Dear All,
I did this :
> v = wool:tension
> spreadLevelPlot(breaks ~ v)
LowerHinge Median UpperHinge Hinge-Spread
B:H 15 17 21 6
A:M 18 21 30 12
A:H 18 24 28 10
B:M 21 28 39 18
B:L 20 29 31 11
A:L 26 51 54 28
Suggested power transformation: -0.2756176
So the ans is approximately = -.2
My query is:corresponding to p , what is the corresponding power
transformation defined as? Is it x^p or (x^p -1) / p ?
Which one of the following is the final transformation ? Please clarify.
> boxplot(breaks ^ -.2 ~v)
> boxplot((breaks^-.2 -1)/(-.2)~v)
>
Best Regards,
Ashim
On Sun, Jan 7, 2018 at 10:59 AM, Ashim Kapoor <ashimkap...@gmail.com> wrote:
> Dear All,
>
> we need to do :
>
> library(car) for the spreadLevelPlot function
>
> I forgot to say that.
>
> Apologies,
> Ashim
>
> On Sun, Jan 7, 2018 at 10:37 AM, Ashim Kapoor <ashimkap...@gmail.com>
> wrote:
>
>> Dear All,
>>
>> I want a transformation which will make the spread of the response at
>> all combinations
>> of 2 factors the same.
>>
>> See for example :
>>
>> boxplot(breaks ~ tension * wool, warpbreaks)
>>
>> The closest I can do is :
>>
>> spreadLevelPlot(breaks ~tension , warpbreaks)
>> spreadLevelPlot(breaks ~ wool , warpbreaks)
>>
>> I want to do :
>>
>> spreadLevelPlot(breaks ~tension * wool, warpbreaks)
>>
>> But I get :
>>
>> > spreadLevelPlot(breaks ~tension * wool , warpbreaks)
>> Error in spreadLevelPlot.formula(breaks ~ tension * wool, warpbreaks) :
>> right-hand side of model has more than one variable
>>
>> What is the corresponding appropriate function for 2 factors ?
>>
>> Many thanks,
>> Ashim
>>
>
>
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