[R-sig-ME] DF in lme
Ben Bolker
bbolker at gmail.com
Thu Mar 17 21:51:55 CET 2011
On 03/17/2011 04:36 PM, Ben Ward wrote:
> On 17/03/2011 17:43, Ben Bolker wrote:
>> [cc'd back to r-sig-mixed models: **please** keep the conversation
>> going in the public forum!]
> Aplogies, I didn't notice the cc had dissapeared.
>> On 03/17/2011 09:57 AM, Ben Ward wrote:
>>> On 17/03/2011 12:01, Ben Bolker wrote:
>>>> On 11-03-17 06:16 AM, i white wrote:
>>>>> Ben
>>>>>
>>>>> Suppose you calculated an average response for each of the 30
>>>>> plates in
>>>>> your experiment, and calculated a standard two-way anova as follows:
>>>>>
>>>>> Source of variation DF
>>>>> Groups 1
>>>>> Lines 4
>>>>> Groups x lines 4
>>>>> Residual 20
>>>>>
>>>>> The F-tests from this anova should agree with the Wald tests from
>>>>> lmer.
>>>>> The residual is based on variation between plates within lines and
>>>>> groups. If I understand the design correctly, the other sources of
>>>>> variation (between disks in plates, between readings within disks)
>>>>> may
>>>>> be of interest but do not feature individually in the testing of
>>>>> groups
>>>>> and lines.
>>>>>
>>>>> When data are balanced, an anova can clarify some of the
>>>>> obscurities of
>>>>> mixed model fitting. Is this a controversial observation on this list?
>>>>>
>>>> I don't disagree.
>>>>
>>>> I'm glad that light seems to be appearing at the end of the tunnel
>>>> for the original poster. I would also say following Murtaugh 2007 (who
>>>> I quote often here) that I think that thinking of the subsampling
>>>> (disks/plates) as being a method for increasing precision of
>>>> measurements, and averaging the values, has advantages in terms of (1)
>>>> simplifying the analysis (and thus lowering the chances of
>>>> mistakes/increasing the chance of detecting them) (2) bringing
>>>> non-normal sample distributions closer to normality by averaging.
>>>> (This
>>>> doesn't work for randomized block designs, or GLMMs, or cases where the
>>>> variation at lower levels of nesting is of intrinsic interest.)
>>>>
>>>> Lineage definitely makes more sense to me as a random effect,
>>>> although there is almost always wiggle room within these definitions
>>>> ...
>>>>
>>>> Murtaugh, Paul A. 2007. “Simplicity and Complexity in Ecological Data
>>>> Analysis.” Ecology 88 (1): 56-62.
>>>> http://www.esajournals.org/doi/abs/10.1890/0012-9658%282007%2988%5B56%3ASACIED%5D2.0.CO%3B2.
>>>>
>>>>
>>>>
>>>>
>>> Hi,
>>>
>>> I averaged the data and performed an ANOVA as suggested, and the
>>> F-values match those of the mixed effects models.
>>> Although when averaged I only have a dataset of 10 values, which is why
>>> myself and my lecturer agreed to taking more measurements into groups,
>>> so we had a larger set to play with, because if too large we could cut
>>> it down or sort it, whereas if I didn't have enough I'd be in trouble.
>>> With only 10 values it's difficult to judge normality on the plots,
>>> although a shapiro-test does result in 0.09, so I'm cautious of what to
>>> make of it, once again transformation by natural log or sqrt doesen't
>>> seem to do much in my case. The heteroscedacity is also more profound in
>>> the anova of averaged data than with the mixed models, I plan to look at
>>> weights options.
>>>
>>> Thanks,
>>> Ben W.
>>>
>> If the data are balanced, then the classical inference based on the
>> averaged data is the same as that based on the nested model. In some
>> sense, you don't "really" have more than 10 independent data points for
>> testing the hypothesis of differences among differences among groups
>> (although each of those data points might be fairly precise because of
>> all the subsampling you have done). The other advantage is that you now
>> have the simple equivalent of a one-way ANOVA: if you are worried about
>> normality, you can fit a robust linear model, or a rank-based test
>> (Wilcoxon etc.), or do a permutation test.
>>
> I understand what you say about me actually only having in effect 10
> datapoints either way. When you say "
>
> (although each of those data points might be fairly precise because of
> all the subsampling you have done)"
>
>
> Do you mean that those "in effect" 10 datapoints in the mixed model
> would be more accurate, than the averaged values and a normal
> anova/t.test/permutation test?
> Or just in general my 10 datapoints are pretty accurate because of the
> subsampling - that they're both as accurate either way?
The latter.
> I suppose that
> doing a non-parametric test on the averaged values, deals with
> inequalities of variance too, where in the mixed model I'd perhaps try
> some weighting option.
Not necessarily: see
<http://en.wikipedia.org/wiki/Mann%E2%80%93Whitney_U>: "Under the null
hypothesis the distributions of both groups are equal"; so you could in
principle reject the null distribution because their distributions were
different (e.g. different variance, skew), even if their location
parameters (mean/median/etc) were the same. (But see farther down on
the page, "Different distributions" for more nuanced interpretations)
> T-tests and Permutation tests agree on estmates
> and p-values, Wilcocxon seemed a long way off from them. R^2 values in
> my mixed model are lower than in the anova, but then I understand R^2 is
> not the best measure, so all in all, I'm not too worried about the
> normality. Depending on what you meant about
>
> (although each of those data points might be fairly precise because of
> all the subsampling you have done)"
>
> I'm considering the permutation test with the 10 datapoints, over the
> mixed model with many, if the points however for the mixed model are
> more accurate than the 10, I'll keep the mixed model.
I'd go with the permutation test for convincing readers/reviewers that
the results were robust; however, at least at some point you might be
interested in saying something about the biological meaning of the
among-lineage variance ...
>
> Thanks,
> Ben.
>
>
>
>>>>
>>>>> Ben Ward wrote:
>>>>>> On 16/03/2011 15:08, Ben Bolker wrote:
>>>>>>> On 11-03-16 03:52 AM, Ben Ward wrote:
>>>>>>>> Hi, I'm using lme and lmer in my dissertation and it's the first
>>>>>>>> time
>>>>>>>> I've used these methods. Taking into account replies from my
>>>>>>>> previous
>>>>>>>> query I decided to go through with a model simplification, and
>>>>>>>> then try
>>>>>>>> to validate the models in various ways and come up with the best
>>>>>>>> one to
>>>>>>>> include in my work, be it a linear mixed effects model or general
>>>>>>>> linear
>>>>>>>> effects model, with log() data or not etc - interestingly it
>>>>>>>> does not
>>>>>>>> seems like doing transofrmations and such makes much difference so
>>>>>>>> far,
>>>>>>>> looking at changes in diagnostic plots and AIC.
>>>>>>> Be careful about comparing fits of transformed and
>>>>>>> non-transformed
>>>>>>> data via AIC/log-likelihood: e.g. see
>>>>>>> <http://www.unc.edu/courses/2006spring/ecol/145/001/docs/lectures/lecture18.htm>.
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> (This does *not* refer to the link function, e.g. the log link
>>>>>>> of the
>>>>>>> Poisson, but to the case where you transform your data prior to
>>>>>>> analysis.)
>>>>>>>
>>>>>>>> Anywho, I simplified to the model using lme (I've pasted it at the
>>>>>>>> bottom). And looking at the anova output the numDF looks right.
>>>>>>>> However
>>>>>>>> I'm concerned about the 342 df in the denDF in anova() and in the
>>>>>>>> summary() output, as it seems to high to me, because at the
>>>>>>>> observation
>>>>>>>> level is too high and pseudoreplicated; 4 readings per disk, 3
>>>>>>>> disks,
>>>>>>>> per plate, 3 plates per lineage, 5 lineages per group, 2 groups so:
>>>>>>>> 4*3*3*5*2=360. If I take this to disk level 3*3*5*2=90, and at dish
>>>>>>>> level it's 3*5*2=30 degrees of freedom for error. And either
>>>>>>>> dish or
>>>>>>>> disk (arguments for both) is the level at which one independant
>>>>>>>> point of
>>>>>>>> datum is obtained, most probably Dish. So I'm wondering if either
>>>>>>>> I'de
>>>>>>>> done something wrong, or I'm not understanding how df are
>>>>>>>> presented and
>>>>>>>> used in mixed models. It's not really explained in my texts, and my
>>>>>>>> lecturer told me I'm working at the edge of his
>>>>>>>> personal/professional
>>>>>>>> experience.
>>>>>>> At what level are Group and Lineage replicated in the model?
>>>>>>> Do you
>>>>>>> have different Groups or Lineages represented on the same disk,
>>>>>>> dish, or
>>>>>>> plate? If you do have multiple Groups and Lineages present at the
>>>>>>> lowest level of replication, then you have a randomized block
>>>>>>> design and
>>>>>>> the degrees of freedom may be higher than you think. If you really
>>>>>>> want
>>>>>>> denominator degrees of freedom and you want them correct, consult an
>>>>>>> experimental design book and figure out what they should be in the
>>>>>>> classical framework ...
>>>>>> I'm now of the opinion that - (Just trying to get my head around
>>>>>> it) -
>>>>>> that I don't have a randomized block design:
>>>>>> I've done a bit like a lenski evolution experiment with my microbes,
>>>>>> which involed two groups, in those two groups i have 5 cultures each,
>>>>>> one group is 5 lineages of bacteria I have been evolving against some
>>>>>> antimicrobial, the other group have not been through this - they are
>>>>>> stock run of the mill organisms. So with those 5 cultures of evolved
>>>>>> bacteria, for each, I'd take some, and spread it on three plates - so
>>>>>> theres no intermingling or randomization/mixing of the cultures: each
>>>>>> gets plated onto a who plate itself three times. Then the three
>>>>>> disks,
>>>>>> loaded with antimicrobial were loaded onto each plate, and they were
>>>>>> incubated, and then I took 4 measurements from each zone that formed
>>>>>> around those disks. The disks all have the same antimicrobial on
>>>>>> them.
>>>>>> So in that way, if what you say by randomized block design is
>>>>>> something like a split plot experiment, where there are several
>>>>>> plots,
>>>>>> and numerous plants, and each one got a different treatment, then I
>>>>>> don't believe my experiment is like that. In my case that would be
>>>>>> like me having different cultures on the same dish, or using disks
>>>>>> with different antimicrobials on, at least I think this is what
>>>>>> you're
>>>>>> asking. In which case Dish is the level at which I get truly
>>>>>> indepentent pieces of data, and 3plates*5lineages*2Groups=30: If I
>>>>>> recode my factor levels then, like so, which I mentioned before as a
>>>>>> possibility:
>>>>>> Diameter<-Dataset$Diameter
>>>>>> Group<-factor(Dataset$Group)
>>>>>> Lineage<-factor(Dataset$Lineage)
>>>>>> Dish<-factor(Dataset$Dish)
>>>>>> Disk<-factor(Dataset$Disk)
>>>>>> lineage<-Group:Lineage
>>>>>> dish<-Group:Lineage:Dish
>>>>>> disk<-Group:Lineage:Dish:Disk
>>>>>>
>>>>>> And then fit the model:
>>>>>>
>>>>>> model<- lme(Diameter~Group*Lineage,random=~1|dish/disk,
>>>>>> method="REML")
>>>>>>
>>>>>> I get the following:
>>>>>>
>>>>>> > summary(model)
>>>>>> Linear mixed-effects model fit by REML
>>>>>> Data: NULL
>>>>>> AIC BIC logLik
>>>>>> 1144.193 1194.346 -559.0966
>>>>>>
>>>>>> Random effects:
>>>>>> Formula: ~1 | dish
>>>>>> (Intercept)
>>>>>> StdDev: 0.2334716
>>>>>>
>>>>>> Formula: ~1 | disk %in% dish
>>>>>> (Intercept) Residual
>>>>>> StdDev: 0.356117 1.079568
>>>>>>
>>>>>> Fixed effects: Diameter ~ Group * Lineage
>>>>>> Value Std.Error DF
>>>>>> t-value
>>>>>> p-value
>>>>>> (Intercept) 15.049722 0.2542337 270 59.19641
>>>>>> 0.0000
>>>>>> Group[T.NEDettol] 0.980556 0.3595407 20 2.72724
>>>>>> 0.0130
>>>>>> Lineage[T.First] -0.116389 0.3595407 20 -0.32372
>>>>>> 0.7495
>>>>>> Lineage[T.Fourth] -0.038056 0.3595407 20 -0.10584
>>>>>> 0.9168
>>>>>> Lineage[T.Second] -0.177500 0.3595407 20 -0.49369
>>>>>> 0.6269
>>>>>> Lineage[T.Third] 0.221111 0.3595407 20 0.61498
>>>>>> 0.5455
>>>>>> Group[T.NEDettol]:Lineage[T.First] 2.275000 0.5084674 20 4.47423
>>>>>> 0.0002
>>>>>> Group[T.NEDettol]:Lineage[T.Fourth] 0.955556 0.5084674 20 1.87929
>>>>>> 0.0749
>>>>>> Group[T.NEDettol]:Lineage[T.Second] 0.828333 0.5084674 20 1.62908
>>>>>> 0.1189
>>>>>> Group[T.NEDettol]:Lineage[T.Third] 0.721667 0.5084674 20 1.41930
>>>>>> 0.1712
>>>>>> Correlation:
>>>>>> (Intr) Gr[T.NED] Lng[T.Frs]
>>>>>> Lng[T.Frt]
>>>>>> Group[T.NEDettol] -0.707
>>>>>> Lineage[T.First] -0.707 0.500
>>>>>> Lineage[T.Fourth] -0.707 0.500 0.500
>>>>>> Lineage[T.Second] -0.707 0.500 0.500
>>>>>> 0.500
>>>>>> Lineage[T.Third] -0.707 0.500 0.500
>>>>>> 0.500
>>>>>> Group[T.NEDettol]:Lineage[T.First] 0.500 -0.707 -0.707
>>>>>> -0.354
>>>>>> Group[T.NEDettol]:Lineage[T.Fourth] 0.500 -0.707 -0.354
>>>>>> -0.707
>>>>>> Group[T.NEDettol]:Lineage[T.Second] 0.500 -0.707 -0.354
>>>>>> -0.354
>>>>>> Group[T.NEDettol]:Lineage[T.Third] 0.500 -0.707 -0.354
>>>>>> -0.354
>>>>>> L[T.S] L[T.T]
>>>>>> Grp[T.NEDttl]:Lng[T.Frs]
>>>>>> Group[T.NEDettol]
>>>>>> Lineage[T.First]
>>>>>> Lineage[T.Fourth]
>>>>>> Lineage[T.Second]
>>>>>> Lineage[T.Third] 0.500
>>>>>> Group[T.NEDettol]:Lineage[T.First] -0.354 -0.354
>>>>>> Group[T.NEDettol]:Lineage[T.Fourth] -0.354 -0.354 0.500
>>>>>> Group[T.NEDettol]:Lineage[T.Second] -0.707 -0.354 0.500
>>>>>> Group[T.NEDettol]:Lineage[T.Third] -0.354 -0.707 0.500
>>>>>> Grp[T.NEDttl]:Lng[T.Frt]
>>>>>> G[T.NED]:L[T.S
>>>>>> Group[T.NEDettol]
>>>>>> Lineage[T.First]
>>>>>> Lineage[T.Fourth]
>>>>>> Lineage[T.Second]
>>>>>> Lineage[T.Third]
>>>>>> Group[T.NEDettol]:Lineage[T.First]
>>>>>> Group[T.NEDettol]:Lineage[T.Fourth]
>>>>>> Group[T.NEDettol]:Lineage[T.Second] 0.500
>>>>>> Group[T.NEDettol]:Lineage[T.Third] 0.500 0.500
>>>>>>
>>>>>> Standardized Within-Group Residuals:
>>>>>> Min Q1 Med Q3 Max
>>>>>> -2.26060119 -0.70948250 0.03630884 0.69899536 3.42475990
>>>>>>
>>>>>> Number of Observations: 360
>>>>>> Number of Groups:
>>>>>> dish disk %in% dish
>>>>>> 30 90
>>>>>>
>>>>>> > anova(model)
>>>>>> numDF denDF F-value p-value
>>>>>> (Intercept) 1 270 39586.82<.0001
>>>>>> Group 1 20 145.07<.0001
>>>>>> Lineage 4 20 4.58 0.0087
>>>>>> Group:Lineage 4 20 5.27 0.0046
>>>>>>
>>>>>> This is closer to what I was expecting in terms of DF: 3 plates*5
>>>>>> lineages=15: 15 samples per group, 15-4(the numDF Lineage)=11,
>>>>>> 11-1(the numDF for Group)= 10 x 2 for the two groups/treatments = 20.
>>>>>> Hopefully I've worked that out correctly, and sombody could tell me
>>>>>> whether . Its' awkward because this experiment is unprecedented at my
>>>>>> uni, it was offered up by a teacher as a topic but then got dropped
>>>>>> due to lack of interest. As it's the first time, myself and my
>>>>>> supervisor were in many ways flying blind. If I remove the Lineage
>>>>>> main effect term, and include it as a random effect, leaving only
>>>>>> group as a fixed effect:
>>>>>> > anova(model2)
>>>>>> numDF denDF F-value p-value
>>>>>> (Intercept) 1 270 8041.429<.0001
>>>>>> Group 1 8 29.469 6e-04
>>>>>>
>>>>>> I get 8DF which by the same reasoning in the above model, is 5-1=4,
>>>>>> 4*2 = 8, so I take that as reassurance my working is correct. I'd
>>>>>> also
>>>>>> like to ask for opinion, on whether it would be advisable to actually
>>>>>> remove lineage as a fixed effect, and include lineage as a random
>>>>>> effect on the slope, rather than intersect which is what I've put all
>>>>>> the others as. I ask this because, whilst I feel whilst lineage might
>>>>>> seem a factor with informative levels( tha's how I first saw them), I
>>>>>> had no way of predicting which ones would show greatest or smallest
>>>>>> differences or how the five factor levels would interact and shape my
>>>>>> data, in that way the factor levels are not really all that
>>>>>> informative at all - they're just numbered as dish and disk are, and
>>>>>> their effects may even be different within my two groups - they don't
>>>>>> really allow any prediction in the same way a factor for different
>>>>>> types of fertiliser would in a plant study would for example, so I'm
>>>>>> thinking maybe it should be a random effect.
>>>>>>
>>>>>> Thank you very much to everyone that's replied to me and assisted me
>>>>>> with this, it's a tough learning curve, but I do think I'm beginning
>>>>>> to grasp how to use lme and lmer for my basic ends. Once I'm
>>>>>> confident
>>>>>> on the above, I'm next considering, whether to try an introduce some
>>>>>> weighting options to see what happens to a small amount of
>>>>>> heterscedacity I have between the two groups.
>>>>>>
>>>>>> Ben W.
>>>>>>>> I've used lmer and the function in languageR to extract p-values
>>>>>>>> without
>>>>>>>> it even mentioning df. Now if the lmer method with pvals.fnc()
>>>>>>>> makes it
>>>>>>>> so as I don't have to worry about these df then in a way it
>>>>>>>> makes my
>>>>>>>> issue a bit redundant. But it is playing on my mind a bit so felt I
>>>>>>>> should ask.
>>>>>>>>
>>>>>>>> My second question is about when I do the equivalent model using
>>>>>>>> lmer:
>>>>>>>> "lmer(Diameter~Group*Lineage+(1|Dish)+(1|Disk), data=Dataset)" -
>>>>>>>> which
>>>>>>>> I'm sure does the same because all my plots of residuals against
>>>>>>>> fitted
>>>>>>>> and such are the same, if I define it with the poisson family,
>>>>>>>> which
>>>>>>>> uses log, then I get a much lower AIC of about 45, compared to over
>>>>>>>> 1000
>>>>>>>> without family defined, which I think defaults to gaussian/normal.
>>>>>>> I don't think you should try to pick the family on the basis of
>>>>>>> AIC --
>>>>>>> you should pick it on the basis of the qualitative nature of the
>>>>>>> data.
>>>>>>> If you have count data, you should probably use Poisson (but you may
>>>>>>> want to add an observation-level random effect to allow for
>>>>>>> overdispersion.) If your response variable is Diameter, it is
>>>>>>> **not** a
>>>>>>> count variable, and you shouldn't use Poisson -- you should use an
>>>>>>> appropriately transformed response variable.
>>>>>> I've tried transforming my response variable in a few ways, like
>>>>>> natural log, sqrt, and (x/1) but they don't really seem to alter the
>>>>>> distribution or shape of my data at all.
>>>>>> Interestingly, if I look at the spread of the data by splitting the
>>>>>> response variable between the two groups, I see much more symmetry -
>>>>>> although still not a nice neat normal distribution, but in Biology
>>>>>> I've been taught never to expect one.
>>>>>>> And
>>>>>>>> my diagnostic plots still give me all the same patters, but just
>>>>>>>> looking
>>>>>>>> a bit different because of the family distribution specified. I
>>>>>>>> then
>>>>>>>> did
>>>>>>>> a model logging the response variable by using log(Diameter),
>>>>>>>> again, I
>>>>>>>> get the same diagnostic plot patterns, but on a different scale,
>>>>>>>> and I
>>>>>>>> get an AIC of - 795.6. Now normally I'd go for the model with the
>>>>>>>> lowest
>>>>>>>> AIC, however, I've never observed this beahviour before, and can't
>>>>>>>> help
>>>>>>>> but think thhat the shift from a posotive 1000+ AIC to a negative
>>>>>>>> one is
>>>>>>>> due to the fact the data has been logged, rather than that the
>>>>>>>> model
>>>>>>>> fitted to log data in this way is genuinley better.
>>>>>>>>
>>>>>>>> Finally, I saw in a text, an example of using lmer but "Recoding
>>>>>>>> Factor
>>>>>>>> Levels" like:
>>>>>>>> lineage<-Group:Lineage
>>>>>>>> dish<-Group:Lineage:Dish
>>>>>>>> disk<-Group:Lineage:Dish:Disk
>>>>>>>> model<-lmer(Diameter~Group+(1|lineage)+(1|dish)+(1|disk)
>>>>>>>>
>>>>>>>> However I don't see why this should need to be done,
>>>>>>>> considering, the
>>>>>>>> study was hieracheal, just like all other examples in that
>>>>>>>> chapter, and
>>>>>>>> it does not give a reason why, but says it does the same job as a
>>>>>>>> nested
>>>>>>>> anova, which I though mixed models did anyway.
>>>>>>> (1|lineage)+(1|dish)+(1|disk)
>>>>>>>
>>>>>>> is the same as
>>>>>>>
>>>>>>> (1|Lineage/Dish/Disk)
>>>>>>>
>>>>>>> (1|Dish) + (1|Disk) is **not** the same as (1|Dish/Disk), if
>>>>>>> Disk is
>>>>>>> not labeled uniquely (i.e. if Dishes are A, B, C, .. and Disks are
>>>>>>> 1, 2,
>>>>>>> 3, ... then you need Dish/Disk. If you have labeled Disks A1,
>>>>>>> A2, ...
>>>>>>> B1, B2, ... then the specifications are equivalent.
>>>>>>>
>>>>>>> For a linear mixed model (i.e. not Poisson counts) you should
>>>>>>> be able
>>>>>>> to run the same model in lmer and lme and get extremely similar
>>>>>>> results.
>>>>>>>
>>>>>>>> Hopefully sombody can shed light on my concerns. In terms of my
>>>>>>>> work
>>>>>>>> and
>>>>>>>> university, I could include what I've done here and be as
>>>>>>>> transparrant
>>>>>>>> as possible and discuss these issues, because log() of the data or
>>>>>>>> defining a distribution in the model is leading to the same
>>>>>>>> plots and
>>>>>>>> conclusions. But I'd like to make sure I come to term with what's
>>>>>>>> actually happening here.
>>>>>>>>
>>>>>>>> A million thanks,
>>>>>>>> Ben W.
>>>>>>>>
>>>>>>>>
>>>>>>>> lme14<- lme(Diameter~Group*Lineage,random=~1|Dish/Disk,
>>>>>>>> data=Dataset,
>>>>>>>> method="REML")
>>>>>>>>
>>>>>>>>> anova(lme14):
>>>>>>>> numDF denDF F-value p-value
>>>>>>>> (Intercept) 1 342 16538.253<.0001
>>>>>>>> Group 1 342 260.793<.0001
>>>>>>>> Lineage 4 342 8.226<.0001
>>>>>>>> Group:Lineage 4 342 9.473<.0001
>>>>>>>>
>>>>>>>>> summary(lme14)
>>>>>>>> Linear mixed-effects model fit by REML
>>>>>>>> Data: Dataset
>>>>>>>> AIC BIC logLik
>>>>>>>> 1148.317 1198.470 -561.1587
>>>>>>>>
>>>>>>>> Random effects:
>>>>>>>> Formula: ~1 | Dish
>>>>>>>> (Intercept)
>>>>>>>> StdDev: 0.1887527
>>>>>>>>
>>>>>>>> Formula: ~1 | Disk %in% Dish
>>>>>>>> (Intercept) Residual
>>>>>>>> StdDev: 6.303059e-05 1.137701
>>>>>>>>
>>>>>>>> Fixed effects: Diameter ~ Group * Lineage
>>>>>>>> Value Std.Error DF
>>>>>>>> t-value
>>>>>>>> p-value
>>>>>>>> (Intercept) 15.049722 0.2187016 342
>>>>>>>> 68.81396
>>>>>>>> 0.0000
>>>>>>>> Group[T.NEDettol] 0.980556 0.2681586 342
>>>>>>>> 3.65662
>>>>>>>> 0.0003
>>>>>>>> Lineage[T.First] -0.116389 0.2681586 342
>>>>>>>> -0.43403
>>>>>>>> 0.6645
>>>>>>>> Lineage[T.Fourth] -0.038056 0.2681586 342
>>>>>>>> -0.14191
>>>>>>>> 0.8872
>>>>>>>> Lineage[T.Second] -0.177500 0.2681586 342
>>>>>>>> -0.66192
>>>>>>>> 0.5085
>>>>>>>> Lineage[T.Third] 0.221111 0.2681586 342
>>>>>>>> 0.82455
>>>>>>>> 0.4102
>>>>>>>> Group[T.NEDettol]:Lineage[T.First] 2.275000 0.3792336 342
>>>>>>>> 5.99894
>>>>>>>> 0.0000
>>>>>>>> Group[T.NEDettol]:Lineage[T.Fourth] 0.955556 0.3792336 342
>>>>>>>> 2.51970
>>>>>>>> 0.0122
>>>>>>>> Group[T.NEDettol]:Lineage[T.Second] 0.828333 0.3792336 342
>>>>>>>> 2.18423
>>>>>>>> 0.0296
>>>>>>>> Group[T.NEDettol]:Lineage[T.Third] 0.721667 0.3792336 342
>>>>>>>> 1.90296
>>>>>>>> 0.0579
>>>>>>>> Correlation:
>>>>>>>> (Intr) Gr[T.NED] Lng[T.Frs]
>>>>>>>> Lng[T.Frt]
>>>>>>>> Group[T.NEDettol] -0.613
>>>>>>>> Lineage[T.First] -0.613 0.500
>>>>>>>> Lineage[T.Fourth] -0.613 0.500 0.500
>>>>>>>> Lineage[T.Second] -0.613 0.500 0.500
>>>>>>>> 0.500
>>>>>>>> Lineage[T.Third] -0.613 0.500 0.500
>>>>>>>> 0.500
>>>>>>>> Group[T.NEDettol]:Lineage[T.First] 0.434 -0.707 -0.707
>>>>>>>> -0.354
>>>>>>>> Group[T.NEDettol]:Lineage[T.Fourth] 0.434 -0.707 -0.354
>>>>>>>> -0.707
>>>>>>>> Group[T.NEDettol]:Lineage[T.Second] 0.434 -0.707 -0.354
>>>>>>>> -0.354
>>>>>>>> Group[T.NEDettol]:Lineage[T.Third] 0.434 -0.707 -0.354
>>>>>>>> -0.354
>>>>>>>> L[T.S] L[T.T]
>>>>>>>> Grp[T.NEDttl]:Lng[T.Frs]
>>>>>>>> Group[T.NEDettol]
>>>>>>>> Lineage[T.First]
>>>>>>>> Lineage[T.Fourth]
>>>>>>>> Lineage[T.Second]
>>>>>>>> Lineage[T.Third] 0.500
>>>>>>>> Group[T.NEDettol]:Lineage[T.First] -0.354 -0.354
>>>>>>>> Group[T.NEDettol]:Lineage[T.Fourth] -0.354 -0.354 0.500
>>>>>>>> Group[T.NEDettol]:Lineage[T.Second] -0.707 -0.354 0.500
>>>>>>>> Group[T.NEDettol]:Lineage[T.Third] -0.354 -0.707 0.500
>>>>>>>> Grp[T.NEDttl]:Lng[T.Frt]
>>>>>>>> G[T.NED]:L[T.S
>>>>>>>> Group[T.NEDettol]
>>>>>>>> Lineage[T.First]
>>>>>>>> Lineage[T.Fourth]
>>>>>>>> Lineage[T.Second]
>>>>>>>> Lineage[T.Third]
>>>>>>>> Group[T.NEDettol]:Lineage[T.First]
>>>>>>>> Group[T.NEDettol]:Lineage[T.Fourth]
>>>>>>>> Group[T.NEDettol]:Lineage[T.Second] 0.500
>>>>>>>> Group[T.NEDettol]:Lineage[T.Third] 0.500 0.500
>>>>>>>>
>>>>>>>> Standardized Within-Group Residuals:
>>>>>>>> Min Q1 Med Q3 Max
>>>>>>>> -2.47467771 -0.75133489 0.06697157 0.67851126 3.27449064
>>>>>>>>
>>>>>>>> Number of Observations: 360
>>>>>>>> Number of Groups:
>>>>>>>> Dish Disk %in% Dish
>>>>>>>> 3 9
>>>>>>>>
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