# Simple Logistic Regression

#### jamesmartinn

##### Member
Five different doses of insecticide were applied under standardized conditions to an insect species. The data are as followed:

Dose (mg/l): 2.6 3.8 5.1 7.7 10.2
Number of insects: 60 60 59 57 60
Number killed: 7 16 20 48 54

I was asked to build a logistic regression model which says the logit of the chance of death changes linearly with the natural logarithm of dose.

I did this in SAS.

I'm asked to give a 95% Likelihood Ratio Confidence Interval for B. Further, translate this interval into an interval for the effect on the odds of death of increasing the dose by 50% (i.e, multiplying the dose factor by 1.5) and interpret. Hint: First translate the multiplying dose factor to the natural log scale.

I'm not sure how to do this, I've attached the appropriate SAS output I think I need to do the question with. I'm thinking take the B estimate (logc in the output is my B estimate) and exponentiate it, then times that value by 1.5. If thats right, what do i do next? same thing to the end points of the confidence interval associated with B?

Thx

#### YaoPau

##### New Member
This is ahead of what I've learned so far, but partially as my trying to help you and partially as my trying to learn myself, here's my stab at it.

I can't figure out what your SAS output means, as I'm getting different numbers. Using a Log regression of ln(Dose) to %Killed, I get %Killed=.0298*4.6535^(ln(Dose)), with a .9645 r^2 and a .1856 standard error.

Using a Linear regression of ln(Dose) to logit(%Killed), I get logit(%Killed)=-5.367+3.26*ln(Dose), with a .956 r^2 and a .4382 standard error.

I'm also not sure what B means. If it's beta coefficient of a standardized regression, I've got .97922 +/- 1.96*.2413. If it's the unstandardized slope, it's 3.26 +/- 1.96*.438205.

So basically, if you increase a dose by 50%, here's how it would play into the Linear regression formula two paragraphs above:

(1) logit(%Killed)=-5.367+3.26*Ln(Dose)

(2) %Killed=exp(-5.367+3.26*Ln(Dose))/(1+exp(-5.367+3.26*Ln(Dose))

(3) x*%Killed=exp(-5.367+3.26*Ln(1.5*Dose))/(1+exp(-5.367+3.26*Ln(1.5*Dose))

(4) x*%Killed=exp(-5.367+1.3218+3.26*Ln(Dose))/(1+exp(-5.367+1.3218+3.26*Ln(Dose))

(5) x=(exp(-5.367+1.3218+3.26*Ln(Dose))/(1+exp(-5.367+1.3218+3.26*Ln(Dose)))/(exp(-5.367+3.26*Ln(Dose))/(1+exp(-5.367+3.26*Ln(Dose)))

And that's where I can't go further, so breaking that down is up to you haha. Hope something there helps/makes sense?

#### jamesmartinn

##### Member
I fitted:

Logit (death/total) = a + b*LN_Dose

I did a natural log transformation for the Dose variable before fitting this.