使用比例作为二项式中的偏移量 (glm) model

Question

I am trying to test the effect of a treatment on the proportions of juveniles in a population of migrating birds. The birds were counted and identified as juveniles or adults daily, but the treatment was only on every second day. Days without treatment were used as a control. The problem is that the proportion of juveniles in the population is expected to be affected not only by the treatment, but also by migration phenology. For example, it is possible that on a given day more juveniles migrated to the study area, and therefor this, and not only the treatment, affected the proportion of juveniles in the population. To account for this problem, I also checked the proportion of juveniles every day at a close by site which was not affected by the treatment (ie, control site). Hence, I have two types of controls. To analyze the data, I thought of using a binomial GLMM, with the proportion of juveniles as the variable of interest, the treatment as a categorical (with or without treatment) explanatory variable and day as a random-intercepts factor, and I use weights to account for the different number of birds in each day, but I am not sure how to input the data from the control site. From what I read, it should be used as an offset, but I am not sure exactly how.

Is the link function affected by the fact it (juveniles prop. at the ctrl. site) is a proportion? Is it better to use a the juveniles prop. at the ctrl. site in an interaction instead of offset (ie, ~ Treatment* Juv.prop.cntrl.site)?

This is the model I have so far, but I am not sure if it makes sense, especially if the offset is set correctly:

glm(Juv.prop.exp.site ~ Treatment + Day, offset = Juv.prop.cntrl.site, weights = Tot.birds.exp.site, data = df, family = Binomial)

Where Juv.prop.exp.site is the number of juveniles divided by the total at this site (juveniles + adults) See the data here: DATA (day starts at 11, because during the first 10 days no birds of that species were observed)

Thanks a lot in advance and happy new year!

Answer 1

Normally, I would suggest that questions regarding statistical analysis are migrated to CrossValidated, where you will get better answers to purely statistical questions. However, in your case, it will help a lot to reshape your data into a tidy format before analysis, which is more of a programming problem.

Essentially, you need one column each for day, site, treatment, number of juveniles, and number of adults. I am assuming that in your data, "V" is the treatment and "X" is the control.

library(tidyverse)

df <- data %>%
  select(1, 2, 4, 5, 8, 9) %>%
  rename_all(~gsub("\\.site", "_site", .x)) %>%
  pivot_longer(1:4, names_sep = "\\.", names_to = c(".value", "Site")) %>%
  mutate(Treatment = ifelse(Site == "Exp_site", Treatment, "X")) %>%
  mutate(Treatment = ifelse(Treatment == "V", "Treatment", "Control")) %>%
  mutate(Site = ifelse(Site == "Exp_site", "Experimental", "Control")) %>%
  rename(Juveniles = Juv, Adults = Ad) %>%
  select(2, 1, 3:5)

This makes your data look like this, and to my mind this is easier to analyse (and to reason about):

df
#> # A tibble: 100 x 5
#>      Day Treatment Site         Juveniles Adults
#>    <int> <chr>     <chr>            <int>  <int>
#>  1    11 Control   Experimental         1      0
#>  2    11 Control   Control              0      0
#>  3    12 Treatment Experimental         2      1
#>  4    12 Control   Control              1      0
#>  5    13 Control   Experimental         2      0
#>  6    13 Control   Control              1      1
#>  7    14 Treatment Experimental         6      3
#>  8    14 Control   Control              4      2
#>  9    15 Control   Experimental         6      4
#> 10    15 Control   Control              1      2
#> # ... with 90 more rows
#> # i Use `print(n = ...)` to see more rows

You can then perform a binomial glm like this, with Treatment and Site as independent variables.

model <- glm(cbind(Juveniles, Adults) ~ Treatment + Site, 
             data = df, family = binomial)

summary(model)
#> Call:
#> glm(formula = cbind(Juveniles, Adults) ~ Treatment + Site, family = binomial, 
#>     data = df)
#> 
#> Deviance Residuals: 
#>     Min       1Q   Median       3Q      Max  
#> -3.4652  -0.6971   0.0000   0.7895   2.9541  
#> 
#> Coefficients:
#>                    Estimate Std. Error z value Pr(>|z|)    
#> (Intercept)          1.0059     0.1461   6.886 5.74e-12 ***
#> TreatmentTreatment   0.3012     0.2877   1.047    0.295    
#> SiteExperimental    -0.1632     0.2598  -0.628    0.530    
#> ---
#> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#> 
#> (Dispersion parameter for binomial family taken to be 1)
#> 
#>     Null deviance: 118.16  on 88  degrees of freedom
#> Residual deviance: 117.07  on 86  degrees of freedom
#> AIC: 244.13
#> 
#> Number of Fisher Scoring iterations: 4