Question

A company maintains three offices in a certain region, each staffed by two employees. Information concerning yearly salaries (1000s of dollars) is as follows:

Office	1	1	2	2	3	3
Employee	1	2	3	4	5	6
Salary	36.7	40.6	37.2	40.6	32.8	36.7

(a) Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary

X.

(Enter your answers for p(x) as fractions.)

x	34.75		36.70	36.95	38.65		40.60
p(x)		1 15				2 15

(b) Suppose one of the three offices is randomly selected. Let X₁ and X₂ denote the salaries of the two employees. Determine the sampling distribution of

X.

(Enter your answers as fractions.)

x	34.75	38.65	38.90
p(x)

(c) How does

E(X)

from parts (a) and (b) compare to the population mean salary μ?

E(X)

from part (a) is  ---Select--- greater than less than equal to μ, and

E(X)

from part (b) is  ---Select--- greater than less than equal to μ.

Answer 1

a)

there are 6 employees of which 2 need to be chosen , so total combinations = 6C2=15

Employee1	Employee2	salary1	salary2	Sample means	P(X)	x̄ * P(X)
1	2	36.7	40.6	38.65	1/15	2.577
1	3	36.7	37.2	36.95	1/15	2.463
1	4	36.7	40.6	38.65	1/15	2.577
1	5	36.7	32.8	34.75	1/15	2.317
1	6	36.7	36.7	36.7	1/15	2.447
2	3	40.6	37.2	38.9	1/15	2.593
2	4	40.6	40.6	40.6	1/15	2.707
2	5	40.6	32.8	36.7	1/15	2.447
2	6	40.6	36.7	38.65	1/15	2.577
3	4	37.2	40.6	38.9	1/15	2.593
3	5	37.2	32.8	35	1/15	2.333
3	6	37.2	36.7	36.95	1/15	2.463
4	5	40.6	32.8	36.7	1/15	2.447
4	6	40.6	36.7	38.65	1/15	2.577
5	6	32.8	36.7	34.75	1/15	2.317
			Total	561.5		37.433

x̄	34.75	35	36.7	36.95	38.65	38.9	40.6
P(X)	2/15	1/15	3/15	2/15	4/15	2/15	1/15

b)

let x1 and x2 be salary of two employees.
office	x1	x2	average salary	P(X)	x̄*P(X)
1	36.7	40.6	38.65	0.333333	12.88333
2	37.2	40.6	38.9	0.333333	12.96667
3	32.8	36.7	34.75	0.333333	11.58333
		Total=			37.43333

each office has equal probability of selection =1/3 =0.333

34.75	38.65	38.65
1/3	1/3	1/3

c)

population mean , µ =Σx/n=    224.6   /   6   =   37.433

E(Xbar) from part a is equal to mu and E(Xbar) from part b is equal to mu