## Estimate the variance of an estimator - r

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Need some help with a homework question.
I suck at math and our prof does not show us how to break down equations into R code at all, nor does she discuss R at all, but expects us to use it. Can someone show me how to work out this problem in R? A classmate and myself have been working on homework all day and we're exhausted and I've got a migraine setting in. Any help is much appreciated!
A population consists of N=10 primary units, each of which consists of Mi=6 secondary units. A two-stage sampling design selects 2 primary units by simple random sampling (without replacement) and 3 secondary units from each selected primary unit, also by simple random sampling. The observed values of the variable of interest are 7, 5, 3 from the first primary unit selected and 4, 2, 3 from the second primary unit selected.
The population mean per secondary unit was 4.
Estimate the variance of the estimator above.
Not sure where to start

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I looked up the documentation. Are they similar to each other? Yes: The divisor is chosen so that R will be in the interval -1 … +1, value 0 indicating completely random grouping. Does 62% mean the groups are different from each other? No, greater than 70% is required. The function displays most important species for each pair of groups. These species contribute at least to 70 % of the differences between groups. There is also this note: The results of simper can be very difficult to interpret. The method very badly confounds the mean between group differences and within group variation, and seems to single out variable species instead of distinctive species (Warton et al. 2012). Even if you make groups that are copies of each other, the method will single out species with high contribution, but these are not contributions to non-existing between-group differences but to within-group variation in species abundance.

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Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers. Questions asking for code must demonstrate a minimal understanding of the problem being solved. Include attempted solutions, why they didn't work, and the expected results. See also: Stack Overflow question checklist Closed 8 years ago. Improve this question More a general question, but since I am using R -> tags My training data set is 15,000 entries big from which around 20 i would like to use for positive data set -> building up the svm. I wanted to use the remaining resampled dataset as my negative dataset, but i was wondering, it might be better to take the same size (around 20) as the negative data set, otherwise it's highly imbalanced? Is there an easy approach to pool then the classifiers (ensemble based) in R after 1000 rounds of resampling? (or even with the e1071 package) Followup question: I would like to calculate a score for each prediction afterwards, is it fine just to take the probabilities times 100?? Thx

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Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers. This question does not appear to be about programming within the scope defined in the help center. Closed 9 years ago. Improve this question I'm not sure if this is a question for stackoverflow, or crossvalidated. I'm looking for away to include covariate measures when calculating the correlation between two measures. For example, Lets say I have 100 samples, for which I have two measurements, x and y. Now lets say I also have a third measure, a covariate (lets say age). I want to measure the correlation between x and y, but I also want to ignore any of that correlation that comes from the covariate, age. If I'm fitting a linear model, I could simply add the term to the model: lm(y~x+age) I know you can't calculate correlation with this kind of model in R (using ~). So I want to know: Does what I'm asking even make sense to do? I suspect it may not. If it does, what R packages should I be using?

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Q1: Relates to Birthday paradox problem As you see in the collision problem section(in wikipedia link above), your question maps exactly. Cast as a collision problem The birthday problem can be generalized as follows: given n random integers drawn from a discrete uniform distribution with range [1,d], what is the probability p(n;d) that at least two numbers are the same? (d=365 gives the usual birthday problem.) You have a range [1,100] from which you select random cards. The probability of collision(two selected cards are the same) is given as p(n;d) = ... Further down, we have formula for average/expected number of selections as Q(100) gives your answer.