[R] summary(object, test=c("Roy", "Wilks", "Pillai", ....) AND ellipse(object, center=....)

[Prof Brian Ripley]

Why do you keep posting something to which you have already received an 
answer?

https://stat.ethz.ch/pipermail/r-help/2008-April/158662.html
answered in
https://stat.ethz.ch/pipermail/r-help/2008-April/158664.html

and repeated at
https://stat.ethz.ch/pipermail/r-help/2008-April/158787.html
and
https://stat.ethz.ch/pipermail/r-help/2008-April/158796.html
(this message).

All of these are with unwrapped lines (I wrapped the version below) and 
contain HTML which the posting guide asked you not to send.

On Sat, 5 Apr 2008, Ray Haraf wrote:

> Dear All,
>
>  I would be very appreciative of your help with the following

But you then showed 0% appreciation of the help you were given!

>  1). I am running multivariate multiple regression through the manova() 
> function (kindly suggested by Professor Venables) and getting two 
> different answers for test=c("Wilks","Roy","Pillai") and 
> tests=c("Wilks","Roy",'"Pillai") as shown below. In the first case 
> (test=c(list)) I got error message which probably means I can only call 
> one test at a time. I thought I could get ride of this by adding "s" to 
> test; in this case (tests=c(list)), I got Pillai test. Does this mean 
> that Pillai would be the default test and summary(manova()) can only 
> post one test at a time?
>
>> summary(manova(cbind(y1, y2) ~ z1, data =
> + ex7.8),test=c("Wilks","Roy","Pillai"))
> Error in match.arg(test) : 'arg' must be of length 1
>> summary(manova(cbind(y1, y2) ~ z1, data =
> + ex7.8),tests=c("Wilks","Roy","Pillai"))
>          Df  Pillai approx F num Df den Df Pr(>F)
> z1         1  0.9375  15.0000      2      2 0.0625 .
> Residuals  3
> ---
> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
>
> 2). My next struggle is to construct prediction ellipse. Both ellipse() 
> and ellipse.lm() are not giving me the solution to "Sampling from 
> multivariate multiple regression prediction regions" posted by Iain 
> Pardoe, Mon May 9 18:43:46 2005. I am working on the same problem and 
> performed all the steps he suggested
>
>> ex7.10 <-
> + data.frame(y1 = c(141.5, 168.9, 154.8, 146.5, 172.8, 160.1, 108.5),
> + y2 = c(301.8, 396.1, 328.2, 307.4, 362.4, 369.5, 229.1),
> + z1 = c(123.5, 146.1, 133.9, 128.5, 151.5, 136.2, 92),
> + z2 = c(2.108, 9.213, 1.905, .815, 1.061, 8.603, 1.125))
>> attach(ex7.10)
>> f.mlm <- lm(cbind(y1,y2)~z1+z2)
>> y.hat <- c(1, 130, 7.5) %*% coef(f.mlm)
>> round(y.hat, 2)
>         y1 y2
> [1,] 151.84 349.63
>> qf.z <- t(c(1, 130, 7.5)) %*%
> + solve(t(cbind(1,z1,z2)) %*% cbind(1,z1,z2)) %*%
> + c(1, 130, 7.5)
>> round(qf.z, 5)
>        [,1]
> [1,] 0.36995
>> n.sigma.hat <- SSD(f.mlm)$SSD # same as t(resid(f.mlm)) %*%resid(f.mlm)
>> round(n.sigma.hat, 2)
>     y1 y2
> y1 5.80 5.22
> y2 5.22 12.57
>> F.quant <- qf(.95,2,3)
>> round(F.quant, 2)
> [1] 9.55
>
>
>> From here how could I calculate a 95% prediction ellipse for y=(y1,y2) 
>> at (z1,z2)=(130,7.5) using either ellipse or ellipse.lm? y1 would be 
>> the x-axis and y2, the y-axis. The center is different from (0,0) and I 
>> don't know what would be the appropriate x (the lm object). Should I 
>> used predicted values or residuals? In both cases I have vectors which 
>> is different from the example given with ellipse.lm
>
> 3). Lastly but not the least, would be too ambitious to draw the axes 
> (i.e, the eigenvalues) to the ellipse?
>
> Thanks and very kind regards,
> Ray
>
> 	[[alternative HTML version deleted]]
>
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-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595