I am a graduate student in plant breeding, and am preparing a description of several varieties. Current conventions are to use an LSD test to identify which varieties are significantly different from a variety of interest. However, given the nature of the situation, we do know that a) all varieties have different yields, and similarly, b) one variety has the highest yield. What would be of interest to a farmer is the percent chance that a given variety has the highest yield in this set of varieties. Is there a test or method that can determine the percent chance that each variety is the highest yielding given our experimental data? I.e. there is a 70% chance that "Variety A" is the highest yielding, 20% chance for "Variety B", 8% for "Variety C", and 2% for "Variety D".
It may well be that there are much smarter solutions - here is just a quick thought:
One could bootstrap the mean values and see how often it happens that the group mean is lower than the bootstrapped mean. The more often this happens, the more likely the group mean is the highest.
Here is an R soultion:
# making up some fake data: 4 groups with 10 values each
g = gl(4,10)
x = rnorm(length(g)) + as.integer(g)*0.5
# getting the groupmeans
m = tapply(x,g,mean)
# 1000 bootstrap samples
v = t(replicate(1000,tapply(sample(x),g,mean)))
# comparisons, which of the groups has a group mean
# larger than the bootstrapped mean
isMax = t(apply(v,1,function(x) m>x))
# the relative frequencies are used as estimates for the probabilities:
colMeans(isMax)
An example result is attached.
[EDIT]
Just recognized: It might be sensible to normalize the results so that the sum is 1.
colMeans(isMax) / sum(isMax) * nrow(isMax)
gives in the shown example 0.367% for "1", 18.2% for "2", 29.0% for "3" and 52.4% for "4".
If I understand this right, which I might not as I am not very comfortable with the "apply" set of functions in R, "colMeans(isMax)" returns the probability that each individual group is higher than the group mean. If this is true, despite their scaling, I don't think the adjusted colMeans give me what I want. In a situation where there four groups, clustering around 1, 10, 11, and 12, where each group is distinct (as generated by the data below):
g = gl(4,10)
x = cbind(rnorm(10,1,0.1),rnorm(10,10,0.1),rnorm(10,11,0.1),rnorm(10,12,0.1))
m = tapply(x,g,mean)
v = t(replicate(1000,tapply(sample(x),g,mean)))
isMax = t(apply(v,1,function(x) m>x))
colMeans(isMax) / sum(isMax) * nrow(isMax)
You get these proportions:
0.0000000 0.3087622 0.3435327 0.3477051
Because all three of the high groups are almost always higher than the mean. However, this doesn't give us an accurate picture of which group is likely the highest (which is clearly the last, given the data).
The specific objective of determining the likelihood of a variety being the highest yielding could have significant benefits in several steps of the breeding process. I do like the idea of bootstrapping, though; I'm wondering if there's a way to modify your method for this purpose...
I would hope that you have a lot of detailed records for things like, rainfall per sight, soil nutrients, light, species and yield. Something you can do is create a multiple linear regression, or something like it, and create a statistical model. Now, let's say you have a simple model like:
Y = B0 + B1*rainfall + B2*Species + B3*rainfall*Species
B0, B1, B2 and B3 are all an "average" value and have a Std Err associated with them. You can optimize this function for rainfall = XXX.XXin/yr. Then you can generate 1000 random values with Bn +/- 2*Std Err and rainfall = XXX.XX +/- YY.YY in/yr and optimize these functions. Now, you can go back and see how many times Species A was optimal then B, then C, etc. This will tell you what the probability a certain species will be best under your given model conditions.
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