Question 1179168: A population consists of the five numbers 2, 3, 6, 8, 11. Consider all possible samples of size two which can be drawn with replacement from this population. Find (a) the mean of the population, (b) the standard deviation of the population, (c) the mean of the sampling distribution of means, (d) the standard deviation of the sampling distribution of means, i.e., the standard error of means.
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
mean of population = 
variance of population = [(2-6)²+(3-6)²+(6-6)²+(8-6)²+(11-6)²]/5 = 10.8
standard deviation of population = √(var) = √(10.8) = 3.286335345
[2,2] mean = 2 var = 0
[2,3] mean = 2.5 var = 0.25
[2,6] mean = 4 var = 4
[2,8] mean = 5 var = 9
[2,11] mean = 6.5 var = 20.25
[3,3] mean = 3 var = 0
[3,6] mean = 4.5 var = 2.25
[3,8] mean = 5.5 var = 6.25
[3,11] mean = 7 var = 16
[6,6] mean = 6 var = 0
[6,8] mean = 7 var = 1
[6,11] mean = 8.5 var = 6.25
[8,8] mean = 8 var = 0
[8,11] mean = 9.5 var = 2.25
[11,11] mean = 11 var = 0
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sums 90 67.5
mean of sample means = 90/15 = 6
variance of sample means = 67.5/5 = 13.5
standard deviation of sample means = √13.5 = 3.674234614
(a) the mean of the population, μ=6
(b) the standard deviation of the population, σ=3.286335345
(c) the mean of the sampling distribution of means, μx̄=6
(d) the standard deviation of the sampling distribution of means, i.e., the
standard error of means. σx̄=3.674234614
Edwin
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