Question

7.43 NBA Champs. Repeat parts (b) and (c) of Exercise 7.41 for samples of size 3. For part (b), use your answer to Exercise 7.13(b).

Short answer

The mean height $\left({\mu }_{\overline{x}}\right)$for samples of size $3$is $78.6$.

Step-by-step solution

Given information

The samples from exercise 7.41 is:

Explanation

Determine the parts (b) of Exercise $7.41$for samples of size $3$. Calculate the mean height $\left({\mu }_{\overline{x}}\right)$for samples of size $3$ .
As a result, the size 3 samples and their means are obtained as given in the table below:

Sample size	Height	Mean(x)
B,W,J	83,76,80	$\frac{83+76+89}{3}=79.67$
B,W,C	83,76,74	$\frac{83+76+74}{3}=77.67$
B,W,H	83,76,80	$\frac{83+76+80}{3}=79.67$
B,J,C	83,80,74	$\frac{83+80+74}{3}=79.00$
B,J,H	83,80,80	$\frac{83+80+80}{3}=81.00$
B,C,H	83,74,80	$\frac{83+74+80}{3}=79.00$
W,J,C	76,80,74	$\frac{76+80+74}{3}=76.67$
W,J,H	76,80,80	$\frac{76+80+80}{3}=78.67$
W,C,H	76,74,80	$\frac{76+74+80}{3}=76.67$
J,C,H	80,74,80	$\frac{80+74+80}{3}=78.00$

Explanation

The number of possible samples $\left(N\right)$of size $3$is $10$. For samples of size $3$ , as illustrated below, the mean of all potential sample means is calculated:
${\mu }_{\overline{x}}=\frac{\sum {\overline{x}}_i}{N}$
$=\frac{79.67+77.67+79.67+79+81+89+76.67+78.67+76.67+78}{10}$
$=\frac{786}{10}$
$=78.6$

As a result, the mean height $\left({\mu }_{\overline{x}}\right)$for samples of size $3$is $78.6$.

Explanation

Calculate the mean height $\left({\mu }_{\overline{x}}\right)$:
The average height of five players in the population is$78.6$ inches.
The population mean is equal to the mean of the sample mean.
That would be to:
${\mu }_{\overline{x}}=\mu$
$=78.6$
Therefore, the mean height $\left({\mu }_{\overline{x}}\right)$for samples of size $3$is $78.6$.