Skip to main content

Samples of 2,8,14,20 with standard error

Short answer

All six samples of size two without replacement are (2,8),(2,14),(2,20),(8,14),(8,20),(14,20), with means 5,8,11,11,14,17. Population mean and mean of sample means both equal 11; SE=(6.7082/√2)√(2/3)=3.8730.

A population consists of the four number 2, 8, 14, 20: i) Write down all possible sample size of two without replacement ii) Verify that the population mean is equal to the mean of the sample mean iii) Calculate the standard error of the sampling distribution of the sample mean

Population: 2, 8, 14, 20 (N = 4), n = 2, without replacement.

i) All possible samples

There are ⁴C₂ = 6 samples.

No.SampleSample mean
1(2, 8)5
2(2, 14)8
3(2, 20)11
4(8, 14)11
5(8, 20)14
6(14, 20)17

ii) Population mean equals mean of sample means

μ = (2 + 8 + 14 + 20)/4 = 11

Sum of sample means = 5 + 8 + 11 + 11 + 14 + 17 = 66

Mean of sample means = 66/6 = 11

Both equal 11, so verified.

iii) Standard error of the sample mean

First the population standard deviation:

μ = Σx/N = 44/4 = 11

σ² = Σ(x − μ)²/N = (81 + 9 + 9 + 81)/4 = 180/4 = 45

σ = √45 = 6.7082

With sampling without replacement from a finite population:

SE = (σ/√n) × √((N − n)/(N − 1)) = (6.7082/√2) × √((4 − 2)/(4 − 1))

= 4.7434 × 0.8165 = 3.8730

(Direct check: the standard deviation of the 6 sample means is 3.8730.)

Answer: SE of sample mean = 3.8730.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Probability and Statistics chapter 4 questions →