Question

There are five sales associates at Mid-Motors Ford. The five associates and the number of cars they sold last week are: Sales Associate Cars Sold Peter Hankish 8 Connie Stallter 6 Juan Lopez 4 Ted Barnes 10 Peggy Chu 6
a. How many different samples of size 2 are possible?

b. List all possible samples of size 2, and compute the mean of each sample.

c. Compare the mean of the sampling distribution of the sample mean with that of the

population.

Answer 1

Answer:

a) There are 10 different samples of size 2.

b) See the explanation section

c) See the explanation section

Step-by-step explanation:

a) We need to select a sample of size 2 from the given population of size 5. We use combination to get the number of difference sample.

$\{ {{5} \atop {2}} \} = \frac{5!}{2!(5-2)!} \\= \frac{5!}{2!3!} \\= \frac{120}{2*6} \\= \frac{120}{12} \\=10$

b) Possible sample of size 2:

Peter Hankish 8 Connie Stallter 6 Juan Lopez 4 Ted Barnes 10 Peggy Chu 6

1. Peter Hankish and Connie Stallter ( Mean = (8 + 6)/2 = 14/2 = 7)
2. Peter Hankish and Juan Lopez (Mean = (8 + 4)/2 = 12/2 = 6)
3. Peter Hankish and Ted Barnes (Mean = (8 + 10)/2 = 18/2 = 9)
4. Peter Hankish and Peggy Chu (Mean = (8 + 6)/2 = 14/2 = 7)
5. Connie Stallter and Juan Lopez (Mean = (6 + 4)/2 = 10/2 = 5)
6. Connie Stallter and Ted Barnes (Mean = (6 + 10)/2 = 16/2 = 8)
7. Connie Stallter and PeggyChu (Mean = (6 + 6)/2 = 12/2 = 6)
8. Juan Lopez and Ted Barnes (Mean = (4 + 10)/2 = 14/2 = 7)
9. Juan Lopez and Peggy Chu (Mean = (4 + 6)/2 = 10/2 = 5)
10. Ted Barnes and Peggy Chu (Mean = (10 + 6)/2 = 16/2 = 8)

c) The mean of the population is:

$mean = \frac{(8+6+4+10+6)}{5} \\= \frac{34}{5} \\= 6.8$

Comparing the mean of the population and the sample; we can say that most of the 2-size sample have their mean higher than that of the population sample. And the variation with the mean is not much. Some sample have their mean greater than population mean, while some sample have their mean greater than the population mean.