Question

Assessing Normality Listed below are the recent salaries (in millions of dollars) of players on the San Antonio Spurs professional basketball team.

a. Do these salaries appear to come from a population that has a normal distribution? Why or why not?

b. Can the mean of this sample be treated as a value from a population having a normal distribution? Why or why not?

12.5 10.4 9.3 7.0 4.0 3.1 2.9 2.1 1.8 1.4 1.1 1.1 0.9

Short answer

a. As the normal quantile plot has points that do not follow a straight-line pattern, the salaries do not appear to come from a normally distributed population.

b. As the sample size is small, and the sample does not come from a normal population, the sample mean salary cannot be considered to come from a normally distributed population.

Step-by-step solution

Given information

A sample showing the recent salaries (in millions of dollars) of players is provided.

Normality of the population

To check if the given sample comes from a normal distributed population, a normal quantile plot is to be constructed.

• If all the points on the plot lie close to a straight-line pattern, the sample points tend to follow a normal distribution.

• If the points on the plot do not lie close to a straight-line pattern and follow some other pattern, the sample points do not follow a normal distribution.

Thus, if the sample observations follow a normal distribution, the population is considered normally distributed.

The following steps are followed to plot a normal quantile plot:

• Let the sample size (n) be equal to 13.
• Tabulate the following values as shown.

Notation	Value
$\frac{1}{2n}$	$$\begin{gathered} \frac{1}{2\left(13\right)}=\frac{1}{26} \\ =0.0385 \end{gathered}$$
$\frac{3}{2n}$	$$\begin{gathered} \frac{3}{2\left(13\right)}=\frac{3}{26} \\ =0.1154 \end{gathered}$$
$\frac{5}{2n}$	$$\begin{gathered} \frac{5}{2\left(13\right)}=\frac{5}{26} \\ =0.1923 \end{gathered}$$
$\frac{7}{2n}$	$$\begin{gathered} \frac{7}{2\left(13\right)}=\frac{7}{26} \\ =0.2692 \end{gathered}$$
$\frac{9}{2n}$	$$\begin{gathered} \frac{9}{2\left(13\right)}=\frac{9}{26} \\ =0.3462 \end{gathered}$$
$\frac{11}{2n}$	$$\begin{gathered} \frac{11}{2\left(13\right)}=\frac{11}{26} \\ =0.4231 \end{gathered}$$
$\frac{13}{2n}$	$$\begin{gathered} \frac{13}{2\left(13\right)}=\frac{13}{26} \\ =0.5000 \end{gathered}$$
$\frac{15}{2n}$	$$\begin{gathered} \frac{15}{2\left(13\right)}=\frac{15}{26} \\ =0.5769 \end{gathered}$$
$\frac{17}{2n}$	$$\begin{gathered} \frac{17}{2\left(13\right)}=\frac{17}{26} \\ =0.6538 \end{gathered}$$
$\frac{19}{2n}$	$$\begin{gathered} \frac{19}{2\left(13\right)}=\frac{19}{26} \\ =0.7308 \end{gathered}$$
$\frac{21}{2n}$	$$\begin{gathered} \frac{21}{2\left(13\right)}=\frac{21}{26} \\ =0.8077 \end{gathered}$$
$\frac{23}{2n}$	$$\begin{gathered} \frac{23}{2\left(13\right)}=\frac{23}{26} \\ =0.8846 \end{gathered}$$
$\frac{25}{2n}$	$$\begin{gathered} \frac{25}{2\left(13\right)}=\frac{25}{26} \\ =0.9615 \end{gathered}$$

• Now, compute the z-scores for the corresponding left-tailed areas under the curve calculated above.

Left-tailedareas	z-scores
0.0385	-1.77
0.1154	-1.20
0.1923	-0.87
0.2692	-0.62
0.3462	-0.40
0.4231	-0.19
0.5000	0.00
0.5769	0.19
0.6538	0.40
0.7308	0.62
0.8077	0.87
0.8846	1.20
0.9615	1.77

• Sort the sample values in ascending order and assign the z-scores to each value as shown.

Sample values (x)	z-scores (y)
0.9	-1.77
1.1	-1.20
1.1	-0.87
1.4	-0.62
1.8	-0.40
2.1	-0.19
2.9	0.00
3.1	0.19
4	0.40
7	0.62
9.3	0.87
10.4	1.20
12.5	1.77

• Mark the values 0, 2, ……., 14 on the horizontal axis and label the axis as ‘Sample Values’.
• Mark the values -*2.00, -1.50, ……., 2.50 on the vertical axis and label the axis as ‘z-score’.
• Plot a dot for the z-score values on the graph corresponding to the sample values.
• Sketch a straight line on the graph.

The normal quantile plot for the sample values looks as shown below.

Analysis of the plot

a.

The graph shows that a lot of points do not lie close to the straight line. This indicates that the population from which the sample is extracted is not normally distributed.

Sampling distribution of the sample mean

b.

If the size of the sample is sufficiently large, the sampling distribution of the sample mean is always considered normally distributed.

Here, the sample size equal to 13 is small (less than 30). Moreover, the sample also does not appear to come from a normally distributed population.

Therefore, the sample mean salary cannot be considered to come from a normally distributed population.