A company maintains three offices in a certain region, each staffed by two emplo

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:

(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.)

(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.)

(c) How does

E(X)

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

E(X)

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

E(X)

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

Office 1 1 2 2 3 3 Employee 1 2 3 4 5 6 Salary   25.7     29.6     26.2     29.6     21.8     25.7   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 Employee 1 Salary 25.7 29.6 26.2 29.6 21.8 25.7 (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. 27.65 29,60 25.70 23,75 25.95 p(x) b) Suppose one of the three offices is randomly selected. Let X1 and X2 denote the sala of the two employees. Determine the samplin distribution of X. Enter your answers a fractions.) 23,75 27,65 27,90 p(x) (c) How does EX) from parts (a) and (b) compare to the population mean salary u? E(X) from part (a) is -Select A, and E(X from part (b) is -Select u.

Explanation / Answer

a) n = 6 , we have to choose 2

6C2 = 15

sorted data = 21.8 ,25.7,25.7 ,26.2,29.7,29.7

we see there are 2 data repeating

so total number of different average = 7

23.75 - 2/15

24 - 1/15

25.7 - 4/15

25.95 - 2/15

27.65 - 2/15

27.9 2/15

29.60 - 2/15

E(Xbar) =

23.75 * 2/15 + 24 * 1/15

+25.7 * 4/15

+25.95 * 2/15

+27.65 * 2/15

+ 27.9 * 2/15

+ 29.60 * 2/15

= 26.43333

b) each one equal probability = 1/3

as offfice 1 has mean = 27.65

office 2 has mean = 27.90

office 3 has mean= 23.75

E(Xbar) = 1/3(27.65+27.90+23.75)= 26.43333

c) . = 1/6(25.7 + 29.6+ 26.2 +29.6 +21.8 + 25.7 ) = 26.43333

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