# [R] Help interpreting data

Chris Ryan cry@n @end|ng |rom b|ngh@mton@edu
Thu Apr 4 18:12:08 CEST 2019

```Your dissertation advisor would probably be the best place to start.

Chris Ryan
--
Sent from my Android device with K-9 Mail. Please excuse my brevity.

On April 3, 2019 7:31:59 PM EDT, Matty A <matirimi using gmail.com> wrote:
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>---------- Forwarded message ----------
>From: matty <matirimi using gmail.com>
>To: r-help using r-project.org
>Cc:
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>Date: Wed, 3 Apr 2019 18:21:28 -0500 (CDT)
>Subject: res intepretation help
>Hi.. im a complete novice and am using R for my dissertation which is a
>meta-analysis... im using res would like some advice on
>understanding my results and what I can do with them.. the original
>data is
>h' values (linearity of social hierarchies) i have 11
>samples with 2 h' values that i am looking at my each res code has only
>one
>h' value the h' value is places where r would be in the code... below
>is a
>few of mu outputs.... thank you
>
>
>> res(1, var.r=NULL, 6, level=95, dig = 4, verbose = TRUE, id=NULL,
>data =
>> NULL) –
>Bonanni et al 2017 La Rustica   Free-ranging submission
>Mean Differences ES:
>
> d [ 95 %CI] = Inf [ NaN , NaN ]
>  var(d) = NaN
>  p-value(d) = NaN
>  U3(d) = 100 %
>  CLES(d) = 100 %
>  Cliff's Delta = 1
>
> Correlation ES:
>
> r [ 95 %CI] = 1 [ NaN , NaN ]
>  var(r) = 0
>  p-value(r) = 0
>
> z [ 95 %CI] = Inf [ Inf , Inf ]
>  var(z) = 0.3333
>  p-value(z) = 0
>
> Odds Ratio ES:
>
> OR [ 95 %CI] = Inf [ NaN , NaN ]
>  p-value(OR) = NaN
>
> Log OR [ 95 %CI] = Inf [ NaN , NaN ]
>  var(lOR) = NaN
>  p-value(Log OR) = NaN
>
> Other:
>
> NNT = 1.25
> Total N = 6
>> res(0.860, var.r=NULL, 6, level=95, dig = 4, verbose = TRUE, id=NULL,
>data
>> = NULL) –
>Bonanni et al 2017 La Rustica Free-ranging aggression
>Mean Differences ES:
>
> d [ 95 %CI] = 3.3706 [ -1.135 , 7.8762 ]
>  var(d) = 3.0722
>  p-value(d) = 0.1125
>  U3(d) = 99.9625 %
>  CLES(d) = 99.1423 %
>  Cliff's Delta = 0.9828
>
> Correlation ES:
>
> r [ 95 %CI] = 0.86 [ -0.1885 , 0.9923 ]
>  var(r) = 0.0136
>  p-value(r) = 0.0752
>
> z [ 95 %CI] = 1.2933 [ -0.1908 , 2.7775 ]
>  var(z) = 0.3333
>  p-value(z) = 0.0752
>
> Odds Ratio ES:
>
> OR [ 95 %CI] = 451.964 [ 0.1276 , 1600648 ]
>  p-value(OR) = 0.1125
>
> Log OR [ 95 %CI] = 6.1136 [ -2.0587 , 14.2859 ]
>  var(lOR) = 10.1071
>  p-value(Log OR) = 0.1125
>
> Other:
>
> NNT = 1.259
> Total N =
>
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