[R-sig-ME] LMM: including ranef or not?
Ben Bolker
bbolker at gmail.com
Mon Jul 21 23:24:51 CEST 2014
If you're going to use ud95 as the response you might as well average
the prop_land values per bird (i.e., aggregate the data down to a single
data record per bird); the within-bird variation in prop_land values
won't affect the model output at all (although the _number_ of
observations per bird will; if you have unbalanced information in this
way, you should incorporate weights proportional to the number of
observations as well).
With only 6 colonies you're going to have some difficulty estimating
the among-colony variance very well; if you end up with zero estimates
of among-colony variance, you might want to use blme or set the colonies
as fixed effects ...
If you use prop_land as the response variable you would indeed want to
put bird_id in as a random effect (and you might consider estimating the
proportional as a binomial (GLMM) response, _if_ you know the total
number of fixes for each bird)
On 14-07-21 10:22 AM, Anna-Marie Corman wrote:
> Dear Christian,
>
> thanks for your answer. So, including the ranef or not depends on the
> response, doesn't it? If I tested the model the other way around
> (prop_land~ud95...) I would have to include bird_id as ranef, because
> there are more than one measurments for each bird, right?
>
> Best,
> Anna
>
> Am 21.07.2014 13:49, schrieb Christian Ritz:
>> Dear Anna,
>>
>> no, with only one measurement per bird there is no need for a
>> bird-specific random effect.
>>
>> Your model looks okay to me.
>>
>> Best wishes Christian
>>
>>
>> On 21-07-2014 13:37, Anna-Marie Corman wrote:
>>> Dear list,
>>>
>>> I want to test whether the UD sizes of several tracked seabird
>>> individuals from 6 different breeding colonies depends on the foraging
>>> target (on land or at sea) as follows:
>>>
>>> mod<-lmer(ud95~prop_land+(1|colony),data=dat);
>>>
>>> I have one UD95 value for each individual, but the proportion of fixes
>>> on land are on a trip basis, i.e. several values for one individual. Do
>>> I need to include bird_id as random factor to exclude pseudo
>>> replication, though the response has only one value per individual???
>>>
>>> Many thanks in advance.
>>>
>>> Best,
>>> Anna
>>>
>>> data:
>>>
>>> bird_id colony FT_id year sex max_speed max_distnest tot_dist tdur mean_dist mean_distnest
>>> 1 HA1_2012 Amrum 1 2012 1 53.65358 36.008004 101.34099 7.5827778 0.4444780 15.867350
>>> 2 HA1_2012 Amrum 2 2012 1 63.88851 69.993403 149.00254 6.1788889 0.8186953 46.947190
>>> 3 HA1_2012 Amrum 3 2012 1 82.68318 70.532407 176.65181 7.7008333 0.7886241 48.160436
>>> 4 HA1_2012 Amrum 4 2012 1 56.25961 5.293994 15.11632 0.8130556 0.6298466 3.710259
>>> 5 HA1_2012 Amrum 5 2012 1 70.04150 71.002162 215.32017 12.6369444 0.5851092 42.360905
>>> 6 HA1_2012 Amrum 6 2012 1 71.40167 71.712123 213.43533 12.1355556 0.5995374 54.878232
>>> mean_speed prop_sea prop_land straightness prop_day prop_night ft_start ft_start_s ft_end ft_end_s
>>> 1 13.01700 0.9912664 0.008733624 0.3553153 0.08296943 0.9170306 21:34:09 77649 05:11:03 18663
>>> 2 25.24134 0.2786885 0.721311500 0.4697464 1.00000000 0.0000000 06:47:53 24473 13:00:35 46835
>>> 3 26.50082 0.1644444 0.835555600 0.3992736 0.92888890 0.0711111 04:28:00 16080 12:12:06 43926
>>> 4 22.27650 0.8000000 0.200000000 0.3502171 0.24000000 0.7600000 04:21:24 15684 05:12:12 18732
>>> 5 19.70902 0.3821138 0.617886200 0.3297516 0.84552850 0.1544715 03:05:05 11105 15:45:16 56716
>>> 6 22.07548 0.2997199 0.700280100 0.3359899 0.91596640 0.0840336 04:02:14 14534 16:12:19 58339
>>>
>>> ud95 ud50 id30 rd30 udoi50_colmean udoi95_colmean idoi30_colmean rdoi30_colmean
>>> 1 460.73 9.76 24.8 64.16 0.002 0.0985 1e-04 0.0059
>>> 2 460.73 9.76 24.8 64.16 0.002 0.0985 1e-04 0.0059
>>> 3 460.73 9.76 24.8 64.16 0.002 0.0985 1e-04 0.0059
>>> 4 460.73 9.76 24.8 64.16 0.002 0.0985 1e-04 0.0059
>>> 5 460.73 9.76 24.8 64.16 0.002 0.0985 1e-04 0.0059
>>> 6 460.73 9.76 24.8 64.16 0.002 0.0985 1e-04 0.0059
>>>
>>>
>>>
>>>
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>>> https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
>>
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