通常情况下，我会建议将有关统计分析的问题迁移到CrossValidated，在那里您将获得纯统计问题的更好答案。然而，在您的情况下，在分析之前将数据重塑为整齐的格式会有很大帮助，这更像是一个编程问题。
基本上，每天、地点、处理、青少年数量和成年数量各需要一列，我假设在您的数据中，“V”是处理，“X”是对照。

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)

这使得你的数据看起来像这样，在我看来，这更容易分析（和推理）：

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

然后，您可以像这样执行二项式glm，其中Treatment和Site作为自变量。

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