Chapter 9.1, Problem 23P

(a)

To determine

Construct a scatter diagram for the data.

Verify the values of ${\displaystyle \sum x}$, ${\displaystyle \sum x^2}$, ${\displaystyle \sum y}$, ${\displaystyle \sum y^2}$, and ${\displaystyle \sum xy}$.

Find the value of r.

Answer to Problem 23P

The scatter diagram for data is,

The value of r is 0.991.

Explanation of Solution

Calculation:

The formula for correlation coefficient is,

$r=\frac{n{\displaystyle \sum xy}-\left({\displaystyle \sum x}\right)\left({\displaystyle \sum y}\right)}{\sqrt{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\sqrt{n{\displaystyle \sum y^2}-{\left({\displaystyle \sum y}\right)}^2}}$

In the formula, n is the sample size.

The variable x denotes average annual hours per person spent in traffic delays and y denotes the average annual gallons of fuel wasted per person in traffic delays.

Step by step procedure to obtain scatter plot using MINITAB software is given below:

• Choose Graph > Scatterplot.
• Choose Simple. Click OK.
• In Y variables, enter the column of x.
• In X variables, enter the column of y.
• Click OK.

The values are verified in the table below,

x	y	$x^2$	$y^2$	xy
28	48	784	2304	1344
5	3	25	9	15
20	34	400	1156	680
35	55	1225	3025	1925
20	34	400	1156	680
23	38	529	1444	874
18	28	324	784	504
5	9	25	81	45
${\displaystyle \sum x}=154$	${\displaystyle \sum y}=249$	${\displaystyle \sum x^2}=3,712$	${\displaystyle \sum y^2}=9,959$	${\displaystyle \sum xy}=6,067$

Hence, the values are verified.

The number of data pairs are $n=8$. The value of correlation r is,

$\begin{array}{c}r=\frac{8\left(6067\right)-\left(154\times 249\right)}{\sqrt{8\left(3712\right)-{\left(154\right)}^2}\sqrt{8\left(9959\right)-{\left(249\right)}^2}}\\ =\frac{48536-38346}{\sqrt{5980}\sqrt{17671}}\\ =\frac{10190}{10,279.7169}\\ =0.991\end{array}$

Hence, the value of r is 0.991.

(b)

To determine

Find the value of $\overline{x}$ and $\overline{y}$ for variables based on averages.

Find the value of $\overline{x}$ and $\overline{y}$ for variables based on single individuals.

Find the value of $s_x$ and $s_y$ for variables based on averages.

Find the value of $s_x$ and $s_y$ for variables based on single individuals.

Identify the set in which standard deviations for x and y is larger.

Explain why standard deviations increase the value of r.

Answer to Problem 23P

The value of $\overline{x}$ for variables based on averages is 19.25.

The value of $\overline{y}$ for variables based on averages is 31.13.

The value of $\overline{x}$ for variables based on single individuals is 20.13.

The value of $\overline{y}$ for variables based on single individuals is 31.88.

The value of $s_x$ for variables based on averages is 10.33.

The value of $s_y$ for variables based on averages is 17.76.

The value of $s_x$ for variables based on single individuals is 13.85.

The value of $s_y$ for variables based on single individuals is 25.18.

The standard deviation for variables based on single individuals is larger.

The values of $s_x$ and $s_y$ are in the denominator making the value of r larger for smaller standard deviations.

Explanation of Solution

Calculation:

Mean and standard deviation for variables based on averages:

Step by step procedure to obtain mean and standard deviation using MINITAB software is given as,

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the column x, y.
• Choose option statistics, and select Mean, standard deviation.
• Click OK.

Output using MINITAB software is given below:

From MINITAB output, the mean of x is 19.25, the mean of y is 31.13, and standard deviation of x is 10.33, the standard deviation of y is 17.76.

Hence, the value of $\overline{x}$ for variables based on averages is 19.25, the value of $\overline{y}$ for variables based on averages is 31.13, the value of $s_x$ for variables based on averages is 10.33, and the value of $s_y$ for variables based on averages is 17.76.

Mean and standard deviation for variables based on single individuals:

Step by step procedure to obtain mean and standard deviation using MINITAB software is given as,

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the column x, y.
• Choose option statistics, and select Mean, standard deviation.
• Click OK.

Output using MINITAB software is given below:

From MINITAB output, the mean of x is 20.13, the mean of y is 31.88, and standard deviation of x is 13.85, the standard deviation of y is 25.18.

Hence, the value of $\overline{x}$ for variables based on single individuals is 20.13, the value of $\overline{y}$ for variables based on single individuals is 31.88, the value of $s_x$ for variables based on single individuals is 13.85, and the value of $s_y$ for variables based on single individuals is 25.18.

It can be observed that value of $s_x\left(=13.85\right)$ for variables based on single individuals is larger than $s_x\left(=10.33 \right)$ for variables based on averages. Similarly, the value of $s_y\left(=25.18\right)$ for variables based on single individuals is larger than $s_y\left(=17.76 \right)$ for variables based on averages. Hence, the standard deviation for variables based on single individuals is larger.

The formula for correlation coefficient r in equation (1) is,

$r=\frac{1}{n-1}{\displaystyle \sum \left[\frac{\left(x-\overline{x}\right)}{s_x}\times \frac{\left(y-\overline{y}\right)}{s_y}\right]}$

In the formula the values of $s_x$ and $s_y$ are in the denominator. If the values of standard deviation are small, then it would increase the value of r. If the values of standard deviation are large, then it would decrease the value of r.

(c)

To determine

Construct a scatter diagram for the second set of data.

Verify the values of ${\displaystyle \sum x}$, ${\displaystyle \sum x^2}$, ${\displaystyle \sum y}$, ${\displaystyle \sum y^2}$, and ${\displaystyle \sum xy}$.

Find the value of r.

Answer to Problem 23P

The scatter diagram for second set of data is,

The value of r is 0.794.

Explanation of Solution

Calculation:

The variable x denotes average annual number of hours lost for the person and y denotes the average annual gallons of fuel wasted for the person.

Step by step procedure to obtain scatter plot using MINITAB software is given below:

• Choose Graph > Scatterplot.
• Choose Simple. Click OK.
• In Y variables, enter the column of x.
• In X variables, enter the column of y.
• Click OK.

The values are verified in the table below,

x	y	$x^2$	$y^2$	xy
20	60	400	3600	1200
4	8	16	64	32
18	12	324	144	216
42	50	1764	2500	2100
15	21	225	441	315
25	30	625	900	750
2	4	4	16	8
35	70	1225	4900	2450
${\displaystyle \sum x}=161$	${\displaystyle \sum y}=255$	${\displaystyle \sum x^2}=4,583$	${\displaystyle \sum y^2}=12,565$	${\displaystyle \sum xy}=7,071$

Hence, the values are verified.

The number of data pairs are $n=8$. The value of correlation r is,

$\begin{array}{c}r=\frac{8\left(7071\right)-\left(161\times 255\right)}{\sqrt{8\left(4583\right)-{\left(161\right)}^2}\sqrt{8\left(12565\right)-{\left(255\right)}^2}}\\ =\frac{56568-41055}{\sqrt{10743}\sqrt{35495}}\\ =\frac{15513}{19527.4879}\\ =0.794\end{array}$

Hence, the value of r is 0.794.

(d)

To determine

Identify the data for averages have a higher correlation coefficient than individual measurements or not.

Mention the reasons why hours lost per individual and fuel wasted per individual might vary more than the same quantities averaged over all the people in a city.

Explanation of Solution

From part (a), the value of r for averages is 0.991. From part (c), the value of r for single observations is 0.794.

It can be observed that, the correlation coefficient for data of averages is larger than correlation coefficient for data of single observations.

The values of standard deviations for variables x and y based on data of averages are smaller when compared to the values of standard deviations for variables x and y based on data of single observations. The smaller standard deviation results in larger value of r, giving a higher correlation coefficient for data of averages.

Also, based on the central limit theorem the distribution of x distribution has the larger standard deviation when compared to $\overline{x}$ distribution.