Hi
I am attempting Anova analysis to compare results from four groups
(Samp1-4) which are lists of intensities from the experiment. I am
doing this by first creating a structured list of the data and then
conducting the ANOVA (Script provided below). Im an R beginner so am
not sure if I am using this correctly. Two major questions I have are:
1)
Is using the code (zzz.aov <- aov(Intensity ~ Group + Error(Sample),
data = zzzanova)) the correct method to calculate the variances between
the four groups (samp 1-4). I am unsure of the inclusion of the error
portion.
2) I beleive this method (aov) assumes equal variances. How can I adjust this to
do an ANOVA with unequal variances
#####SCRIPT STARTS
#Creates a structured list suitable for ANOVA analysis
# Intensity Group (1,2,3,4) Sample(1:62)
"zzzanova" <-
structure(list(Intensity = c(t(Samp1), t(Samp2), t(Samp3), t(Samp4)),
Group = structure(c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,
3,3,3,3,3,3,3,3,3,3,3,3,3,3,
4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4), .Label = c("Group1",
"Group2", "Group3", "Group4"), class =
"factor"),
Sample = structure(c( 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
30, 31, 32, 33, 34, 35, 36, 37, 38, 39,
40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62)
))
, .Names = c("Intensity",
"Group", "Sample"), row.names =
c("1", "2", "3", "4", "5",
"6", "7", "8", "9", "10",
"11", "12", "13", "14", "15",
"16", "17", "18", "19", "20",
"21", "22", "23", "24", "25",
"26", "27", "28", "29", "30",
"31", "32", "33", "34", "35",
"36", "37", "38", "39", "40",
"41", "42", "43", "44", "45",
"46", "47", "48", "49", "50",
"51", "52", "53", "54", "55",
"56", "57", "58", "59", "60",
"61","62"),class = "data.frame")
#Conducts the ANOVA for that PCI
zzz.aov <- aov(Intensity ~ Group + Error(Sample), data = zzzanova)
#####SCRIPT ENDS
THANKS IN ADVANCE
Your line of code: zzz.aov <- aov(Intensity ~ Group + Error(Sample), data = zzzanova) indicates that you are trying to do a repeated measures ANOVA, not just an ANOVA. The Error(Sample) term in your expression indicates that Sample is a within subjects factor, which I presume is not the case. The way I understood your data, the Sample column is just an id number for the subjects - is that right? If you just want to compare the means of Intensity for the the four Groups, then just do: zzz.aov <- aov(Intensity ~ Group, data = zzzanova) You may want to consult this resource for explicit examples of ANOVA in R: http://www.personality-project.org/R/r.anova.html Ravi Kulkarni -- View this message in context: http://n4.nabble.com/Help-with-ANOVA-in-R-tp1585992p1586936.html Sent from the R help mailing list archive at Nabble.com.
> > Your line of code: > > zzz.aov <- aov(Intensity ~ Group + Error(Sample), data = zzzanova) > > indicates that you are trying to do a repeated measures ANOVA, not just an > ANOVA. The Error(Sample) term in your expression indicates that Sample is a > within subjects factor, which I presume is not the case. > The way I understood your data, the Sample column is just an id number for > the subjects - is that right? > > If you just want to compare the means of Intensity for the the four Groups, > then just do: > > zzz.aov <- aov(Intensity ~ Group, data = zzzanova) > > You may want to consult this resource for explicit examples of ANOVA in R: > > http://www.personality-project.org/R/r.anova.html > > Ravi KulkarniNo, Error(Sample) does NOT indicate a repeated-measures ANOVA. This use of Error() is, in fact, correct for this between design, and labels the error partition, although not doing so will still return the same result with a generic name for the error partition. However, slightly more complicated designs (say, one between and one within factor) will *require* the Sample ID vector (as factor) to assign the partitions. [i.e., with s as the factor of IDs, and P as the repeated measure factor, and A as the between factor, then aov(dv~A*P+Error(s/P)) is the correct specification, although I prefer, and instruct my students, to expand explicitly in the aov call, what the call itself expands to: aov(dv~A+P+A:P+Error(s+s:P)), making it clear that there will be two partitions: s and s:P---s being used to test A, and s:P being used to test P and A:P]. -- Please avoid sending me Word or PowerPoint attachments. See <http://www.gnu.org/philosophy/no-word-attachments.html>
Hello everyone:
I'm a new member of this group.
Following the question(2) of Amit Patel-7,
I know "oneway.test" has the option of
"var.equal=F".
Maybe it's can be the answer of this question!
But, I have a question about
"oneway.test",
How can I get "SS", and "MS"
information from
"oneway.test"?
R code and the results are as follows.
Thank you very much. :)
> anova(lm(BackCalac~factor(Assay),data=Control))
Analysis of Variance Table
Response: BackCalac
Df Sum Sq Mean Sq F value Pr(>F)
factor(Assay) 4 270.846 67.711 56.219 1.345e-10 ***
Residuals 20 24.088 1.204
> oneway.test(BackCalac~factor(Assay), var.equal=T,data=Control)
One-way analysis of means
data: BackCalac and factor(Assay)
F = 56.2191, num df = 4, denom df = 20, p-value = 1.345e-10
> oneway.test(BackCalac~factor(Assay), var.equal=F,data=Control)
One-way analysis of means (not assuming equal variances)
data: BackCalac and factor(Assay)
F = 92.8834, num df = 4.000, denom df = 9.625, p-value = 1.165e-07
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