# Tag Info

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Yes, every statistic has a sampling distribution (though some may be degenerate). What would they be like? The sampling distribution of a statistic - just as with the mean - will in general depend on the population distribution you start with (and the sample size, naturally). As an example, in a random sample from a normal distribution, the sample variance ...

4

To help make the ideas clear, I will use capital letters for random variables. Everything follows from the restriction $\sum W_i=n,$ because that implies this sum has zero variance. Since each $W_i$ is a Bernoulli variable, $$\operatorname{Var}(W_i) = \frac{n(N-n)}{N^2}.$$ Computing the variance of the sum and assuming, as is the case with simple random ...

3

If you have a biased coin with probabilities $P(H)=0.6$ and $P(T)=0.4$, selecting the most probable one would be returning $H$ every time. Sampling from this distribution would return $H$ 60% of the times and $T$ 40% of the times. When doing next word prediction, selecting the most probable word all the time might result in getting sequences that are not ...

2

Computers can only do pseudo random sampling directly from a Uniform Distribution. Sampling from any other distribution requires some numerical transformations, such as Inverse Transform Sampling. This method, however, only allows to sample from distribution that have a defined Cumulative Distribution Function that can be inverted - and this is the case for ...

2

This question is really about R syntax and not about the normal distribution or sampling. So I guess it's 'off topic' here, and may be closed. I'm not quite sure what your difficulty is. If you set the same seed before each program, then you should fill an $n\times p$ matrix by columns with exactly the same normal variates. n=5; p=2 set.seed(713) #...

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I would rather not suggest this course of action since, all chains starting from the very same point, $X^{(2000)}$ say, these chains need run long enough to cancel the dependence on that starting point and recover simulations from the target. For instance, if one removes some burnin part (20%? 50%?) from all 10 parallel chains, this would come closer to a ...

2

Posting a phone number on a board in select departments at a unversity asking for volunteers would likely result in candidates who are healthy, above average in intellect, motivation, character and as such, perhaps representative of future above average wage earners, homeowners, etc., and depending on the international student population present at the ...

2

When you take a sample from a population, you are gathering information about the population, which you might use for making a confidence interval or for testing a hypothesis about population parameters (perhaps the population mean $\mu).$ When you 're-sample' as in bootstrapping, you are analyzing data already taken from a population. Re-sampling does not ...

2

Yes, as every statistic is a function of you sample (which are random variables) they will have a distribution. It might not be as easy to deduce the distribution as with the sample mean.

1

Beta distributions have support $(0,1).$ Here is the density function for $\mathsf{Beta}(2,4),$ which has $\mu=1/3,\sigma^2=2/63.$ [Using R.] curve(dbeta(x,2,4), 0, 1, ylab="PDF", main="BETA(2,4)") abline(h=0, col="green2") Averages of even small samples from this distribution are nearly normal. The boundedness prevents extreme ...

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Based on some of the methods whuber used in his answer, I decided to derive this covariance again, but this time in the way that I originally attempted the derivation (starting by showing the covariance as an expectation and going from there). I get the right answer now: $$Cov\left(\sum_{i=1}^{N}{w_ix_i},\sum_{i=1}^{N}{w_iy_i}\right)$$ =\ E\left[\left(\...

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Sampling from finite populations The sampling theory for finite populations is usually applied to objects of a non-random, determined nature, for example all adults aged 18 or older living in a country at a given date. All is based on the probability $\pi_i$ that element $i$, $i=1,\dots,N$, is selected for the sample, and the probability $\pi_{ij}$ that ...

1

No, the answer is "It depends". If the experiment consists of choosing $k$ balls without replacement from a set of $n$ balls numbered from $1$ to $n$, then the outcome (or sample values if you prefer) must be a set of distinct numbers from $\{1,2,\cdots, n\}$ and in this case, the sample values are not independent. If the experiment consists of ...

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Comment continued, to show graphs of some specific distributions. Scenario 1. Urn with 5 red chips and 10 blue ones. Sample 4 chips at random with replacement. Then the number $X$ of red chips drawn is $\mathsf{Binom}(n=4, p=1/3),$ so that $E(X) = np = 4/3; Var(X) = np(1-p) = 4(1/3)(2/3) = 8/9 = 0.8889.$ x=0:4; pdf.b = dbinom(x, 4, 1/3) mean = sum(x*pdf.b); ...

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You have a positive class prior of $0.0015$, which is way to low to train a classifier from. You need at least an excess of $100$ positive cases, your most rare class. You will be able to start with a few (most) discriminative features. Next, a support vector machine is not a suited classifier for such a skewed prior distribution. A recommended approach is ...

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A small population with twice as many children as adults. Suppose there are 4 each of children of ages 5, 10, and 15; and that there are 3 each of adults of ages 25 and 45. That means the average weight in the population of $36$ is $[4(20+35+50) + 3(90+85)]/18 = 52.5$kg. What sample size? Also, suppose we are take a random sample of $n = 3$ from this ...

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This looks like, stratified two-phase cluster sampling. You've stratified into urban/rural and then you select blocks (clusters). You then sample households within each block which means you are doing two-phase cluster sampling (one-phase would mean you just take all the households in the block). The simplest way to determine survey weights would be to just ...

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