How do I generate a range of specific numbers given mean? - r

Question

Given a vector of specified values, for example:
x = c(4.0, 3.7, 3.3, 3.0, 2.7, 2.3, 2.0, 1.7, 1.3, 1.0)
I would like to create a new vector of any length comprised only of values in x sampled randomly, that will result in the combined mean of 3.15. I have tried using the rnorm() function but however, I can only generate random numbers that equate to the mean of 3.15 and not the specified values I wanted. Could anyone point me in the correct direction?

Answer 1

The problem with your question is that there are an infinite number of ways to sample from
x = c(4.0, 3.7, 3.3, 3.0, 2.7, 2.3, 2.0, 1.7, 1.3, 1.0)
to get a mean of roughly 3.15, you just have a specify a probability for each value.
Doing
n = 20
sample(x, n, replace = TRUE)
assumes each value is equally likely and you would get a mean close to 2.5. But if you re-weight the probabilities, you can get closer to what you want. One way of doing this might be
p = 1/(x - 3.15)^2 # or try p = 1/abs(x - 3.15)
sample(x, n, replace = TRUE, prob = p)
where p weights values closer to 3.15 higher so these are more likely to be accepted. It isn't perfect (meaning the true expected value is something like 3.12 and most values are just 2.7, 3.0 and 3.3), but then again there isn't a single solution.

Answer 2

Here's my brute force method:
samp315<-function(n=20, desmean=3.15, distance=0.001) { # create a function with default n=20 and range 3.149-3.151
x<- c(4.0, 3.7, 3.3, 3.0, 2.7, 2.3, 2.0, 1.7, 1.3, 1.0)
samp<-0 # reset samp to 0
i<-0 # reset my counter to zero
while (!between(mean(samp),desmean-distance,desmean+distance) & i<1000000) { # the following will run continuously until a sample (samp) with a mean that falls within the specified range is found, OR until 1 million attempts have been made
samp<-sample(x,n,replace=TRUE) # try to generate a sample of n times from the list of values (x)
i=i+1 # add to counter towards 1 million
}
ifelse(i<1000000,samp,warning("Couldn't find an appropriate sample, please select a lower n, a desired mean closer to 2.5, or a greater distance")) # if the while loop ended because the counter reached a million, exit with an error, otherwise, return the contents of samp.
}
Now, every time you do samp315():
eg<-samp315()
mean(eg)
[1] 3.15
eg
[1] 3.0 3.7 3.0 3.7 3.3 3.7 3.3 3.3 4.0 1.0 1.7 3.0 2.0 4.0 3.7 3.7 2.3 3.3 4.0 3.3
If you want a sample of different length, just place whatever number you wish inside samp315(). The larger the number, however, the longer it will take to find a sample that will get your desired mean.
You can also change your desired mean by setting desmean, and play around with the range by changing distance to whatever the distance (+/-) from your desired mean can be. The default is n=20, with the range from 3.149 to 3.151.
To avoid an infinite loop for highly unlikely combinations of n and range, I set a maximum of 1m samples, after which the function quits with a warning.

Answer 3

As #mickey pointed out, we can weight the probability of each item according to how far it is from the mean. However, that does not quite work, because there are more elements in x lower than than the desired mean, which skews the sampling towards them. We can account for this by adjusting the probabilites relative to how many elements are higher or lower than the desire mean:
x = c(4.0, 3.7, 3.3, 3.0, 2.7, 2.3, 2.0, 1.7, 1.3, 1.0)
n = 100000
xbar=3.15
xhi = x[which(x>xbar)]
xlo = x[which(x<xbar)]
probhi = 1/(xhi-xbar)
problo = 1/(xbar-xlo)
probhi = probhi * length(problo) / length(probhi)
n=1e5
set.seed(1)
y = sample(x, size = n, replace = TRUE, prob = c(probhi,problo))
mean(y)
# [1] 3.150216