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Sampling distribution: population 5,6,9,12

Short answer

Population mean μ = (5+6+9+12)/4 = 8, σ = √7.5 ≈ 2.7386. Mean of sampling distribution of mean = 8. Standard error for sampling without replacement: SE = (σ/√n)√((N-n)/(N-1)) = (2.7386/√2)√(2/3) ≈ 1.5811.

A population consists of 5, 6, 9, 12. Consider all possible samples of size two which can be drawn without replacement from this population. Find i) Population mean and population standard deviation. ii) Mean of sampling distribution of mean. iii) Standard error of sampling distribution of mean.

Population: 5, 6, 9, 12 (N = 4); samples of size n = 2 without replacement, ⁴C₂ = 6 samples.

i) Population mean and population standard deviation

μ = Σx/N = 32/4 = 8

σ² = Σ(x − μ)²/N = (9 + 4 + 1 + 16)/4 = 30/4 = 7.5

σ = √7.5 = 2.7386

ii) Mean of the sampling distribution of the mean

All samples:

No.SampleSample mean
1(5, 6)5.5
2(5, 9)7
3(5, 12)8.5
4(6, 9)7.5
5(6, 12)9
6(9, 12)10.5

Sum of sample means = 5.5 + 7 + 8.5 + 7.5 + 9 + 10.5 = 48

Mean of sample means = 48/6 = 8

This equals the population mean μ = 8.

iii) Standard error of the sampling distribution of the mean

With sampling without replacement from a finite population:

SE = (σ/√n) × √((N − n)/(N − 1)) = (2.7386/√2) × √((4 − 2)/(4 − 1))

= 1.9365 × 0.8165 = 1.5811

(Direct check: the standard deviation of the 6 sample means is 1.5811.)

Answer: μ = 8, σ = 2.7386; mean of sample means = 8; SE = 1.5811.

✓ Answer checked by Claude Sonnet 5.5

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