MrAnon9 (02-08-2012)
Sorry for any confusion I brought - I realize now it might have sounded like a suggestion. I was questioning noetsi because I don't think it's a good idea.
Since you have multiple observations at each point you could try calculating the logit of the estimated probabilities and then plotting those against dose. If logistic regression is a good fit then these should be approximately linear.
MrAnon9 (02-08-2012)
"Very few theories have been abandoned because they were found to be invalid on the basis of empirical evidence...." Spanos, 1995
If there aren't that many observations though it might make sense to smooth the estimated probabilities using some sort bayesian estimator. Something like (x+1)/(n+2) or (x+.5)/(n+1) would probably work well enough.
Hi again, thanks for the help. I have done two plots from the data..
Are they relevant plots for deciding if the logistic model is suitable and if so, what can I deduce from the two graphs?
I think for plot 2, it needs to be a straight line relationship? Is this correct and what can I conclude from the first plot?
Should I plot against dose or log dose, does it matter?
Thanks in advance guys.
Why are you plotting the log of the estimated probability in the second plot? We would want a plot of the logit of the estimated probability to analyze if a logistic seems alright.
MrAnon9 (02-09-2012)
I don't get it
According to the theory, g(proportion)=Xb where g(proportion) is the logit function. Therefore, if your data is a case for logistic function then the logit of the proportion will be linear with predictors i.e. dose. Therefore, Dason asked you to check how does the logit of the proportion(i.e. log(p/1-p) fare when plotted against dose.
MrAnon9 (02-09-2012)
But what is p then? The actual logistic function? That would mean I would have to calculate the coefficients..?
Logit(x) = Log(x/(1-x)) where Log is the natural log. In your case you would stick the estimated probabilities in for x. So if you had 6 success out of 8 trials you would calculate x= 6/8 = .75. So Logit(.75) = Log(.75/.25)
MrAnon9 (02-09-2012)
Thanks for the quick response. I need to do this by tomorrow :P
Is that not what I did though? Since log ( ri/ni / 1- ri/ni ) = log (ri / (ni-ri) ) right?
where ri is number of beetles killed for each dose and ni is the total for each dose, hence ni-ri is the survival
plotted graphs wrong anyway, but still not sure if what i plotted is right (previous reply)
You would probably want to plot dose on the x axis (instead of log(dose)) but yeah - I didn't put as much thought into it as I probably should have - seems like what you have should work.
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