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I need to do a logistic regression using R on my data. My response variable (y) is survival at weaning (surv=0; did not surv=1) and I have several independent variables which are binary and categoricals in nature.

I am following some examples on this website http://www.ats.ucla.edu/stat/r/dae/logit.htm and trying to run some models.

Running the model:

> mysurv2 <- glm(surv~as.factor(PTEM) + as.factor(pshiv) + as.factor(presp) + 
                 as.factor(pmtone), family=binomial(link="logit"), data=ap)
> summary(mysurv2)

Call:
glm(formula = surv ~ as.factor(PTEM) + as.factor(pshiv) + as.factor(presp) + 
    as.factor(pmtone), family = binomial(link = "logit"), data = ap)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-2.2837  -0.5121  -0.5121  -0.5058   2.0590  

Coefficients:
                   Estimate Std. Error z value Pr(>|z|)    
(Intercept)        -0.01135    0.23613  -0.048  0.96166    
as.factor(PTEM)2   -0.74642    0.24482  -3.049  0.00230 ** 
as.factor(PTEM)3   -1.95401    0.23259  -8.401  < 2e-16 ***
as.factor(pshiv)2  -0.02638    0.06784  -0.389  0.69738    
as.factor(presp)2   0.74549    0.10532   7.079 1.46e-12 ***
as.factor(presp)3   0.66793    0.66540   1.004  0.31547    
as.factor(pmtone)2  0.54699    0.09678   5.652 1.58e-08 ***
as.factor(pmtone)3  1.82337    0.75409   2.418  0.01561 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 7892.6  on 8791  degrees of freedom
Residual deviance: 7252.8  on 8784  degrees of freedom
  (341 observations deleted due to missingness)
AIC: 7268.8

Number of Fisher Scoring iterations: 4

Adding the na.action=na.pass at the end of the model gave me an error message. I thought that this would take care NA's in my independent variables.

> mysurv <- glm(surv~as.factor(PTEM) + as.factor(pshiv) + as.factor(presp) + 
                as.factor(pmtone), family=binomial(link="logit"), data=ap, 
                na.action=na.pass)
Error: NA/NaN/Inf in foreign function call (arg 1)

Since this is my first time to venture into logistic regression, I am wondering whether there is any package in R that would be more suitable?

I am also tryng to understand the regression coefficients. The independent variables used in the model are:

  1. rectal temperature:

    • (PTEM)1 = newborns with rectal temp. below 35.4 0C
    • (PTEM)2 = newborns with rectal temp. between 35.4 to 36.9 0C
    • (PTEM)3 = newborns with rectal temp. above 37.0 0C
  2. shivering:

    • (pshiv)1 = newborns that were not shivering
    • (pshiv)2 = newborns that were shivering
  3. respiration:

    • (presp)1 = newborns with normal respiration
    • (presp)2 = newborns with slight respiration problem
    • (presp)3 = newborns with poor respiration
  4. muscle tone:

    • (pmtone)1 = newborns with normal muscle tone
    • (pmtone)2 = newborns with moderate muscle tone
    • (pmtone)1 = newborns with poor muscle tone

Looking at the coefficients, I got the following:

                   Estimate Std. Error z value Pr(>|z|)    
(Intercept)        -0.01135    0.23613  -0.048  0.96166    
as.factor(PTEM)2   -0.74642    0.24482  -3.049  0.00230 ** 
as.factor(PTEM)3   -1.95401    0.23259  -8.401  < 2e-16 ***
as.factor(pshiv)2  -0.02638    0.06784  -0.389  0.69738    
as.factor(presp)2   0.74549    0.10532   7.079 1.46e-12 ***
as.factor(presp)3   0.66793    0.66540   1.004  0.31547    
as.factor(pmtone)2  0.54699    0.09678   5.652 1.58e-08 ***
as.factor(pmtone)3  1.82337    0.75409   2.418  0.01561 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

In my other analysis, I found that newborns:

a) with higher rectal temperature
b) do not shiver
c) good respiration and
d) good muscle tone at birth were more likely to survive.

I am a bit confused with the coefficients I am getting above. I am wondering whether whether I am not interpreting the results correctly or is it something else?

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    $\begingroup$ What specifically are you confused about? $\endgroup$ – mark999 May 22 '11 at 10:54
  • $\begingroup$ Isn't it a question for the sister site, SO? It seems to be less a question of statistics and more of programming language... $\endgroup$ – Manoel Galdino May 22 '11 at 23:44
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I think you're confused because you defined survival at weaning as surv=0 rather than surv=1. In your model, negative coefficients indicate high odds of survival (low odds of surv=1).

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  • $\begingroup$ you are absolutely right. I was looking at surv=0 and not surv=1. Thanks for the help! $\endgroup$ – baz May 24 '11 at 4:21
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The lrm function in the R rms package is devoted to binary and ordinal logistic regression, and my help, once you understand the rms documentation. Detailed case studies using rms may be found in course notes at http://biostat.mc.vanderbilt.edu/rms. However there are more important issues. Categorizing continuous variables leads to erroneous conclusions, underfitting, and residual confounding (here, temperature and perhaps respiratory rate).

One R coding convention suggestion: Set up your data frame the way you want to treat the variables in later modeling steps. For example if variables are really factors, give them descriptive levels and make them factors once and for all. The rms package almost requires this.

If you do use rms you will find its summary, Predict, plot, contrast, and nomogram functions handy for interpreting the model, once the model properly uses continuous variables.

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  • $\begingroup$ I am following that course notes and it a big help. However, I need to do: a "backward elimination" to get my best model $\endgroup$ – baz May 25 '11 at 2:32
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    $\begingroup$ Why would you need to do that? That will invalidate betas, SEs, P-values, and confidence limits. It can be somewhat safe if you set alpha for keeping a variable to 0.5. You can do that with the rms fastbw function. Better still is to set alpha=1.0 (don't drop any variables). Who suggested that stepwise variable selection without penalization is a good idea? $\endgroup$ – Frank Harrell May 25 '11 at 2:51
  • $\begingroup$ Well, it is something that I thought would work and was not considering what you just mentioned. The whole idea is this: (i) run the full model, (ii) keep only predictors that are significant (table of coefficients), (iii) test for the overall effect of each predictor and between each level (to get some ideas of their effect (iv) get the odd ratios $\endgroup$ – baz May 25 '11 at 3:06
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    $\begingroup$ One way to understand the issue is to consider a different model, the ordinary multiple regression model. This model involves a residual variance. Stepwise variable selection results in an underestimate of this variance. Two unbiased estimators for the variance have been developed: (1) fit a full pre-specified model and compute the residual variance; or (2) run stepwise selection or other data mining procedure and use Ye's generalized degrees of freedom (d.f.) to estimate the "effective d.f." of the modeling process (close to # candidate variables), then use it to estimate the variance. $\endgroup$ – Frank Harrell May 25 '11 at 12:40
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Two small points that haven't been addressed yet:

Regarding your difficulty with NA's, you don't need a different package (although there's nothing wrong with using rms instead). Note that ?na.pass does nothing to your dataset, so the NA's are still there. You may want to try na.omit.

I agree with Frank Harrell about categorizing your variables, but in addition to that, shouldn't whether or not a baby is shivering be highly correlated with its temperature? I suspect that's why you unexpectedly don't find shivering to be 'significant' here, even though rectal temp is.

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