Question

Suppose that a population consists of the six values 4, 8, 5, 3, 8, and 4. (a) Find the population mean and variance. (b) Calculate the sampling distribution of the mean of a sample of size 3 by displaying all possible such samples (chosen without replacement). (c) Use the results from (b) to find the mean and variance of the sampling distribution. (d) Compare the answers from (c) with those obtained from the general formulas for E(X) and Var(X) derived in class

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Answer 1

(a)

X = (4, 8, 5, 3, 8, 4)

Mean of X = (4 + 8 +5 + 3 + 8 + 4)/6 = 5.33

Variance of X is,

=  [(4 - 5.33)² + (8 - 5.33)² + (5 - 5.33)² + ( 3 - 5.33)² + ( 8 - 5.33)² + ( 4- 5.33)² ]/5

= 4.67

(b)

All possible samples of size 3 are

4, 8, 5, Mean = 5.67

4,8,3  Mean = 5

4,8,8  Mean = 6.67

4,8,4  Mean = 5.33

4,5,3  Mean = 4

4,5,8  Mean = 5.67

4,5,4  Mean = 4.33

4,3,8  Mean = 5

4,3,4  Mean = 3.67

4,8,4  Mean = 5.33

8,5,3  Mean = 5.33

8,5,8  Mean = 7

8,5,4  Mean = 5.67

8,3,8  Mean = 6.33

8,3,4  Mean = 5

8,8,4  Mean = 6.67

5,3,8  Mean = 5.33

5,3,4  Mean = 4

5,8,4  Mean = 5.67

3,8,4   Mean = 5

So, the sampling distribution of the mean of a sample of size 3 is,

(5.67, 5, 6.67, 5.33, 4, 5.67, 4.33, 5, 3.67, 5.33, 5.33, 7, 5.67, 6.33, 5, 6.67, 5.33, 4, 5.67, 5)

(d)

Similarly as in part (a), the mean of Y = (5.67, 5, 6.67, 5.33, 4, 5.67, 4.33, 5, 3.67, 5.33, 5.33, 7, 5.67, 6.33, 5, 6.67, 5.33, 4, 5.67, 5) is,

Mean = 5.3335

Var = 0.8195397

(d)

Mean of sampling distribution of mean = Mean of population = 5.33

Variance of sampling distribution of mean = Variance of population / n = 4.67 / 6 = 0.7783333

These results are approximately equalt to that in part (c)