[R] Profile confidence intervals and LR chi-square test

[Henric Nilsson]

On 2006-11-14 00:41, Inman, Brant A. M.D. skrev:
> System: R 2.3.1 on Windows XP machine.

Time to upgrade!

> 
> I am building a logistic regression model for a sample of 100 cases in
> dataframe "d", in which there are 3 binary covariates: x1, x2 and x3.

Please provide a reproducible example (as suggested by the posting guide).

> 
> ----------------
> 
>> summary(d)
>  y      x1     x2     x3    
>  0:54   0:50   0:64   0:78  
>  1:46   1:50   1:36   1:22  
> 
>> fit <- glm(y ~ x1 + x2 + x3, data=d, family=binomial(link=logit))
> 
>> summary(fit)
> 
> Call:
> glm(formula = y ~ x1 + x2 + x3, family = binomial(link = logit), 
>     data = d)
> 
> Deviance Residuals: 
>     Min       1Q   Median       3Q      Max  
> -1.6503  -1.0220  -0.7284   0.9965   1.7069  
> 
> Coefficients:
>             Estimate Std. Error z value Pr(>|z|)  
> (Intercept)  -0.3772     0.3721  -1.014   0.3107  
> x11          -0.8144     0.4422  -1.842   0.0655 .
> x21           0.9226     0.4609   2.002   0.0453 *
> x31           1.3347     0.5576   2.394   0.0167 *
> ---
> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
> 
> (Dispersion parameter for binomial family taken to be 1)
> 
>     Null deviance: 137.99  on 99  degrees of freedom
> Residual deviance: 120.65  on 96  degrees of freedom
> AIC: 128.65
> 
> Number of Fisher Scoring iterations: 4
> 
>> exp(fit$coef)
> (Intercept)         x11         x21         x31 
>   0.6858006   0.4429233   2.5157321   3.7989873 
> ---------------
> 
> After reading the appropriate sections in MASS4 (7.2 and 8.4 in
> particular), I decided to estimate the 95% confidence intervals for the
> odds ratios using the profile method implemented in the "confint"
> function. I then used the "anova" function to perform the deviance
> chi-square tests for each covariate.
> 
> ---------------
>> ci <- confint(fit); exp(ci)
> Waiting for profiling to be done...
>                 2.5 %    97.5 %
> (Intercept) 0.3246680  1.413684
> x11         0.1834819  1.048154
> x21         1.0256096  6.314473
> x31         1.3221533 12.129210
> 
>> anova(fit, test='Chisq')
> Analysis of Deviance Table
> 
> Model: binomial, link: logit
> 
> Response: y
> 
> Terms added sequentially (first to last)
               ^^^^^^^^^^^^
Hence, your use of the `anova' function doesn't return tests 
corresponding to the CIs computed above.

> 
> 
>      Df Deviance Resid. Df Resid. Dev P(>|Chi|)
> NULL                    99    137.989          
> x1    1    5.856        98    132.133     0.016
> x2    1    5.271        97    126.862     0.022
> x3    1    6.212        96    120.650     0.013
> ----------------
> 
> My question relates to the interpretation of the significance of
> variable x1.  The OR for x1 is 0.443 and its profile confidence interval
> is 0.183-1.048.  If a type I error rate of 5% is assumed, this result
> would tend to suggest that x1 is NOT a significant predictor of y.

This is also suggested by the Wald test computed by the `summary' function.

> However, the deviance chi-square test has a P-value of 0.016, which
> suggests that x1 is indeed a significant predictor of y. How do I

That p-value corresponds to adding x1 to a model containing only the 
intercept term.

> reconcile these two differing messages? I do recognize that the upper

Generally, in order to compute the LR test for the null hypothesis of 
some subset of the parameters being equal to zero, you need to 
explicitly fit both the restricted and the unrestricted model and 
compare them using the `anova' function.

Also, see FAQ 7.18.


HTH,
Henric



> bound of the confidence interval is pretty close to 1, but I am certain
> that some journal reviewer will point out the problem as inconsistent.
> 
> Brant Inman
> 
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