In: Statistics and Probability
Suppose your population of interest is composed of ages of all 4 children in a family. You write the population values (1,3,5,7) on slips of paper. You then randomly select 2 slips of paper, with replacement.
(a) List all possible samples of size n = 2 and calculate the mean of each.
(b) Create the sampling distribution of the sample means.
(c) Determine the mean, variance, and standard deviation of the sample means.
(d) Compare your results with the mean, variance, and the standard deviation of the original) population. What do you conclude?
Solution:
Population values are 1,3,5,7
Mean = = (1+3+5+7)/4= 16/4 =4
variance ==[ (1-4)2+(3-4)2+(5-4)2+(7-4)2 ]/4=20/4=5
N=4
a)Draw samples size 2 with replacement and calculate the mean of each sample
Sr.no. |
Samples | means(m) |
1 | 1,1 | 1 |
2 | 1,3 | 2 |
3 | 1,5 | 3 |
4 | 1,7 | 4 |
5 | 3,1 | 2 |
6 | 3,3 | 3 |
7 | 3,5 | 4 |
8 | 3,7 | 5 |
9 | 5,1 | 3 |
10 | 5,3 | 4 |
11 | 5,5 | 5 |
12 | 5,7 | 6 |
13 | 7,1 | 4 |
14 | 7,3 | 5 |
15 | 7,5 | 6 |
16 | 7,7 | 7 |
Total | xbar=64 |
Now,
b) sampling distribution of the sample means
m | P(m) | m.P(m) | m2P(m) |
1 | 1/16 | 1/16 | 1/16 |
2 | 2/16 | 4/16 | 8/16 |
3 | 3/16 | 9/16 | 27/16 |
4 | 4/16 | 16/16 | 64/16 |
5 | 3/16 | 15/16 | 75/16 |
6 | 2/16 | 12/16 | 72/16 |
7 | 1/16 | 7/16 | 49/16 |
Total | 1 | 4 | 18.5 |
c) From above mean(m)= 4
Var(m) = 18.5 - 42
var(m)= 2.5
Standard deviation(m)= sqrt(2.5) = 1.58
d) from the above result,
The sample mean is an unbiased estimator of the population mean.
The variance of the sample mean is a biased estimator of the population variance.
The standard deviation of the sample mean is a biased estimator of the population standard deviation .