Question

In: Statistics and Probability

The following data represent the number of days absent per year in a population of six...

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).

Solutions

Expert Solution

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.


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