Respuesta :
Solution :
Given :
Player A B C D E
Height [tex]$76$[/tex] [tex]$78$[/tex] [tex]$79$[/tex] [tex]$81$[/tex] 86
The population mean = [tex]$\frac{76+78+79+81+86}{5}$[/tex]
= 80
The sample of the size two :
Pair Heights [tex]$\text{Sample Mean}$[/tex] [tex]$\text{Difference}$[/tex] from population mean
(A,B) [tex]$(76,78)$[/tex] [tex]$77$[/tex] [tex]$-3$[/tex]
(A,C) (76,79) 77.5 -2.5
(A,D) (76,81) 78.5 -1.5
(A,E) (76,86) 81 1
(B,C) (78,79) 78.5 -1.5
(B,D) (78,81) 79.5 -0.5
(B,E) (78,86) 82 2
(C,D) (79,81) 80 0
(C,E) (79,86) 82.5 2.5
(D,E) (81,86) 83.5 3.5
The sample mean are arranged between (77, 83.5)
The standard deviation of the sample means = STDEV.P(C10:C19) = 2.09
We can thus see that the sample means are spread ± 3.5 inches from the population mean.
Sample size of four
Group Heights Sample mean Difference from population mean
(A,B,C,D) (76,78,79,81) 78.5 -1.5
(A,B,C,E) (76,78,79,86) 79.75 -0.25
(A,B,D,E) (76,78,81,86) 80.25 0.25
(A,C,D,E) (76,79,81,86) 80.5 0.5
(B,C,D,E) (78,79,81,86) 81 1
The sample mean are ranged between the (78.5, 81).
The standard deviation of the sample means = STDEV.P(H15:H19) = 0.85
Thus we can see that sample mean are spread ± 1.5 inches from the population mean.