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So I have two datasets from two different groups (blackbirds and song thrushes), with the same 4 variables (Foraging, Singing, Alarm, Aggressive). I've been struggling to find a good way to compare the data in these different sets, the sampling that was conducted is called scan sampling, so it's just a recording of observations of behaviour an individual is performing. What would be an appropriate way to compare these two sets with the hypothesis that both species share similar time budgets?

Rows account for the number of observations in total over a period of 10 mins so there isn't really a sample size for this.

Blackbirds

blackbirds <- structure(list(Foraging = c(54L, 24L, 2L, 10L, 10L, 7L, 4L, 21L, 
15L, 14L, 8L, 0L, 7L, 12L, 3L, 5L, 41L, 15L, 12L, 6L, 23L, 17L, 
10L, 17L, 11L, 3L, 0L, 2L, 12L, 7L, 14L, 3L), Singing = c(12L, 
4L, 4L, 3L, 2L, 4L, 1L, 4L, 5L, 5L, 2L, 0L, 0L, 6L, 1L, 6L, 11L, 
3L, 8L, 2L, 10L, 8L, 4L, 3L, 3L, 2L, 2L, 7L, 2L, 5L, 7L, 3L), 
    Alarm = c(2L, 2L, 0L, 0L, 1L, 0L, 1L, 1L, 3L, 0L, 2L, 0L, 
    1L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 2L, 0L, 0L, 2L, 2L, 1L, 1L, 
    1L, 0L, 0L, 2L, 0L), Aggressive = c(1L, 0L, 0L, 0L, 0L, 0L, 
    0L, 1L, 1L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 
    0L, 0L, 0L, 1L, 0L, 0L, 0L, 2L, 0L, 0L, 0L)), .Names = c("Foraging", 
"Singing", "Alarm", "Aggressive"), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", 
"14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24", 
"25", "26", "27", "28", "29", "30", "31", "32"))

Song thrushes

song_thrushes <- structure(list(Foraging = c(6L, 4L, 4L, 2L, 4L, 2L, 4L, 3L, 8L, 
11L, 11L, 3L, 2L, 7L, 4L, 6L, 7L, 9L, 0L, 5L, 11L, 2L, 5L, 6L, 
7L, 7L, 8L, 9L, 5L, 1L, 0L, 2L), Singing = c(4L, 3L, 3L, 1L, 
1L, 2L, 1L, 1L, 3L, 3L, 4L, 3L, 6L, 2L, 0L, 2L, 3L, 2L, 7L, 2L, 
4L, 0L, 5L, 2L, 4L, 0L, 2L, 2L, 3L, 0L, 2L, 2L), Alarm = c(0L, 
0L, 1L, 0L, 0L, 1L, 0L, 1L, 2L, 1L, 0L, 1L, 0L, 2L, 0L, 0L, 1L, 
0L, 2L, 0L, 0L, 0L, 1L, 1L, 2L, 1L, 0L, 1L, 1L, 0L, 0L, 0L), 
    Aggressive = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 2L, 0L, 
    0L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 
    0L, 0L, 1L, 1L, 0L, 0L)), .Names = c("Foraging", "Singing", 
"Alarm", "Aggressive"), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", 
"14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24", 
"25", "26", "27", "28", "29", "30", "31", "32"))

I've read about one-way ANOVA tests and Pearson correlations, but I'm not sure how to compare the proportions between the blackbirds and song thrushes. I'm really new to R so forgive me for being so lost.

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  • $\begingroup$ Are row number 5 in blackbirds and row number 5 in song thrushes paired data or unpaired data? Were the sounds counted at the same time of the year and samy time of the day with the same wheather when they are in the same row number or are these just arbitrary? $\endgroup$ – Bernhard Oct 19 '17 at 13:02
  • $\begingroup$ Yes all data was collected simultaneously as the type of sampling used was called scan sampling. The row is just a total of observations made in 10 minutes at that specific location, so they're just a tally of counts.I have all of the recordings for (morning/evening, location point etc.) but wasn't sure if it was necessary to include all of that since I'm only really assessing the proportions of time allocated to the specific behaviours. So ultimately what I'm looking for is how to compare the proportions with a valid statistical test. $\endgroup$ – jdsurfer Oct 19 '17 at 20:36
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As these are count data and there "is not really a sample size for this", a poisson model would be neat. Note however, that even a simple t-test will find significant differences in two out of four column pairs:

> t.test(blackbirds$Foraging, song_thrushes$Foraging, paired=TRUE)

Paired t-test

data:  blackbirds$Foraging and song_thrushes$Foraging
t = 3.4877, df = 31, p-value = 0.001481
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
  2.906607 11.093393
sample estimates:
mean of the differences 
                  7 

> t.test(blackbirds$Singing, song_thrushes$Singing, paired=TRUE)

Paired t-test

data:  blackbirds$Singing and song_thrushes$Singing
t = 3.346, df = 31, p-value = 0.002159
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 0.7321223 3.0178777
sample estimates:
mean of the differences 
              1.875 

> t.test(blackbirds$Alarm, song_thrushes$Alarm, paired=TRUE)

    Paired t-test

data:  blackbirds$Alarm and song_thrushes$Alarm
t = 1.1909, df = 31, p-value = 0.2427
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.1558696  0.5933696
sample estimates:
mean of the differences 
            0.21875 

> t.test(blackbirds$Aggressive, song_thrushes$Aggressive, paired=TRUE)

    Paired t-test

data:  blackbirds$Aggressive and song_thrushes$Aggressive
t = -0.2544, df = 31, p-value = 0.8009
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.2817822  0.2192822
sample estimates:
mean of the differences 
               -0.03125 

Link to Poisson regression: https://www.cs.princeton.edu/~bee/courses/lec/lec_jan29.pdf

However, with only 32 observations and the mean number of alarm and aggressive calls well below 1, chances are, that a more complex model will not find that much more information to reveal (mean Aggressive count is .25 vs .28 in n=32!)

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  • $\begingroup$ Thank you very much for your reply! It's a lot of help. Would a chi-square analysis work as well? $\endgroup$ – jdsurfer Oct 21 '17 at 1:15
  • $\begingroup$ Chi-Squared-Tests are made for count data and you could use them to compare the sum of all foraging calls in blackbirds to those of all song_thruses but that would not account for the paired nature of the data. In a sum of all foraging sounds of all blackbirds, sunny days would be overrepresented over rainy days, if birds sing more often on sunny than on rainy days. If for any reason, my proposal of t-tests does not answer the question, I would advise you to specify the question. $\endgroup$ – Bernhard Oct 22 '17 at 12:37

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