Question

1. The following table shows different baskets of black pens:

Baskets	Number of baseballs (Population)
1	45
2	27
3	45
4	50
5	54

(a) List all samples of size 2, and compute the mean of each sample.

(b) Compute the mean of the distribution of the sample mean and the population mean. Compare the two values.

(c) Compare the dispersion in the population with that of the sample mean.

Answer 1

(a) The table is given below.

Samples (Basket Nos.)	Baseballs in 1st Basket	Baseballs in 2nd basket	Mean
1,2	45	27	36
1,3	45	45	45
1,4	45	50	47.5
1,5	45	54	49.5
2,3	27	45	36
2,4	27	50	38.5
2,5	27	54	40.5
3,4	45	50	47.5
3,5	45	54	49.5
4,5	50	54	52

(b) Mean of the distribution of the sample mean = mean of the "Mean" column table (pasted above) = 44.2
Population mean = (45 + 27 + 45 + 50 + 54)/5 = 44.2
The population mean and the mean of the sampling distribution of the mean are the same.

(c) The population dispersion = population SD = SD of the data set (45, 27, 45, 50, 54) = = 9.239.
The dispersion of the sampling distribution of the sample mean = sample SD of the "Mean" column of the table (pasted above) = 5.964.
The sample means are less than the population means.