Thread,
When discussing statistical tests with my students, I emphasize clarity.
I have noticed that once they clearly define what they mean by
"compare," the method becomes obvious to them.
In this question, it is very easy to compare 84.42477 and 76.86037 --
the first is larger than the second -- however, this is really not the
question being asked. What is the question being asked? Is the first
(statistically) significantly larger than the second? Well, what is the
standard error at those points? Is the first (practically) significantly
larger than the second? That depends on the goals of the research. Both
questions are important. Neither question can be answered by the
question posed.
~ Ole
On 3/31/2012 9:01 PM, AbouEl-Makarim Aboueissa wrote:
What I need is to compare these predicted values on both lines:
predict(m1, newdata = data.frame(x = x))
1 2 3 4 5
84.42477 84.26280 81.34729 60.29085 27.89632
predict(m2, newdata = data.frame(x = x))
1 2 3 4 5
76.86037 76.69170 73.65554 51.72772 17.99261
for example: compare 84.42477 versus 76.86037 , and 84.26280 versus
76.69170 and so on.
with many thanks
abou
==========================
AbouEl-Makarim Aboueissa, Ph.D.
Associate Professor of Statistics
Graduate Program Coordinator
Department of Mathematics& Statistics
University of Southern Maine
96 Falmouth Street
P.O. Box 9300
Portland, ME 04104-9300
USA
Tel: (207) 228-8389
Fax: (207) 780-5607
Email: [email protected]
[email protected]
Office: 301C Payson Smith
"R. Michael Weylandt"<[email protected]> 3/31/2012 9:13 PM>>>
Again, you need to clarify what you mean by compare, but here's a
better way to do predict from a linear model -- my first guess was
simple interpolation (not very good):
m1<- lm(y ~ x, data = data.frame(y = y1, x = x1))
m2<- lm(y ~ x, data = data.frame(y = y2, x = x2))
predict(m1, newdata = data.frame(x = x))
predict(m2, newdata = data.frame(x = x))
Perhaps you want an ANOVA (?anova) test to compare the models: if you
want to compare the predicted values at x directly, you could use
interval="predict" for the different models and see if they overlap at
a given confidence level.
Michael
PS -- Please do cc the list: I'm not a statistics expert by any means
and I'd hate for what I say to go unreviewed by actual experts.
On Sat, Mar 31, 2012 at 8:19 PM, AbouEl-Makarim Aboueissa
<[email protected]> wrote:
I need to compare the value of y1-hat versus y2-hat corresponding to these
values of x<-c(0.10,0.20,2.0,15.0,35.0)
e.g. for x=0.10 compare y1-hat(x=0.10) versus y2-hat(x=0.10) and so on.
thanks
abou
"R. Michael Weylandt"<[email protected]> 3/31/2012 8:10 PM>>>
? approxfun
f1<- approxfun(x1,y1)
f2<- approxfun(x2,y2)
f1(x) - f2(x)
But I'm not sure what you mean be compare...do you have a particular
test in mind?
Michael
On Sat, Mar 31, 2012 at 8:04 PM, AbouEl-Makarim Aboueissa
<[email protected]> wrote:
Dear All:
I am trying to compare points on two regression lines given some specific
values of X. I am not sure how to do it in R and/or in SAS. Can someone help
me, and show me how to conduct such test in R.
You can use the following data for both lines.
### data for line 1:
### =========
x1<-c(0.0,0.25,2.5,25.0,50.0)
y1<-c(100.0,79.0,74.0,36.0,8.0)
### data for line 2:
### =========
x2<-c(0.0,0.25,2.5,25.0,50.0)
y2<-c(100.0,79.0,55.0,18.0,2.0)
### these are the given values of X:
### ===================
x<-c(0.10,0.20,2.0,15.0,35.0)
thank you very much in advance
abou
==========================
AbouEl-Makarim Aboueissa, Ph.D.
Associate Professor of Statistics
Graduate Program Coordinator
Department of Mathematics& Statistics
University of Southern Maine
96 Falmouth Street
P.O. Box 9300
Portland, ME 04104-9300
USA
Tel: (207) 228-8389
Fax: (207) 780-5607
Email: [email protected]
[email protected]
Office: 301C Payson Smith
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