SOLUTION: A population consists of five observations 1,2,3,4,5. Draw all possible samples of size 2 with replacement. Find the mean of the sampling distribution of the variances. Compare i

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Question 1121012: A population consists of five observations 1,2,3,4,5. Draw all possible samples of
size 2 with replacement. Find the mean of the sampling distribution of the
variances. Compare it with the variance of the population.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Note: These are samples of size n=2 so we divide by n-1 = 2-1 or 1, not n=2

sample           means                        variances
------------------------------------------------------------------
  1,1       (1+1)/2 = 1.0      [(1-1.0)²+(1-1.0)²]/1 = 0.0/1 = 0.0
  1,2       (1+2)/2 = 1.5      [(1-1.5)²+(2-1.5)²]/1 = 0.5/1 = 0.5
  1,3       (1+3)/2 = 2.0      [(1-2.0)²+(3-2.0)²]/1 = 2.0/1 = 2.0
  1,4       (1+4)/2 = 2.5      [(1-2.5)²+(4-2.5)²]/1 = 4.5/1 = 4.5
  1,5       (1+5)/2 = 3.0      [(1-3.0)²+(5-3.0)²]/1 = 8.0/1 = 8.0
  2,1       (2+1)/2 = 1.5      [(2-1.5)²+(1-1.5)²]/1 = 0.5/1 = 0.5
  2,2       (2+2)/2 = 2.0      [(2-2.0)²+(2-2.0)²]/1 = 0.0/1 = 0.0
  2,3       (2+3)/2 = 2.5      [(2-2.5)²+(3-2.5)²]/1 = 0.5/1 = 0.5
  2,4       (2+4)/2 = 3.0      [(2-3.0)²+(4-3.0)²]/1 = 2.0/1 = 2.0
  2,5       (2+5)/2 = 3.5      [(2-3.5)²+(5-3.5)²]/1 = 4.5/1 = 4.5
  3,1       (3+1)/2 = 2.0      [(3-2.0)²+(1-2.0)²]/1 = 2.0/1 = 2.0
  3,2       (3+2)/2 = 2.5      [(3-2.5)²+(2-2.5)²]/1 = 0.5/1 = 0.5
  3,3       (3+3)/2 = 3.0      [(3-3.0)²+(3-3.0)²]/1 = 0.0/1 = 0.0
  3,4       (3+4)/2 = 3.5      [(3-3.5)²+(4-3.5)²]/1 = 0.5/1 = 0.5
  3,5       (3+5)/2 = 4.0      [(3-4.0)²+(5-4.0)²]/1 = 2.0/1 = 2.0
  4,1       (4+1)/2 = 2.5      [(4-2.5)²+(1-2.5)²]/1 = 4.5/1 = 4.5
  4,2       (4+2)/2 = 3.0      [(4-3.0)²+(2-3.0)²]/1 = 2.0/1 = 2.0
  4,3       (4+3)/2 = 3.5      [(4-3.5)²+(3-3.5)²]/1 = 0.5/1 = 0.5
  4,4       (4+4)/2 = 4.0      [(4-4.0)²+(4-4.0)²]/1 = 0.0/1 = 0.0
  4,5       (4+5)/2 = 4.5      [(4-4.5)²+(5-4.5)²]/1 = 0.5/1 = 0.5
  5,1       (5+1)/2 = 3.0      [(5-3.0)²+(1-3.0)²]/1 = 8.0/1 = 8.0
  5,2       (5+2)/2 = 3.5      [(5-3.5)²+(2-3.5)²]/1 = 4.5/1 = 4.5
  5,3       (5+3)/2 = 4.0      [(5-4.0)²+(3-4.0)²]/1 = 2.0/1 = 2.0
  5,4       (5+4)/2 = 4.5      [(5-4.5)²+(4-4.5)²]/1 = 0.5/1 = 0.5
  5,5       (5+5)/2 = 5.0      [(5-5.0)²+(5-5.0)²]/1 = 0.0/1 = 0.0
------------------------------------------------------------------
                                                        sum = 50.0
      
There are 25 samples, so mean of variances = 50.0/25 = 2      

Population: 1,2,3,4,5

The mean of the population is m = (1+2+3+4+5)/5 = 15/5 = 3

Note: Since this is a population, we will divide by n=5, (not n-1):

       x    (x-m)²
       ---------------
       1    (1-3)² = 4
       2    (2-3)² = 1
       3    (3-3)² = 0
       4    (4-3)² = 1
       5    (5-3)² = 4
       ---------------
                  5)10
                    ---
                     2 = variance of population

So they are the same, both 2.

Edwin