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

[Prof Brian Ripley]

Your problem is the interpretation of anova(): it is a sequential test and 
x1 is the first term.  Using dropterm() would give you the correct LR 
test.

However, you also have a Wald test given by the line

> x11          -0.8144     0.4422  -1.842   0.0655 .

which is not significant at the 5% level.  The correct LRT would be 
expected to be more accurate, and your inversion of the profile likelihood 
is just a way to compute the LRT.


On Mon, 13 Nov 2006, Inman, Brant A.   M.D. wrote:

>
> System: R 2.3.1 on Windows XP machine.
>
> 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.
>
> ----------------
>
>> 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)
>
>
>     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.
> 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
> reconcile these two differing messages? I do recognize that the upper
> 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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>

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595