26 CHAPTER 2 Sampling from Populationswould get exactly the population mean, which is 31.1667. This result
can be proven mathematically. In this case, since we know what the
fi nite population is, we can calculate the 15 possible sample means and
take their average to verify that it equals the population mean. Recall
the 15 possible samples are:{A, B, C, D} with sample mean = (26 + 1 7 + 4 5 + 70)/4 = 39.50
{A, B, C, E} with sample mean = (26 + 1 7 + 4 5 + 32)/4 = 30.00
{A, B, C, F} with sample mean = (26 + 1 7 + 4 5 + 9)/4 = 24.25
{A, B, D, E} with sample mean = (26 + 1 7 + 7 0 + 32)/4 = 36.25
{A, B, D, F} with sample mean = (26 + 1 7 + 7 0 + 9)/4 = 30.50
{A, B, E, F} with sample mean = (26 + 1 7 + 3 2 + 9)/4 = 21.00
{A, C, D, E} with sample mean = (26 + 4 5 + 7 0 + 32)/4 = 43.25
{A, C, D, F} with sample mean = (26 + 4 5 + 7 0 + 9)/4 = 43.25
{A, C, E, F} with sample mean = (26 + 4 5 + 3 2 + 9)/4 = 28.00
{A, D, E, F} with sample mean = (26 + 7 0 + 3 2 + 9)/4 = 34.25
{B, C, D, E} with sample mean = (17 + 4 5 + 7 0 + 32)/4 = 41.00
{B, C, D, F} with sample mean = (17 + 4 5 + 7 0 + 9)/4 = 35.25
{B, C, E, F} with sample mean = (17 + 4 5 + 3 2 + 9)/4 = 25.75
{B, D, E, F} with sample mean = (17 + 7 0 + 3 2 + 9)/4 = 32.00
{C, D, E, F} with sample mean = (45 + 7 0 + 3 2 + 9)/4 = 39.00In this case, the largest mean is 43.25, and the smallest is 21.00, and
the value closest to the population mean is 34.25. This shows that the
estimate has a lot of variability. To verify the property of unbiasedness,
we need to average these 15 estimates and verify that the average is
33.1667. This goes as follows:The expected value of the averages is (39.5 + 30.0 + 24.25 +
36.25 + 30.5 + 21.0 + 43.25 + 37.5 + 28.0 + 34.25 + 41.0 +
35.25 + 25.75 + 32.0 + 39.0)/15 = 497.50/15 = 33.1667 rounded
to four decimal places.In Section 2.5 , we will generate bootstrap samples. In bootstrap-
ping, we do simple random sampling with replacement. So this can be