can be calculated by finding the mean of all possible sample means.
First, we list all possible samples of size two that can be drawn without replacement from the population {4,8,12,16,20,24}:
- Sample 1: (4,8)
- Sample 2: (4,12)
- Sample 3: (4,16)
- Sample 4: (4,20)
- Sample 5: (4,24)
- Sample 6: (8,12)
- Sample 7: (8,16)
- Sample 8: (8,20)
- Sample 9: (8,24)
- Sample 10: (12,16)
- Sample 11: (12,20)
- Sample 12: (12,24)
- Sample 13: (16,20)
- Sample 14: (16,24)
- Sample 15: (20,24)
Next, we calculate the mean of each sample:
- Sample 1 mean: (4+8)/2 = 6
- Sample 2 mean: (4+12)/2 = 8
- Sample 3 mean: (4+16)/2 = 10
- Sample 4 mean: (4+20)/2 = 12
- Sample 5 mean: (4+24)/2 = 14
- Sample 6 mean: (8+12)/2 = 10
- Sample 7 mean: (8+16)/2 = 12
- Sample 8 mean: (8+20)/2 = 14
- Sample 9 mean: (8+24)/2 = 16
- Sample 10 mean: (12+16)/2 = 14
- Sample 11 mean: (12+20)/2 = 16
- Sample 12 mean: (12+24)/2 = 18
- Sample 13 mean: (16+20)/2 = 18
- Sample 14 mean: (16+24)/2 = 20
- Sample 15 mean: (20+24)/2 = 22
Finally, we calculate the mean of all sample means:
(6+8+10+12+14+10+12+14+16+14+16+18+18+20+22) / 15 = 14
Therefore, the mean of the sampling distribution of means is 14.
A population consists of six numbers 4,8,12,16,20,24. Consider all samples of size two which can be drawn without replacement from this population.
i. The mean of the sampling distribution of means
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