Question

The following table provides the starting players of a basketball team and their heights

Player	A	B	C	D	E
Height (in.)	75	77	79	82	87

a. The population mean height of the five players is _____ .

b. Find the sample means for samples of size 2.
A, B: x¯ = ___ .
A, C: x¯ = ___ .
A, D: x¯¯ = ___ .
A, E: x¯ = ____ .
B, C: x¯¯ = ____ .
B, D: x¯ = _____ .
B, E: x¯ = ____.
C, D: x¯ = ___ .
C, E: x¯ = ____ .
D, E: x¯ = ____ .

c. Find the mean of all sample means from above:
x¯ = ____ .

Answer 1

The population comprises of five basketball players A,B,C,D and E and their heights in inches are 75, 77, 79, 82 and 87.

a)

Thus, the population mean is (75 + 77 + 79 + 82 + 85) / 5 = 79.6

b)

We have 5 population units and we want to make samples of size 2 from them. Thus the total number of such samples is   = 10.

The samples are (A, B) , (A, C) , (A, D) , (A, E) , (B, C) , (B, D) , (B, E) , (C, D) , (C, E) , (D, E).

The respective sample means are

A,B : ( 75 + 77)/2 = 76

A,C : (75 + 79)/2 = 77

A,D : (75 + 82)/2 = 78.5

A,E : (75 + 85)/2 = 80

B,C : (77 + 79)/2 = 78

B,D : (77 + 82)/2 = 79.5

B,E : (77 + 85)/2 = 81

C.D : (79 + 82)/2 = 80.5

C,E : (79 + 85)/2 = 82

D,E : (82 + 85)/2 = 83.5

c)

The means of the sample means are

(76 + 77 + 78.5 + 80 + 78 + 79.5 + 81 + 80.5 + 82 + 83.5)/10 = 79.6