Re: 2 obvious

*To*: mathgroup at smc.vnet.net*Subject*: [mg115231] Re: 2 obvious*From*: Darren Glosemeyer <darreng at wolfram.com>*Date*: Tue, 4 Jan 2011 18:50:01 -0500 (EST)

On 12/23/2010 2:55 AM, Francisco Gutierrez wrote: > Dear Group: > Ok, here comes a really silly question. After v. 7, the statistical capacities of Mathematica have been substantially boosted. However, I haven't been able to interpret the simple command LogitModelFit. THe problem is that the documentation only offers examples with two independent variables. I have not managed to find how can n more independent variables can be inserted so that the command works. Can somebody send me a simple and clear example? > > Let this be a pretext to thank all the incredibly useful help the members of this list have generously provided to me and others. Happy holidays, > Fg > > > > Hi Francisco, The data argument is like for other fitting functions. The independent (predictor) variables are the first n-1 elements in each data point and the last element in the data point is a dependent (response) variable. Here is a list of 5 data points each having two predictors and one response. In[1]:= data = {{10, 4, 0.26}, {8, 3, 0.04}, {2, 0, 0.17}, {4, 8, 0.09}, {9, 4, 0.83}}; This treats the predictors as x and y in the fitting In[2]:= lm = LogitModelFit[data, {x, y}, {x, y}]; and here we get the fitted function and a table of parameter information for the fitting. In[3]:= lm[{"BestFit", "ParameterTable"}] 1 Out[3]= {-------------------------------------, 2.384 - 0.236963 x + 0.0664776 y 1 + E > Estimate Standard Error z\[Hyphen]Statistic P\[Hyphen]Value} 1 -2.384 3.41536 -0.698022 0.485163 x 0.236963 0.388785 0.609496 0.542196 y -0.0664776 0.529535 -0.125539 0.900097 I hope this helps. Darren Glosemeyer Wolfram Research