Your outcome variable is binary. So normality of the residual is meaningless. Use other diagnostic tools such as:

1. Can calculate the quasilikelihood which is the counterpart of AIC. But be cautious as this metric is highly sensitive with departures from the model 2. The ROC curve and the area under the curve can be used to see the overall performance of the model 3. Calibration and Hosmer-Lemeshow test can be used to check how well the logistic regression model fits the data 4. One can also use the cross-validation technique to calculate the misclassification error rate to check and compare several models.

All the best.

If your response variable is "proportion" type, you can try beta regression.

The null deviance is the “badness of fit” with no predictors but the mean (the intercept). The residual deviance is after taking account of the specified predictors. The difference is distributed (approximately) as a chi square distribution with degree of freedom equal to the number of extra parameters. Clearly your predictors do not reduce the badness of fit by much. This also shows up in the p values.

did you expect to have 180 observations dropped due to missing values, that is bulk of your observations are being disregarded, they are not being used.

how are you passing the denominator of the proportion ( success count out of total count) in the model specification?

Dear Kelvyn Jones , thank you for your answer. I try to understand all you said, but there are still some questions:

- you said: 'Clearly your predictors do not reduce the badness of fit by much.' I still don't understand how you can see this.

- Your first question: yes, I expected 180 observations to be dropped out, because I did not have the temperatures from all the nests.

- Your second question I don't really understand...

Badness of fit: Testing reduction in the deviance

Null deviance: 20.175 on 43 degrees of freedom

Residual deviance: 19.569 on 38 degrees of freedom

So the reduction in the badness of fit is tiny: 20.175 -19.569 , with change of df of 43-38, that is 5 extra predictors; 0.05 p value is a cutoff of 11.070 , so the reduction could have occurred quite frequently by chance

https://online.stat.psu.edu/stat504/node/157/

See also significance testing chapter in Book Developing multilevel models for analysing contextuality, he...

although this is a different context, the principles are the same.

Second question: a proportional response variable is made up of the ratio of two numbers ; the numerator is the number of successes, while the denominator is (number of success plus number of failures). This is important as you can expect greater inherent variation with smaller denominators.See my responses here https://www.researchgate.net/post/what_is_the_difference_between_linear_regression_and_logistics_regression

Have a look at the response to this question:

https://stats.stackexchange.com/questions/26762/how-to-do-logistic-regression-in-r-when-outcome-is-fractional-a-ratio-of-two-co

" The glm function in R allows 3 ways to specify the formula for a logistic regression model.

The most common is that each row of the data frame represents a single observation and the response variable is either 0 or 1 (or a factor with 2 levels, or other variable with only 2 unique values).

Another option is to use a 2 column matrix as the response variable with the first column being the counts of 'successes' and the second column being the counts of 'failures'.

You can also specify the response as a proportion between 0 and 1, then specify another column as the 'weight' that gives the total number that the proportion is from (so a response of 0.3 and a weight of 10 is the same as 3 'successes' and 7 'failures')." They use weight where I use denominator as that is what it is.

If the denominators are all 1's you are performing a Bernoulli analysis, if they are variable , then it is a binomial regression. Since you have a proportion for a response, I am presuming that you have varying denominators. This sets it out clearly

Neil Wrigley The Use of Percentages in Geographical Research

AreaVol. 5, No. 3 (1973), pp. 183-186 (4 pages)Published By: The Royal Geographical Society (with the Institute of British Geographers)DOI: 10.2307/20000750https://www.jstor.org/stable/20000750

a percentage is a closed ratio proportion multiple by 100.

This is a good guide to GLM

GLIM for Ecologists (Ecological Methods and Concepts) Hardcover – 9 Sept. 1993

by Mick Crawley (Author)

Thanks again Kelvyn Jones for putting energy in my question.

So the larger the difference between the null deviance and the residual deviance, the better the fit of the model?

I discovered that I did not clearly my definition of proportional data in my question. The hatching success does not say 1 or 0 (i.e. success or no success), but is determined by dividing the number of hatchlings by the number of eggs per nest. Each nest has thus a success value BETWEEN 0 and 1.

Q1 : Yes

Q2 you need to include the number of eggs per nest, the denominator - I am not a R user , so I do not know the proper syntax.

But this looks useful

https://stats.stackexchange.com/questions/345502/glm-binomiallogit-with-weights-tested

"[...] but is determined by dividing the number of hatchlings by the number of eggs per nest. Each nest has thus a success value BETWEEN 0 and 1."

If you have the number of successes, say N_Success, and the number of eggs in the nest, say N_Eggs (both being variables in the data.frame dd), the model would be:

glm(formula = N_Success ~ Species + Location + `Average temperature` +

`emergence tunnel (cm)`, family = binomial("logit"), weights = N_Eggs,

data = dd)

The same result is obtained when the numbers of successed and failures are given as a two-column data.frame for the response:

glm(formula = cbind(N_Success , N_Eggs-N-Success) ~ Species + Location

+ `Average temperature` + `emergence tunnel (cm)`,

family = binomial("logit"), data = dd)

Here are more details:

https://stats.stackexchange.com/questions/322038/input-format-for-response-in-binomial-glm-in-r