I have collected data on population sex ratios i.e. sexes (male or female) at adult age; my sample size is ~1500.

I want to see if my sex ratio (no. of males/(no. of males+no. of females) differs significantly from the theoetical expectation of 0.5. I was reading Box 3.3 and it says that there are two distinct questions to ask:

"first, does the sex ratio differ from the assumed binomial distribution; and second does it differ from some expected value, e.g. 0.5"

To see if it differed from a binomial distribution, I ran a null logistic regression model (e.g. glm(sex~1, family=binomial) and compared the null deviance to the chi-squared distribution with the null degrees of freedom. I also ran a binomial test e.g. binom.test(no. of males, no. of males and females, p=0.5); both were significant.

But I don't understand why these tests are different (I get different p-values - much lower on the former). What is a null binomial model doing that a binomial test isn't? What would it mean if the GLM was non-siginficiant but the binomial test was? and vice-versa?


1 Answer 1


Let's take a look at this with some data

N <- 1500

Here, the population average for one of the sexes is 45%. The data sex come from a binomial distribution.

First, the GLM

glm(formula = cbind(sex, N - sex) ~ 1, family = binomial(link = "logit"))

Deviance Residuals: 
[1]  0

            Estimate Std. Error z value Pr(>|z|)   
(Intercept) -0.13889    0.05176  -2.683  0.00729 **
Signif. codes:  
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 0  on 0  degrees of freedom
Residual deviance: 0  on 0  degrees of freedom
AIC: 9.7603

Number of Fisher Scoring iterations: 2

It is important to be clear on what the deviance tells us. I'm not going to get into the hypothesis testing details, but will approach it from a high level. The residual deviance is comparing our model to a saturated model (essentially, a model with one parameter per observation. Not super useful since we only have one observation and no predictors). If our residual deviance is not too big, then we can conclude that there appears to be evidence that our model is nearly as good as a saturated model. That is good because often times the model we have in hand is much simpler than a saturated model, and thus we should prefer the model in hand because it is simpler (thus more parsimonious).

The null deviance is tells us the deviance for a model which only has one intercept. If you take the difference between the residual and null deviance and compare that difference to an appropriate chi-square distribution, then that tells us how our model in hand compares to a model with no predictors. Our hypothesis is often that our predictors tell us something about the outcome, and so we want the difference between residual and null deviance to be quite large.

In this example, the deviance is uninteresting because a) we have no predictors, and b) we have one observation.

So far as the difference between the two tests, they are testing different things. binom.test tests the proportion parameter for your data. The deviance goodness of fit tests are comparing your model to two very different models (a saturated and null model). A similar test to prop.test with the GLM is the wald test for the intercept. Let's go back to our data and use prop.test.

    1-sample proportions test without continuity

data:  sex out of N, null probability 0.5
X-squared = 7.2107, df = 1, p-value = 0.007247
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
 0.4402119 0.4906319
sample estimates:

And now let's look at our intercept from our GLM on the appropriate scale

>sigmoid<-function(x) 1/(1+exp(-x))
> sigmoid(coef(model))

Our intercept returns the same estimate of the population proportion, and the estimate is significant as per the output.

  • $\begingroup$ You could also simulate x<-rbinom(N, 1, 0.45) and fit the model glm(x~1, familty = binomial). This model would have 1500 Bernoulli observations. It would be interesting if you could explain how/if this differs from the "aggregated" model, so to speak. $\endgroup$ Commented Nov 17, 2019 at 8:02

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