title: Interpreting Weibull parameters from survreg
body: <p>I am trying to generate an inverse Weibull distribution using parameters estimated from survreg in R. By this I mean I would like to, for a given probability (which will be a random number in a small simulation model implemented in MS Excel), return the expected time to failure using my parameters. I understand the general form for the inverse Weibull distribution to be:</p>
  
  <pre><code>X=b[-ln(1-rand())]^(1/a)
  </code></pre>
  
  <p>where a and b are shape and scale parameters respectively and X is the time to failure I want. My problem is in the interpretation of the intercept and covariate parameters from survreg. I have these parameters, the unit of time is days:</p>
  
  <pre><code>             Value   Std. Error    z    p
  (Intercept)     7.79    0.2288  34.051  0.000
  Group 2        -0.139   0.2335  -0.596  0.551
  Log(scale)     0.415    0.0279  14.88   0.000
  Scale= 1.51 
  
  Weibull distribution
  Loglik(model)= -8356.7   Loglik(intercept only)= -8356.9 
  Chisq = 0.37 on 1 degrees of freedom, p= 0.55 
  Number of Newton-Raphson Iterations: 4 
  n=1682 (3 observations deleted due to missing values)
  </code></pre>
  
  <p>I have read in the help files that the coefficients from R are from the "extreme value distribution" but I'm unsure what this really means and how I get 'back to' the standard scale parameter used directly in the formulae. Using b=7.79 and a=1.51 gives nonsensical answers. I really want to be able to generate a time for both the base group and 'Group 2'. I also should note that I did not perform the analysis myself and cannot interrogate the data further.</p>