In: Statistics and Probability
The following data represent the number of days absent per year in a population of six employees in a small company: 1 3 6 7 9 10
a. Compute the population mean.
b. Compute the population standard deviation.
c. Assuming that you sample without replacement, select all possible samples of size n=2 and construct the sampling distribution of the mean.
d. Compute the mean of all the sample means. How does the mean of the sample means and the population mean compare?
e. Compute the standard deviation of all the sample means. How does the standard deviation of the sample means and the population standard deviation compare?
f. Repeat parts (c) – (e) for all possible samples of size n=3. How do the results compare with those of sample size n=2? Comment on what happened as the sample size increased (from 2 to 3).
Following table shows the calculations for population mean and population standard deviation:
X | (X-mean)^2 | |
1 | 25 | |
3 | 9 | |
6 | 0 | |
7 | 1 | |
9 | 9 | |
10 | 16 | |
Total | 36 | 60 |
(a)
The population mean is
(b)
The population standard deviation is
(c)
Sine samples are drawn without replacement so possible number of samples are C(6,2) = 15
Since all samples are equally likely so probability of getting each sample is 1/15 = 0.06667
Following table shows all the possible samples and sample means:
Samples | Sample mean, xbar | P(xbar) | |
1 | 3 | 2 | 0.06667 |
1 | 6 | 3.5 | 0.06667 |
1 | 7 | 4 | 0.06667 |
1 | 9 | 5 | 0.06667 |
1 | 10 | 5.5 | 0.06667 |
3 | 6 | 4.5 | 0.06667 |
3 | 7 | 5 | 0.06667 |
3 | 9 | 6 | 0.06667 |
3 | 10 | 6.5 | 0.06667 |
6 | 7 | 6.5 | 0.06667 |
6 | 9 | 7.5 | 0.06667 |
6 | 10 | 8 | 0.06667 |
7 | 9 | 8 | 0.06667 |
7 | 10 | 8.5 | 0.06667 |
9 | 10 | 9.5 | 0.06667 |
Now we need to sum the probabilities of samples that have same sample mean. Following table shows the sampling distribution of sample mean:
Sample mean, xbar | P(xbar) |
2 | 0.06667 |
3.5 | 0.06667 |
4 | 0.06667 |
4.5 | 0.06667 |
5 | 0.13334 |
5.5 | 0.06667 |
6 | 0.06667 |
6.5 | 0.13334 |
7.5 | 0.06667 |
8 | 0.13334 |
8.5 | 0.06667 |
9.5 | 0.06667 |
d)
Following table shows the calculation for mean of sampling distribution of sample means:
Sample mean, xbar | P(xbar) | xbar*P(xbar) |
2 | 0.06667 | 0.13334 |
3.5 | 0.06667 | 0.233345 |
4 | 0.06667 | 0.26668 |
4.5 | 0.06667 | 0.300015 |
5 | 0.13334 | 0.6667 |
5.5 | 0.06667 | 0.366685 |
6 | 0.06667 | 0.40002 |
6.5 | 0.13334 | 0.86671 |
7.5 | 0.06667 | 0.500025 |
8 | 0.13334 | 1.06672 |
8.5 | 0.06667 | 0.566695 |
9.5 | 0.06667 | 0.633365 |
Total | 6.0003 |
So,
The mean of the sample means and the population mean are equal.