search for: males2a

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2006 Sep 21
1
survival function with a Weibull dist
...of this function should be two: lambda = z (presumably "Value" in R for each treatment) alpha = k (scale in R) Survival function (S): S(t)= exp-(zt)^k I don't quite understand how to use the intercept. First I thought adding it to each other treatment value, for example: Survival Males2a (t) = exp ? ((intercept + 0) * t) ^ k Survival Males2b (t) = exp ? ((intercept + zb)* t) ^ k But the curves I get using this interpretation are not similar to my data Does anyone use this function and could help me to interpret the results? many thanks Anaid