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Sampling distribution of mean; standard error

Short answer

The sampling distribution of the sample mean is the probability distribution of means from all possible samples of size n. Its standard error is sigma/√n, and for finite population sampling without replacement the variance is (sigma²/n)((N−n)/(N−1)). Example: population 2,4,6, n=2 gives SE 1.1547.

What do you mean by sampling distribution of a sample mean and its standard error? Explain with example. What would be the variance of sampling distribution of mean, if sample is taken from finite population?

Sampling distribution of the sample mean: if all possible samples of size n are drawn from a population and the mean x̄ of each is calculated, the probability distribution of these x̄ values is called the sampling distribution of the sample mean.

Its properties (population mean μ, standard deviation σ, sample size n):

  • Mean: E(x̄) = μ
  • Variance: σ²/n (infinite population or sampling with replacement)
  • If the population is normal, or n is large (CLT), x̄ is normal.

Standard error (SE): the standard deviation of the sampling distribution of the mean:

SE(x̄) = σ/√n

It measures how much sample means vary from sample to sample; a larger n gives a smaller SE.

Example

Population: 2, 4, 6 (μ = 4, σ = 1.6330). Take all samples of size n = 2 with replacement (9 samples):

SampleMeanSampleMeanSampleMean
(2,2)2(4,2)3(6,2)4
(2,4)3(4,4)4(6,4)5
(2,6)4(4,6)5(6,6)6

Mean of x̄ values = 36/9 = 4 = μ.

Standard deviation of x̄ values = 1.1547, which equals σ/√n = 1.6330/√2 = 1.1547.

Variance when sampling from a finite population

If the sample of size n is drawn without replacement from a finite population of size N, the variance is reduced by the finite population correction factor:

Var(x̄) = (σ²/n) × ((N − n)/(N − 1))

and SE(x̄) = (σ/√n) √((N − n)/(N − 1)).

The factor is nearly 1 when the sample is a small part of the population (n/N < 0.05), so it is then ignored.

✓ Answer checked by Claude Sonnet 5.5

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