Background Info:     ȳ=(72+74+7 6+77+81+84+85+91)/8= 80.25...

Question

# Background Info:     ȳ=(72+74+7 6+77+81+84+85+91)/8= 80.25...

Background Info:

ȳ=(72+74+7

6+77+81+84+85+91)/8= 80.25

Population variance: σ2=N1​∑i=1N​(Yi​−μ)2

N= population size= 8

Yi= population values

μ= population mean

μ=872+74+76+77+81+84+85+91​

μ=8640​

μ=80

σ2=81​[(72−80)2+(74−802)+............+(91−80)2]

σ2=81​[64+36+................+121]

σ2=8288​

σ2=36

(i)Variance of the sample means: σ^2(Ȳ) = σ^2 / n

σ^2(Ȳ) = 1.25 / 2 = 0.625

Sample means: Ȳ1 = 2, Ȳ2 = 2.5, Ȳ3 = 3, Ȳ4 = 2.5, Ȳ5 = 3, Ȳ6 = 3.5, Ȳ7 = 3.5, Ȳ8 = 4

Sample variance: s^2 = 1.25

Sampling variance of the sample means:

Σ(Ȳ - Ȳ)^2 / (S-1) = [(2 - 3)^2 + (2.5 - 3)^2 + (3 - 3)^2 + (2.5 - 3)^2 + (3 - 3)^2 + (3.5 - 3)^2 + (3.5 - 4)^2 + (4 - 4)^2] / (8-1)

= 2.125 / 7

= 0.3036

(ii) estimated sampling variance: V(Ȳ) - s = σ^2 / n * (1 - n / N)

V(Ȳ) - s = 1.25 / 2 * (1 - 2 / 8)

= 0.46875

Evaluating this result to the sampling variance of the sample means obtained in (i), we see a difference.

(iii) Evaluating the result from (i) to the sampling variance of the mean estimated directly from one sample realization of size n=2, we can see that:

V(s) = [(1/2-1)/(2)] * 2.375 = 0.3958

In conclusion, our results demonstrate that the sampling variance of the mean for samples of size n=2 is higher than the population element variance and the variance of the mean for larger samples.

This highlights the need for larger sample sizes in order to reduce sampling variability and get more accurate estimates of population parameters.

Question:

e) Answer question (c) for simple random sample sizes of size 3.

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