Differences between SPSS and R on probit analysis

fm1 <- glm(affected/total ~ log(dose), family=binomial(link = probit), data=finney71[finney71$dose != 0, ])

Hi,

I'm working on the effects of alternative larvicides on Aedes aegypti. Right now, I am doing a binary mortality response with a single explanatory variable (dose) on 4 concentrations of one larvicide (+ control). Our university is fond of SPSS, and I have learned to conduct the basic probit model with it, including a natural logarithm transformation on my dosis data.
Not so long ago, I've started working with R, and through a combination of the 'glm' and 'dose.p' functions, I get the same slope and intercept, as well as LD50 calculations. Nevertheless, the standard errors and Z-scores calculated through the Probit model in SPSS comes out completely different in R. Additionally, the 95% confidence intervals for the LD50 come out very differently between the two programs. I really don't have a clue on how I am getting the same slopes, intercepts and LD50's, but totally different SE, Z, and 95% CI. Can anybody help me so I can get the same results in R??

I'll pass you the script and hypothetical data:

dose <- c(6000, 4500, 3000, 1500, 0)
total <- c(100, 100, 100, 100, 100)
affected <- c(91, 82, 69, 49, 0)

finney71 <- data.frame(dose, total, affected)

fm1 <- glm(affected/total ~ log(dose),
family=binomial(link = probit), data=finney71[finney71$dose != 0, ])

xp1 <- dose.p(fm1, p=c(0.5,0.9))
xp.ci <- xp1 + attr(xp1, "SE") %*% matrix(qnorm(1 - 0.05/2)*c(-1,1), nrow=1)
EAUS.Aa <- exp(cbind(xp1, attr(xp1, "SE"), xp.ci[,1], xp.ci[,2]))
dimnames(EAUS.Aa)[[2]] <- c("LD", "SE", "LCL","UCL")

So, this is the regression results I get with R:
summary(fm1)

Deviance Residuals:
1 2 3 4
0.06655 -0.02814 -0.06268 0.03474

Coefficients:
Estimate Std. Error z value
(Intercept) -6.8940 10.7802 -0.640
log(dose) 0.9333 1.3441 0.694
Pr(>|z|)
(Intercept) 0.522
log(dose) 0.487

(Dispersion parameter for binomial family taken to be 1)

Null deviance: 0.513878 on 3 degrees of freedom
Residual deviance: 0.010356 on 2 degrees of freedom
AIC: 6.5458

Number of Fisher Scoring iterations: 5

And the LD50 and CI transformed:

print(EAUS.Aa)
LD SE LCL UCL
p = 0.5: 1614.444 3.207876 164.3822 15855.91
p = 0.9: 6373.473 3.764879 474.1600 85669.72

These are the values I get on SPSS (just replacing the values on R output) :

Coefficients:
Estimate Std. Error z value
(Intercept) -6.8940 1.082 -6.373
(dose) 2.149 0.311 6.918

And the LD50 and CI transformed:

LD LCL UCL
p = 0.5: 1614.444 1198.932 1953.120
p = 0.9: 6373.473 5145.767 9013.354

So, please if somebody can help me with this, I'd be grateful. If working with those functions won't do it, I'll use another, the one you recommend.

Thank you very much!


Best wishes,

Bianca



PD. I've already googled it but there's no satisfactory answer.



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