Thank you for the clarification.
On Sat, Oct 23, 2021 at 11:42 PM Bert Gunter wrote:
>
> You appear to be bombarding the list with statistics questions. Please note
> per the posting guide linked below:
>
> "Questions about statistics: The R mailing lists are primarily intended for
> questions
Thank you. Sorry for the fuzziness of the question but I find it
difficult to give a proper definition of the problem. I have given a
graphical rendering on this post
https://www.researchgate.net/post/How_to_find_95_CI_of_a_matrix_of_classification_data
As you can see in the figure, there are dots
Hi Luigi,
I may be missing the point, but:
matrix((z1+z2+z3)/3,ncol=10)
gives you the mean rating for each item, and depending upon what
distribution you choose, the confidence intervals could be calculated
in much the same way.
Jim
On Sun, Oct 24, 2021 at 7:16 AM Luigi Marongiu wrote:
>
>
You appear to be bombarding the list with statistics questions. Please note
per the posting guide linked below:
"*Questions about statistics:* The R mailing lists are primarily intended
for questions and discussion about the R software. However, questions about
statistical methodology are
Hello,
I am using the ksvm function from the library kernlab to generate an
SVM classification. I am running the model with k-mean
cross-validation, thus obtaining different accuracy.
Is it possible to merge the different models obtained with the
separate data set to generate a kind of median
Hello,
I have a series of classifications of the same data. I saved this
classification in a single dataframe (but it could be a list). X and Y
are the variable and Z is the classification by three raters. `I` is
the individual identifier of each entry:
```
z1 = c(0,0,0,0,0,1,0,0,0,2,
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