[R] Differences between SPSS and R on probit analysis

[Biank M]

Yes I got the warning message from 'glm' and ignored it, I will never do it again.
On the other hand,  I think that including the 'weights' argument improved the results ( I don't get a warning message anymore) and now they are closer to the SPSS results, thought they are not exactly the same (as I expected).

Do you have any ideas on why this is happening or do you have an alternative function combination that would give me the same SPSS results?

Thank you very much for your help!

Bianca.

________________________________
De: William Dunlap <wdunlap at tibco.com>
Enviado: viernes, 24 de febrero de 2017 14:29
Para: Biank M
Cc: r-help at r-project.org
Asunto: Re: [R] Differences between SPSS and R on probit analysis

Another model specification equivalent to
    cbind(afflicted, total-afflicted) ~ ...
is the ratio you had accompanied by the total as the 'weights' argument
    afflicted/total ~ ..., weights=total
Bill Dunlap
TIBCO Software
wdunlap tibco.com


On Fri, Feb 24, 2017 at 12:01 PM, William Dunlap <wdunlap at tibco.com> wrote:
> Did you not get a warning from glm, such as the following one?
>> fm1 <- glm(affected/total ~ log(dose), family=binomial(link = probit), data=finney71[finney71$dose != 0, ])
> Warning message:
[[elided Hotmail spam]]
> Do not ignore warnings.
>
> The left hand side of the formula should a matrix containing the counts
> of the afflicted and non-afflicted:
>    cbind(affected, total-affected)
> not the fraction of the total that were afflicted.  Then you would get
>
> Coefficients:
>             Estimate Std. Error z value Pr(>|z|)
> (Intercept)  -6.8940     1.0780  -6.395 1.60e-10 ***
> log(dose)     0.9333     0.1344   6.944 3.82e-12 ***
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
>
> On Thu, Feb 23, 2017 at 12:26 PM, Biank M <bianca12_domi at hotmail.com> wrote:
>> Hi,
>>
>> I'm working on the effects of alternative larvicides on Aedes aegypti. Right now, I am doing a binary mortality response with a single explanatory variable (dose) on 4 concentrations of one larvicide (+ control). Our university is fond of SPSS, and I have learned to conduct the basic probit model with it, including a natural logarithm transformation on my dosis data.
>> Not so long ago, I've started working with R, and through a combination of the 'glm' and 'dose.p' functions, I get the same slope and intercept, as well as LD50 calculations. Nevertheless, the standard errors and Z-scores calculated through the Probit model in SPSS comes out completely different in R. Additionally, the 95% confidence intervals for the LD50 come out very differently between the two programs. I really don't have a clue on how I am getting the same slopes, intercepts and LD50's, but totally different SE, Z, and 95% CI. Can anybody help me so I can get the same results in R??
>>
>> I'll pass you the script and hypothetical data:
>>
>> dose <- c(6000, 4500, 3000, 1500, 0)
>> total <- c(100, 100, 100, 100, 100)
>> affected <- c(91, 82, 69, 49, 0)
>>
>> finney71 <- data.frame(dose, total, affected)
>>
>> fm1 <- glm(affected/total ~ log(dose),
>> family=binomial(link = probit), data=finney71[finney71$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.
>>
[[elided Hotmail spam]]
>>
>>
>> Best wishes,
>>
>> Bianca
>>
>>
>>
>> PD. I've already googled it but there's no satisfactory answer.
>>
>>
>>
>>         [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help

thz.ch/mailman/listinfo/r-help>
stat.ethz.ch
The main R mailing list, for announcements about the development of R and the availability of new code, questions and answers about problems and solutions using R ...



>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.

	[[alternative HTML version deleted]]