameter estimates with the best fit. I was told there is no feasible solution. How to properly use Solver? Thanks Barbo "Mike Middleton" wrote: Barbo - What is the problem? There is no problem. You should expect a difference. Your Solver method (which I prefer) minimizes sum of squared deviations between actual S and fitted S. The other method uses transformed values, so it does not yield a better fit. (Excel's trendline features use transformations to fit the logarithmic, power, and exponential functions. The approach appears to be a computational convenience. Better fits are obtained using Solver.) - Mike http://www.MikeMiddleton.com "Barbo" wrote in message ... I have a survival dataset (see below). I want to fit to Weibull function S(t)=EXP(-alpha*Time^gamma). If I use Excel Solver, alpha=0.00367 and gamma=2.32. If I use a method tranforming probability to Ln(Ln(1/S(t)) which will be a linear function of Ln (t). But I got a different alpha and gamma. When I compared two alphas adn gammas, the Solver results had a better goodness of fit. What is the problem? THanks Month Probability of survival 0 1 1 1 2 0.92 3 0.90 4 0.88 5 0.83 6 0.76 7 0.74 8 0.69 9 0.57 10 0.48 11 0.39 12 0.27 13 0.27 14 0.13 . |
#4
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Weibull function in Excel
Barbo -
How to properly use Solver? Solver's success with nonlinear functions may depend on the initial values for the changing cells. For simple functions, like the Weibull, you may have some idea of reasonable initial values. - Mike http://www.MikeMiddleton.com "Barbo" wrote in message ... Hi Mike, thanks for the reply. So if I want to find a function with best fit, adding a trendline is not an proper method. I got the weibull function estimates from someone else and then try to replicate the results in Excel. When I used Solver, most of the time I can get the parameter estimates with the best fit. I was told there is no feasible solution. How to properly use Solver? Thanks Barbo "Mike Middleton" wrote: Barbo - What is the problem? There is no problem. You should expect a difference. Your Solver method (which I prefer) minimizes sum of squared deviations between actual S and fitted S. The other method uses transformed values, so it does not yield a better fit. (Excel's trendline features use transformations to fit the logarithmic, power, and exponential functions. The approach appears to be a computational convenience. Better fits are obtained using Solver.) - Mike http://www.MikeMiddleton.com "Barbo" wrote in message ... I have a survival dataset (see below). I want to fit to Weibull function S(t)=EXP(-alpha*Time^gamma). If I use Excel Solver, alpha=0.00367 and gamma=2.32. If I use a method tranforming probability to Ln(Ln(1/S(t)) which will be a linear function of Ln (t). But I got a different alpha and gamma. When I compared two alphas adn gammas, the Solver results had a better goodness of fit. What is the problem? THanks Month Probability of survival 0 1 1 1 2 0.92 3 0.90 4 0.88 5 0.83 6 0.76 7 0.74 8 0.69 9 0.57 10 0.48 11 0.39 12 0.27 13 0.27 14 0.13 . |
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