In: Statistics and Probability
Solution:
We are given the population valuess as {-2,-1,0,1,2} population size N=5 and sample size n=3. Thus, the number of possible samples which can be drawn without replacement is
This is for taking 2 samples
The sampling distribution of the sample mean is
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Sample Sample Sample Sample Sample Sample Sample Sample No. Values Mean No. Values Mean 1 -1.5 -1 0 -1 7 0.5 N -2 -2,-1 -2,0 -2,1 -2,2 -1,0 8 -1,1 -1,2 0,1 0,2 1, 2 -0.5 0 0.5 4 1 0 10 5 -0.5 1.5
x_barf -1.5 f(x_bar) x_bar*f(x_bar) 0.10 0.1 1 -1 1 0.10 0.1 -0.5 2 0.4 0.20 0.20 0 2 0.4 0.5 2 0.10 0.2 1 1 0.10 0.1 1.5 1 0.10 0.1 10
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For 3 samples
Sample no. | Sample Values | Sample Mean | Sample no. | Sample Values | Sample Mean |
1 | -2,-1,0 | -1 | 6 | 0,1,2 | 1 |
2 | -2,-1,1 | -2/3 | 7 | -2,0,1 | -1/3 |
3 | -2,-1,2 | -1/3 | 8 | -2,0,2 | 0 |
4 | -1,0,1 | 0 | 9 | -1,1,2 | 2/3 |
5 | -1,0,2 | 1/3 | 10 | -2,1,2 | 1/3 |
The sampling distribution of the sample mean is
X_bar | f | f(x_bar) | x_bar*f(xbar) |
-1 | 1 | 0.1 | 0.1 |
-2/3 | 1 | 0.1 | 0.1 |
-1/3 | 2 | 0.2 | 0.4 |
0 | 2 | 0.2 | 0.4 |
1/3 | 2 | 0.2 | 0.4 |
2/3 | 1 | 0.1 | 0.1 |
1 | 1 | 0.1 | 0.1 |