RE: Factor analysis - which package is best for Windows?

2001-09-06 Thread Magill, Brett

MVA comes with R base.  However, it is a seperate library.  Libraries that
are not sent with base are available in Windows binaries on CRAN, but you do
not have to worry about that for MVA.

Type:

library()

and you will get a list of the available packages.  To make MVA available
(i.e. load it), type:

library(mva)

then you can ask for, for example:

help (factanal)



-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, September 05, 2001 5:42 PM
To: [EMAIL PROTECTED]
Subject: Re: Factor analysis - which package is best for Windows?


[EMAIL PROTECTED] (Magill, Brett) wrote in message
news:[EMAIL PROTECTED]...
 Also check out R, a GNU implementation of the S language, most prominently
 known through its use in S-Plus.  R is a fully featured statisitical
 programming environment.  In its MVA (Multivariate) package, it includes
 routines for factor analysis using maximum liklihood estimation with
varimax
 and promax rotations.
 

I have installed R1.3.0 on  my Windows system and have noted that MVA
is an add-on.  The FAQ tells how to obtain these add-ons but only for
UNIX.  Is this add-on actually available for Windows?  If so, how do I
obtain it?

Thanks,
Peter


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RE: Factor analysis - which package is best for Windows?

2001-08-30 Thread Magill, Brett

Also check out R, a GNU implementation of the S language, most prominently
known through its use in S-Plus.  R is a fully featured statisitical
programming environment.  In its MVA (Multivariate) package, it includes
routines for factor analysis using maximum liklihood estimation with varimax
and promax rotations.

R is open-source, which means that it is frequently updated and, most
importantly, it can be downloaded free of charge.  The only downside (to
some) is that at this stage of its development R is completely
command-prompt driven.  However, I find the R language intuitive and easy to
learn.

http://www.r-project.org


-Original Message-
From: Aron Landy [mailto:[EMAIL PROTECTED]]
Sent: Thursday, August 30, 2001 6:33 AM
To: [EMAIL PROTECTED]
Subject: Re: Factor analysis - which package is best for Windows?


I have tried it and it is amazing. A bargain ;)


Richard Wright [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 KyPlot runs under Windows, is freeware and gives you several factor
 analysis algorithms to choose from.

 http://www.rocketdownload.com/Details/Math/kyplot.htm







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RE: book on elaboration and regression

2001-05-29 Thread Magill, Brett

Rosenberg, Morris. 1968. The Logic of Survey Analysis. New York: Basic Books

An older book, but nice treatment of the elaboration model using tables.
Might be hard to find now however.  I think it is in the process of being
updated by another author.  


-Original Message-
From: John Hendrickx [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, May 29, 2001 10:12 AM
To: [EMAIL PROTECTED]
Subject: book on elaboration and regression


I'll be giving a course next fall on the basics of multivariate analysis 
and I'm looking for some texts. The students will be familiar with basic 
descriptive statistics, correlation and regression. I want to teach them 
how other variables can affect such bivariate relationships, e.g. 
spurious, interpreted, suppressor, interaction effects. That's basically 
it: how a third variable can affect an observed bivariate relationship 
and how this should be taken into account in the conceptual model and the 
analysis. This principle will be illustrated using crosstables, 
percentages and association measures on the one hand and regression 
models on the other.

Now comes the question: can anyone suggest some good texts on these 
subjects? I'm looking for a text on elaboration using tables on the one 
hand and on elementary linear regression on the other. The course is for 
students of International Management at the University of Nijmegen in the 
Netherlands. It'll be taught in English though, but books related to 
business and economics are preferred. 

Advance thanks for any help,
John Hendrickx 


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RE: Question

2001-05-11 Thread Magill, Brett

Don and Dennis,

Thanks for your comments, I have some points and futher questions on the
ussue below.

For both Dennis and Don:  I think the option of aggregating the information
is a viable one.  Yet, I cannot help but think there is some way to do this
taking into account the fact that there is variation within organizations.
I mean, if I have a organizational salary mean of .70 (70%) with a very tiny
s.d. it is different than a mean of .70 with a large s.d.  Should be some
way to account for this.  In addition, the problems with aggregation are
well documented and I believe in gereneral suggest that aggregated results
overestimate relationships.


Don:  I suggested that the problem was not a traditional multilevel problem.
Perhaps I am wrong, but here is where I thought the difference was.
Typically, say in a classroom problem, I want to assess the effect of
classroom characterisitcs (student/teacher ratio, teacher experience, etc.)
which are constant within classrooms on say student performance, which
varies within classroom across individuals.  The difference between this and
the problem I presented is that the OUTCOME is a contextual variable.  That
is, rather than individual-level varaition, the outcome caries only at the
organizational level.  Perhaps this can be modeled with MLMs, but it is
certainly different than the typical problem.

With regard to independence, I am talking about the independence of the
X2's.  That is X2-1 is not independent of X2-2 and X2-4 is not independent
of X2-5.  This is because these cases come from the same organization.  So,
if we simply regressed Y~X2, not accounting for X1 in the model, this causes
problems for ANOVA and regression, the GLM family more generally.  The lack
of independence here is exactly the reason for repeated measures and MLM
more generally, no?

Perhaps I am making to much of the issue, but the data structure is one that
I have not encountered before and I found it something of an interesting and
challenging problem, just hoping I might learn something along the way.
Would appreciate any comments on my comments above.

Oh, and just so there is no confusion, the data below I constructed.  It
reflects that structure of the data and nature of the relatinoship, but I
generated this data set.  In addition, the real thing does include variables
such as tenure, previous experience, etc. that are also used as covariates
at the individual level.  Of course, this also means that these would need
be aggregated as well if that approach is taken.

Best

 IDX1  X2  Y
 1 1   0.700.40
 2 1   0.800.40
 3 1   0.650.40
 4 2   1.200.25
 5 2   1.100.25
 6 3   0.900.30
 7 4   0.500.50
 8 4   0.600.50
 9 4   0.700.50


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Question

2001-05-10 Thread Magill, Brett

A colleague has a data set with a structure like the one below:

ID  X1  X2  Y
1   1   0.700.40
2   1   0.800.40
3   1   0.650.40
4   2   1.200.25
5   2   1.100.25
6   3   0.900.30
7   4   0.500.50
8   4   0.600.50
9   4   0.700.50

Where X1 is the organization.  X2 is the percent of market salary an
employee within the organization is paid--i.e. ID 1 makes 70% of the market
salary for their position and the local economy.  And Y is the annual
overall turnover rate in the organization, so it is constant across
individuals within the organization.  There are different numbers of
employee salaries measured within each organization. The goal is to assess
the relationship between employee salary (as percent of market salary for
their position and location) and overall organizational turnover rates.

How should these data be analyzed?  The difficulty is that the data are
cross level.  Not the traditional multi-level model however.  That there is
no variance across individuals within an organization on the outcome is
problematic.  Of course, so is aggregating the individual results.  How can
this be modeled both preserving the fact that there is variance within
organizations and between organizations.  I suggested that this was a
repeated measures problem, with repeated measurements within the
organization, my colleague argued it was not. Can this be modeled
appropriately with traditional regression models at the individual level?
That is, ignoring X1 and regressing Y ~ X2.  It seems to me that this
violates the assumption of independence.  Certainly, the percent of market
salary that an employee is paid is correlated between employees within an
organization (taking into account things like tenure, previous experience,
etc.).

Thanks


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Statistical Notation

2001-04-20 Thread Magill, Brett

Does anyone know of a resource that lists symbols often used in statistics
and probability.  What I am looking for is something with the symbol, its
name, and some common uses. In particular, I would like web sources, but I
would be grateful for any suggestions.

Best,

Brett  


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FW: Regression with repeated measures

2001-02-28 Thread Magill, Brett

These both sound to me as if multi-level models would be appropriate to
handle the type of data to which you are referring.  

Look at this site for some basic info on multi-level models (MLM):

http://www.ioe.ac.uk/multilevel/ 


Interested in learning more... then dowload this classic text on MLM for
free:

http://www.arnoldpublishers.com/support/goldstein.htm


Finally, If you decide this method is what you are looking for, then have a
look at the following text that describes Linear MLM or as they call it
Hierarchichal Linear Models (HLM)--the multilevel equivalent of linear
regression:

Bryk,A.S., and Raudenbush,S.W. (1992). Hierarchical Linear Models. Newbury
Park, Sage.


-Original Message-
From: Rich Strauss [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 28, 2001 2:40 PM
To: [EMAIL PROTECTED]
Subject: Re: Regression with repeated measures


I don't have an answer, but I'm very glad this question was asked because
I'm having a similar problem.  I have 14 grids, values from which are to be
used as the dependent variable in a regression.  Each 6x6 grid consists of
36 observation points.  Their are some fairly strong spatial correlations
among the values at each grid, so I certainly can't treat them as if they
were independent, yet reducing each grid to a single mean value (the other
extreme) seems like a foolish waste of power.  I'm trying to figure out how
to use all of the observations, but also use the estimated spatial
autocorrelations to weight them in the regression.  (The design was
originally created to answer a very different question, which is how I got
into this mess.)

I hope that there's a single answer to both of our questions.

Rich Strauss

At 10:54 AM 2/28/01 -0600, Michael M. Granaas wrote:

I have a student coming in later to talk about a regression problem.
Based on what he's told me so far he is going to be using predicting
inter-response intervals to predict inter-stimulus intervals (or vice
versa).

What bothers me is that he will be collecting data from multiple trials
for each subject and then treating the trials as independent replicates.
That is, assuming 10 tials/S and 10 S he will act as if he has 100
independent data points for calculating a bivariate regression.
 
Obviously these are not independent data points.
 
Is the non-independence likely to be severe enough to warrant concern?
 
If yes, is there some method that will allow him to get the prediction
equation he wants?
 
Thanks
 
Michael



Dr Richard E Strauss
Biological Sciences  
Texas Tech University   
Lubbock TX 79409-3131

Email: [EMAIL PROTECTED]  (formerly [EMAIL PROTECTED])
Phone: 806-742-2719
Fax: 806-742-2963 



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RE: Sample size question

2001-02-23 Thread Magill, Brett

G*Power is a powere analysis package that is freely available.  You can
download it at:

http://www.psychologie.uni-trier.de:8000/projects/gpower.html 

You can calculate a sample size for a given effect size, alpha level, and
power value. 


-Original Message-
From: Scheltema, Karen [mailto:[EMAIL PROTECTED]]
Sent: Friday, February 23, 2001 10:07 AM
To: [EMAIL PROTECTED]
Subject: Sample size question


Can anyone point me to software for estimating ANCOVA or regression sample
sizes based on effect size?

Karen Scheltema
Statistician
HealthEast
Research and Education
1700 University Ave W
St. Paul, MN 55104
(651) 232-5212   fax (651) 641-0683
[EMAIL PROTECTED]



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RE: NY Times on statisticians' view of election

2000-11-17 Thread Magill, Brett

It has created controversy, as witnessed by the replies it has
generated, therefore it is controversial.


I am not sure why the results that were presented need to be terribly
controversial.  Democratic supporters tend to be minority, older, poorer,
and less educated than their republican counterparts.  I would suggest
(perhaps revealing my bias) that this is because the democratic party has
done more to protect the interests of disenfranchised groups.  On the other
hand, the republican party leans toward favoring the wealthy and business
interest who are normatively better educated, white, and definition of
higher economic status.  

Even if these broad generalizations are not the case generally, it is
certainly true that the stage has been set in this manner during this
presidential campaign.  If this influenced voters, it makes sense that we
would find these demographic differences in the presidential vote.

It would be interesting to see the means presented previously with these
demographic characteristics controlled.  I cannot imagine that there would
be differences.  I do not believe anyone truly believes that party
affiliation somehow affects literacy.  On the other hand, other
characteristics associated with literacy (education, economic status, age,
race) tend to influence party selection.  Thus, I suggest that literacy
problems that manifest in ballots inherently favor republicans and in doing
so wholly disenfranchise a large number of already disadvantaged voters (my
partisan statement).  


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RE: Polls: Errors on Prime Time - NOT AN ERROR

2000-11-13 Thread Magill, Brett

I believe the issue is that the questionable balloting method was used in
predominantly democratic districts and therefore disproportionately affected
democratic voters, i.e. Gore supporters.  Furthermore some have argued that
they did in fact ask for a new ballot, which was denied.

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Monday, November 13, 2000 11:58 AM
To: [EMAIL PROTECTED]
Subject: Re: Polls: Errors on Prime Time - NOT AN ERROR


Are you saying that only Gore supporters could not figure out the
ballot?  Plus, only Gore voters were too timid to ask for assistance
or for a new ballot?  :-))   Could it be that they are complaining ex
post facto when confronted with an unpopular result?  :-) Apparently,
upon leaving the polling place, they at first must have "misled" the
exit pollsters!  Only later after the polls closed they remembered
their "error."  Hm.

On Mon, 13 Nov 2000 08:42:36 -0500, SSCHEINE [EMAIL PROTECTED] wrote:

Actually, the exiting polls got it right!!! Remember, a lot of people
left the polling booth thinking that they had voted for Gore, when they
had actually messed up their ballot. Based on who they thought that they
had voted for, they informed the exit pollers who called it for Gore.

Sam 
 
**
Samuel M. Scheiner
Div. Envir. Biol. (Rm 635) National Science Foundation
4201 Wilson Blvd.  Arlington, VA 22230
Tel: 703-292-7189  Fax: 703-292-9065
Email: [EMAIL PROTECTED]



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RE: 120 subjects on 120 occassion: a model ?

2000-10-13 Thread Magill, Brett

I don't know enough about time series really to provide much advice.
However, I have seen methods by which a slope was calculated across time for
each subject with the first measurement as the incercept (within subjects).
Subsequently, the individual slope was regressed on other factors.  Thus,
answering the question what factors (X) influence the rate of
change/direction across time in Y.



-Original Message-
From: MJ Ray [mailto:[EMAIL PROTECTED]]
Sent: Friday, October 13, 2000 8:15 AM
To: [EMAIL PROTECTED]
Subject: Re: 120 subjects on 120 occassion: a model ?


"Gaj Vidmar" [EMAIL PROTECTED] writes:
 there seems to be no word from professional statisticians yet, so here's
an
 addenum.

This message was posted in many places, so presumably we will get a
summary of responses if we care?

My own suggestion (mangled by a bad emailer) was to use vector time
series methods, but this could lead to a fairly large computation
probelm without extra information.  I wasn't able to recommend a very
good specialist reference off the top of my head, though.

MJR


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Nested Models and HLM

2000-09-08 Thread Magill, Brett

Is there a difference between a Nested Model in general and what is referred
to as a hierarchical linear model?

Thanks,

Brett Magill


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Regression and Correlation (Was Correlation)

2000-05-19 Thread Magill, Brett

I am no statistician, so let me make sure I am understanding what you are
saying.  Your point is that you may have an identical regression equation
despite the fact that the correlation may vary depending on the amount of
variation in X.  If this is your point, I agree and recognize this--r is a
measure of the fit about the regression line.

Nonetheless, regression and correlation are the same in the bivariate case
with the exception of scale.  In a bivariate regression, the standardized
Beta coefficient is equal to the Pearson r.  As with any standardization, it
removes the scale of the variation and the result is that the slope
describes the relationship or B = r.

Brett

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Friday, May 19, 2000 11:43 AM
To: [EMAIL PROTECTED]
Subject: Re: Correlation


Magill, Brett [EMAIL PROTECTED] wrote:
Mike,

In the bivariate case, regression and correlation are identical.

This is false.  Correlation is the measure of the
proportion of the variance of one variable explained by a
linear function of the other in a joint distribution, while
linear regression is the linear relation itself.  One can
have non-linear versions as well.

If in fact E(Y|X) = aX + b, this will also be the case no
matter how selection is made on X, whereas the correlation
can vary greatly.


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RE: Correlation

2000-05-17 Thread Magill, Brett

Mike,

In the bivariate case, regression and correlation are identical.  Assuming
you want to select one of your proxy measures to use in place of the
expensive 'true" measure, run the regression models--"true" measure
regressed on each of the "different techniques". The r's that you will get
can be interpreted the same as the correlation coefficient you would
calculate.  Of course, r^2 is the coefficient of determination--the amount
of total variance in the dependent variable attributable to variation in the
independent variable.  Compare these across your proxy measures of the
"true" score and pick the best one.  You also then have your prediction
model.  Model fit and the like can be assessed in the typical manner for
regression. 

-Original Message-
From: mbattagl [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, May 17, 2000 12:02 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Correlation


I have data that measures light intensity with a number of different 
techniques.  One of the measurements (a direct measurement and "true" 
measurement of light intensity) involves lots of time, labor, and expense.  
The other techniques are more practical in the sense of time and labor, but 
are indirect measurements (based on canopy structure (density, location of 
holes in the canopy, etc).  My goal is to determine if the indirect 
measurements are valid estimates of the direct measurements.  However, I
would 
also like to predict light intensity based on the indirect methods.

I can see two methods of analysis for this situation: correlation and 
regression.  It seems that correlation would be the best option to validate 
the measurements to each other.  If the measurements are correlated then the

use of regression analysis would yield a prediction equation.

For the correlation analysis, I can use Pearson or Spearman analysis.  To
use 
Pearson, the variables should be normally distributed.  However, I have read

that the distribution for correlation should be bivariate normally 
distributed. I understand how to test for normality with a univariate normal

distribution but have no idea how to test for bivariate normal distribution.
I 
am using the SAS program to do my analysis.  Does anyone know how to test
for 
bivariate normal distribution?

If the variables are bivariate normally distributed then I use Pearson, but
if 
they are not normally distributed I use Spearman.  Is this correct?

The regression analysis is also somewhat confusing.  Regression analysis is 
based on the fact that the Y (dependent variable) is random and the X 
(independent variable) is fixed with no error.  For my case, both X and Y
are 
random and have some measurement error.  Is it correct to use simple linear 
regression for this analysis or is there another type of analysis to obtain 
predictions?

I apologize for such a long post, but I have been struggling with this 
analysis for sometime and the more information I obtain from Statistics
books, 
the more confused I get.

Thanks in advance, Mike



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RE: Blackjack problem

2000-04-28 Thread Magill, Brett

Clip from earlier message...

"The Player may choose to play exactly the same rules
as the Dealer is REQUIRED to play; or the Player may choose some of the
other
options. Since the Player has more choices or options in play than does the
Dealer, why does the Dealer have the statistical advantage?  It seems to me
the
Player would have the advantage."



Doesn't the law of large numbers figure in here somewhere too:

1.  The probability of winning with the house strategy is known a priori and
it is optimal (as someone else pointed out).
2.  An individual playing with this same strategy may win or lose more or
less in the short run.
3.  With the volume of games the house plays, the empirical probability will
approach the a priori probability in the long run--to the house's advantage.

Simplistic and poorly articulated I am sure, but I think it captures the
essence of the mechanism at work here.


"The Player may choose to play exactly the same rules
as the Dealer is REQUIRED to play; or the Player may choose some of the
other
options. Since the Player has more choices or options in play than does the
Dealer, why does the Dealer have the statistical advantage?  It seems to me
the
Player would have the advantage."


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RE: Power for Pilot Studies

2000-03-17 Thread Magill, Brett

Seems to me that the notion of power in a pilot study is moot.  Typically, a
pilot study is a test of the research methodology and instruments.  As such,
your sample size is a pragmatic decision and should consist of enough
observations to test your design and research instruments. For instance, if
you are planning to assess the psychometric properties of an instrument that
you are using, you need enough observations to meet the requirements of the
statistical tests you chose, such as Cronbach's Alpha or a Principle
Components.  Power has nothing to do with this decision.   


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FW: could someone help me with this intro to stat. problem

1999-12-08 Thread Magill, Brett

Mike,

With randomization pre, it is not necessary to take a pre-intervention
measurement. Test the difference in confidence following the training.  If
it is significant, there is a difference.  Decide what direction it is in
and attribute the difference to the training. You can make this attribution
because of random assignment even without pre-measure.

-Original Message-
From: Mike Wogan [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, December 08, 1999 2:16 PM
To: Luv 2 muah 143
Cc: [EMAIL PROTECTED]
Subject: Re: could someone help me with this intro to stat. problem


On 8 Dec 1999, Luv 2 muah 143 wrote:

 5 of 10 volunteers are randomly selected to receive self-defense training.
The
 other 5 receive no training.  At the end of the training period, all
subjects
 complete a self-confidence questionnaire.  
 
 a.)  Is there a difference in self-confidence between the 2 groups
(p.01)?
 
 
 b.)  What are the effects of self-defense traing on self-confidence (I'm
 assuming a two-tailed test?).  Explain analysis
 
 Please help, I can't figure it out...my mind has gone blank

Without a pre-test measure of self-confidence, taken prior to the
training, even if there is a significant difference post-training, it's
not possible to tell whether the difference is the result of the training 
or was there to begin with.  

If there is a pre-post measurement of self-confidence, then you need a
mixed model Anova, with Training vs. No Training as the between groups
factor and Pre-Post as the within groups factor.

Mike