A population consist of four values. 12,12,14,16

Question

• i have to list all samples of size 2 and compute the mean of each sample

 

• Compute the mean of the distribution of the sample mean and population mean

 

• Compare the dispersion in the population with that of the sample mean 

Answer 1

There are six samples of size 2.

 

12,12 with mean 12

12,14  (two samples) with mean 13

12,16  (two samples) with mean 14

14,16 with mean 15

 

The mean of these sample means is (12+13+13+14+14+15)/6 = 13.5

 

The population mean is (12+12+14+16)/4 = 13.5

 

This is the same, as expected, as the sample mean x-bar is known to be an unbiased estimator of the population mean.

 

The standard deviations:

 

For sample means it is

 

√{[(12 - 13.5)² + 2(13 - 13.5)² + 2(14 - 13.5)² + (15 - 13.5)²] / 5} = √1.1 ≈1.05

 

For the population data it is

 

√{[2(12 - 13.5)² + (14 - 13.5)² + (16 - 13.5)²] / 3} = √(7/3) ≈ 1.53

 

The former is smaller. We expect this since bigger samples should result in more accurate estimates for the mean.