Weibull function in Excel

[Barbo]

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

 

[Mike Middleton]

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
(e-mail address removed)

 

[Barbo]

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]

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
(e-mail address removed)