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Introduction

I'm doing a small pilot research in bird aggression in a colonising frontier regarding their breeding ground.

Background

The study was conducted over multiple years, presenting the colonising (south) and settled (north) collared flycatcher males with conspecific and pied flycatcher males. Scoring their behaviour based upon a quantifiable set aggressive actions. They have found this Island 60 years ago and are steadily spreading from one point in their breeding ground and pushing their relative pied flycatcher away from the more insect bearing territories. In previous studies it was shown that more aggressive males are at the front of such colonising action. In the north sites there is a near 100% collared and the south still has a mixed population.

Hypothesises

In the south location male collared flycatcher will act with higher aggression towards both species. Males will react in the north relatively more to conspecifics than they would in the south.

Problem

After having scored all the interactions I'm now at a loss at what test to use to present the data. Many people give different advice for lm or simple Anova etc. I have been learning R and statistics at the same time but many terms still confuse me and questions and answers found on the internet I found difficult to interpret to my data.

Question

What test out of the following three could be best used to show that there is or is not a statistical significance?

  • Anova(lm(score~dummy_species*location))
  • summary(aov(score~dummy_species*location))
  • summary(lm(score~dummy_species*location))

Data structure

The data is unfortunately unbalanced.

The amount of conspecific trials was 104 of which 77 were in the northern test area and 27 in the south. Similarly of the 50 pied flycatcher dummy tests 36 were in the north and 14 in the south.

'data.frame':   154 obs. of  8 variables:
 $ location        : Factor w/ 2 levels "N","S": 1 1 1 1 1 1 1 1 2 1 ...
 $ score           : int  1 4 0 1 1 8 9 9 4 3 ...
 $ dummy_species   : Factor w/ 2 levels "CF","PF": 1 1 2 2 1 1 1 1 1 2 ...

Location is north and south

Score is 0 to 7 with 7 being the highest score for aggressive behaviour. Dummy species represents conspecific and heterospecific types. So dependent variable has two two level independent factors

model.tables(aov(scoreCF$score~scoreCF$location),"means")

Tables of means
Grand mean
2.993506

 dummy_species 
     CF    PF
  3.529  1.88

rep 104.000 50.00

 location 
      N      S
      2.742  3.686
rep 113.000 41.000

 dummy_species:location 
         location
dummy_species N     S    
      CF   3.19  4.48
      rep 77.00 27.00
      PF   1.81  2.07
      rep 36.00 14.00

TukeyHSD(aov(score~dummy_species*location))

Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = score ~ dummy_species * location)

$dummy_species
       diff       lwr        upr     p adj
PF-CF -1.648846 -2.613568 -0.6841239 0.0009332

$location
     diff         lwr    upr     p adj
S-N 0.9440487 -0.07800284 1.9661 0.0699746

$`dummy_species:location`
           diff        lwr        upr     p adj
PF:N-CF:N -1.389250 -2.8774793 0.09898005 0.0766924
CF:S-CF:N  1.286676 -0.3619293 2.93528192 0.1824646
PF:S-CF:N -1.123377 -3.2649782 1.01822492 0.5246337
CF:S-PF:N  2.675926  0.7993571 4.55249475 0.0016744
PF:S-PF:N  0.265873 -2.0557788 2.58752484 0.9908082
PF:S-CF:S -2.410053 -4.8376320 0.01752615 0.0524523

Results

Anova(lm(score~dummy_species*location))
Anova Table (Type II tests)

Response: score
                    Sum Sq  Df F value    Pr(>F)  
dummy_species            93.91   1 11.6673 0.0008186 ***
location                 26.82   1  3.3326 0.0699100 .  
dummy_species:location    6.98   1  0.8675 0.3531437    
Residuals              1207.39 150         

summary(aov(score~dummy_species*location))*

                    Df Sum Sq Mean Sq F value   Pr(>F)    
dummy_species            1   91.8   91.80  11.405 0.000933 ***
location                 1   26.8   26.82   3.333 0.069910 .  
dummy_species:location   1    7.0    6.98   0.868 0.353144    
Residuals              150 1207.4    8.05          

summary(lm(score~dummy_species*location))
Call:
lm(formula = score ~ dummy_species * location)

Residuals:
    Min      1Q  Median      3Q     Max 
-4.4815 -2.1948 -0.8056  2.1280  6.9286 

Coefficients:
                      Estimate Std. Error t value Pr(>|t|)   
(Intercept)                 3.1948     0.3233   9.881   <2e-16 ***
dummy_speciesPF            -1.3892     0.5728  -2.425   0.0165 *  
locationS                   1.2867     0.6346   2.028   0.0444 *  
dummy_speciesPF:locationS  -1.0208     1.0960  -0.931   0.3531    

Residual standard error: 2.837 on 150 degrees of freedom
Multiple R-squared:  0.09423,   Adjusted R-squared:  0.07611 
F-statistic: 5.202 on 3 and 150 DF,  p-value: 0.001909

Ideally given the time investment (in the field and behind the screen) I would love to have it that male aggression is likely influenced by both location and species. But only if the lm approach would be relevant.

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What test out of the following three could be best used to show that there is or is not a statistical significance?

  • Anova(lm(score~dummy_species*location))
  • summary(aov(score~dummy_species*location))
  • summary(lm(score~dummy_species*location))

The first and last items are running the same test. Anova() and summary() just change how the result is presented from the same underlying test: lm(...) for "linear model".

The second item using aov(...) is not appreciably different. It just uses a different function to achieve the same results.

I think the lm(...) call is okay and the results are sufficient from the call to anova(lm(...)) to gauge your results. Your ANOVA table is:

dummy_species            93.91   1 11.6673 0.0008186 ***
location                 26.82   1  3.3326 0.0699100 .  
dummy_species:location    6.98   1  0.8675 0.3531437    
Residuals              1207.39 150  

The result of interest to the hypothesis you wish to reject is the dummy_species:location interaction in the ANOVA table; P=0.353 which is not statistically significant. This implies that your data do not support suggesting that aggression-differential is different in the Northern birds compared to Southern birds.

The result does suggest that birds in the north and south show more aggression towards members of their own species than other species (P<.001)

And though it does not reach statistical significance, a follow-up study (or a balanced study) might support that northern birds are slightly less aggressive than southern birds, generally. This test was probably most adversely affected by your design imbalance. You were very unbalanced across North-South.

But conspecific aggression levels do not vary between the north and south or; conversely, northern aggression level does not clearly differ from southern aggression level based on whether they are presented with CF or PF competitors.

Some of this could be a weakness in the 0-7 scale but I doubt you would increase your power accounting for this further (colloquially, you are probably borrowing power from the continuous model). In any case, because of the scale, you should check your diagnostic plots to make sure you do not have some bad assumptions (heteroscedasticity and non-normality).

So in conclusion:

HA: Settled CF birds will react with more aggression to CF birds than PF birds in all locations. (You did not ask about this one)

P<0.001 - Reject H0 -- The experiment supports this conclusion.

HA: In the south location male collared flycatcher will act with higher aggression towards both species.

P=0.0699 - Do not reject H0 -- The experiment does not support your conclusion that "In the south location male collared flycatcher will act with higher aggression towards both species."

HA: Males will react in the north relatively more to conspecifics than they would in the south.

P=0.3531 - Do not reject H0 -- The experiment does not support your conclusion that "Males will react in the north relatively more to conspecifics than they would in the south."

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  • $\begingroup$ Dear StatNoodle, I thank you greatly for taking the time to write such a well guided reply. In the meantime I've tried an Ordinal Logistic Regression fit but it is out of my scope for comprehension. This is for a bachelor thesis and will suffice with the help you have just given. I would really like to express my gratitude once again. The study design was indeed a pilot and I had to merge data from previous years in order to increase sample size. But the lacking statistical knowledge was more exhausting than all those early sunrises. Again thank you very much! $\endgroup$ – Salvuryc Jan 5 '16 at 21:19
  • $\begingroup$ P.S. Thank you for phrasing the hypothesis so well. $\endgroup$ – Salvuryc Jan 5 '16 at 21:22
  • $\begingroup$ I suspect you could run a continuation-ratio ordinal logistic regression for your data, rather easily, but I'm guessing about how you achieved your aggression score. I would not (and did not) encourage you to do so for this data or for this purpose. The discrete scale does not bother me in this study. It is too bad you did not sample southern birds more; I think there is a north-south difference but I do not think it clearly interacts with intra- versus inter- species competition. Good project, though. $\endgroup$ – StatNoodle Jan 6 '16 at 20:35
  • $\begingroup$ Thank you kindly. The project is part of a much much bigger study and this was one of the data collected during many field seasons. However when many people are doing many things it could have gone more coördinated. However there are trial repeats on the same and several days so after these videos would be scored and my statistical knowledge increases who knows what could be found. Now it will just be a focus on first day first trial encounters. Again thank you for your help, much joy was had. $\endgroup$ – Salvuryc Jan 7 '16 at 10:47

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