Q. 7.43 - Elementary Statistics

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).

Step-by-Step Solution

Answer

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

Step 1: Given information

The samples from exercise 7.41 is:

Step 2: 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	$\frac{80+74+80}{3}=78.00$

Step 3: 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 $78.6$.

Step 4: 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 $78.6$.