Chapter 4.1, Problem 23P

(a)

To determine

The scatter plot, whether the provided values of ${\displaystyle \sum x}$, ${\displaystyle \sum y}$, ${\displaystyle \sum x^2}$ $,{\displaystyle \sum y^2}$, ${\displaystyle \sum xy}$ are correct or not, and the correlation coefficient.

Answer to Problem 23P

Solution: The provided values, that is, ${\displaystyle \sum x}=154$, ${\displaystyle \sum y=249}$, ${\displaystyle \sum x^2=3712}$, ${\displaystyle \sum y^2}=9959$ ${\displaystyle \sum xy}=6067$ are correct and the value of r is 0.991.

Explanation of Solution

Given: The provided table consists of values of x and y, where x represents the average annual hours spent by a person in traffic delay, y represents the average annual gallons of fuel wasted per person due to traffic delay. The data consists of 8 data pairs, thus n is 8.

Calculation: Follow the steps given below in MS Excel to obtain the scatter plot of the data.

Step 1: Enter the data into an MS Excel sheet. The screenshot is given below.

Step 2: Select the data and click on ‘Insert’. Go to charts and select the chart type ‘Scatter’.

Step 3: Select the first plot and then click ‘add chart element’ provided in the left corner of the menu bar. Insert the ‘Axis titles’ and ‘Chart title’. The scatter plot for the provided data is shown below:

To calculate ${\displaystyle \sum x}$, ${\displaystyle \sum y}$, ${\displaystyle \sum x^2}$ $,{\displaystyle \sum y^2}$ and ${\displaystyle \sum xy}$, it is easy to form the data in a table of five columns. The table is given 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
${\displaystyle \sum x}=154$	${\displaystyle \sum y=249}$	${\displaystyle \sum x^2=3712}$	${\displaystyle \sum y^2}=9959$	${\displaystyle \sum xy}=6067$

The provided values, ${\displaystyle \sum x}=154$, ${\displaystyle \sum y=249}$, ${\displaystyle \sum x^2=3712}$, ${\displaystyle \sum y^2}=9959$ and ${\displaystyle \sum xy}=6067$ have been verified.

Now, the value of $r$ can be calculated by using the formula below:

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

Substituting the values in the above formula. Thus:

$\begin{array}{c}r=\frac{8\left(6067\right)-\left(154\right)\left(249\right)}{\sqrt{\left(8\right)\left(3712\right)-{\left(154\right)}^2}\sqrt{\left(8\right)\left(9959\right)-{\left(249\right)}^2}}\\ \approx 0.991\end{array}$

Therefore, the correlation coefficient is 0.991.

(b)

To determine

The averages $\overline{x},\overline{y}$ and the standard deviations $s_x,s_y$ for both the data sets, the comparison between the standard deviations of both the samples, and the reason behind the tendency of an increase in the value of r for smaller standard deviations $s_x\text{ and }{s}_y$.

Answer to Problem 23P

Solution: The values for data set 1 are $\overline{x}=19.25,\overline{y}=31.13,s_x\approx 10.33$ and $s_y=17.76$.

The values for data set 2 are $\overline{x}=20.13,\overline{y}=31.87,s_x\approx 13.84$ and $s_y=25.18$.

Explanation of Solution

Given: The provided table consists of values of x and y, where x represents the average annual hours spent by a person in traffic delay, y represents the average annual gallons of fuel wasted per person due to traffic delay.

The second table consists of x and y values where, x represent the annual hours lost by a person spent in traffic delay, y represents the annual gallons of fuel wasted by that person in traffic delay.

The data sets consist of 8 data pairs, thus n is 8 for both the data sets.

The provided values of data set 1 are, ${\displaystyle \sum x}=154$, ${\displaystyle \sum y=249}$, ${\displaystyle \sum x^2=3712}$, ${\displaystyle \sum y^2}=9959$ ${\displaystyle \sum xy}=6067$.

The provided values of data set 2 are, ${\displaystyle \sum x}=161$, ${\displaystyle \sum y=255}$, ${\displaystyle \sum x^2=4583}$, ${\displaystyle \sum y^2}=12565$ ${\displaystyle \sum xy}=7071$.

Calculation:

The value of $\overline{x}$ for data set 1 can be calculated as follows:

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{154}{8}\\ =19.25\end{array}$

The value of $\overline{y}$ for data set 1 can be calculated as follows:

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{249}{8}\\ =31.125\end{array}$

The standard deviation of x for data set 1 can be calculated as,

$\begin{array}{c}{s}_x=\sqrt{\frac{{\displaystyle \sum x^2-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}}}{n-1}}\\ =\sqrt{\frac{3712-\frac{{154}^2}{8}}{8-1}}\\ \approx 10.33\end{array}$

The standard deviation of $y$ for data set 1 can be calculated as,

$\begin{array}{c}{s}_y=\sqrt{\frac{{\displaystyle \sum y^2-\frac{{\left({\displaystyle \sum y}\right)}^2}{n}}}{n-1}}\\ =\sqrt{\frac{9959-\frac{{249}^2}{8}}{8-1}}\\ \approx 17.76\end{array}$

The value of $\overline{x}$ for data set 2 can be calculated as follows:

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{161}{8}\\ =20.13\end{array}$

The value of $\overline{y}$ for data set 2 can be calculated as follows:

$\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{255}{8}\\ =31.87\end{array}$

The standard deviation of $x$ for data set 2 can be calculated as,

$\begin{array}{c}{s}_x=\sqrt{\frac{{\displaystyle \sum x^2-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}}}{n-1}}\\ =\sqrt{\frac{4583-\frac{{161}^2}{8}}{8-1}}\\ \approx 13.84\end{array}$

The standard deviation of $y$ for data set 2 can be calculated as,

$\begin{array}{c}{s}_y=\sqrt{\frac{{\displaystyle \sum y^2-\frac{{\left({\displaystyle \sum y}\right)}^2}{n}}}{n-1}}\\ =\sqrt{\frac{12565-\frac{{255}^2}{8}}{8-1}}\\ \approx 25.18\end{array}$

For the second data set, that is, for the variables based on single individuals, the standard deviations $s_x\text{ and }{s}_y$ are larger.

The values $s_x\text{ and }{s}_y$ are in the denominator in the formula for calculating r. Dividing by smaller values of $s_x\text{ and }{s}_y$ tends to increase the value of r.

(c)

To determine

The scatter plot, whether the provided values of ${\displaystyle \sum x}$, ${\displaystyle \sum y}$, ${\displaystyle \sum x^2}$ $,{\displaystyle \sum y^2}$, ${\displaystyle \sum xy}$ are correct or not, and the correlation coefficient.

Answer to Problem 23P

Solution: The provided values, that is, ${\displaystyle \sum x}=161$, ${\displaystyle \sum y=255}$, ${\displaystyle \sum x^2=4583}$, ${\displaystyle \sum y^2}=12565$ ${\displaystyle \sum xy}=7071$ are correct and the value of r is 0.794.

Explanation of Solution

The provided table consists of values of x and y, where x represents the average annual hours spent by a person in traffic delay, y represents the average annual gallons of fuel wasted per person due to traffic delay.

The data sets consist of 8 data pairs, thus n is 8.

Calculation: Follow the steps given below in MS Excel to obtain the scatter plot of the data.

Step 1: Enter the data into an MS Excel sheet. The screenshot is given below.

Step 2: Select the data and click on ‘Insert’. Go to charts and select the chart type ‘Scatter’.

Step 3: Select the first plot and then click ‘add chart element’ provided in the left corner of the menu bar. Insert the ‘Axis titles’ and ‘Chart title’. The scatter plot for the provided data is shown below:

Calculation: The calculation for ${\displaystyle \sum x}$, ${\displaystyle \sum y}$, ${\displaystyle \sum x^2}$ $,{\displaystyle \sum y^2}$ and ${\displaystyle \sum xy}$ is shown 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
${\displaystyle \sum x}=161$	${\displaystyle \sum y=255}$	${\displaystyle \sum x^2=4583}$	${\displaystyle \sum y^2}=12565$	${\displaystyle \sum xy}=7071$

The provided values, ${\displaystyle \sum x}=161$, ${\displaystyle \sum y=255}$, ${\displaystyle \sum x^2=4583}$, ${\displaystyle \sum y^2}=12565$ and ${\displaystyle \sum xy}=7071$ have been verified.

Now, the value of $r$ can be calculated by using the formula below:

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

Substituting the values in the above formula. Thus:

$\begin{array}{c}r=\frac{8\left(7071\right)-\left(161\right)\left(255\right)}{\sqrt{\left(8\right)\left(4583\right)-{\left(161\right)}^2}\sqrt{\left(8\right)\left(12565\right)-{\left(255\right)}^2}}\\ \approx 0.794\end{array}$

Therefore, the correlation coefficient is 0.794.

(d)

To determine

Comparison between the values of r that are calculated in part (a) and part (c), whether the data for average have a higher correlation coefficient than the data for individual measurement or not, and the reason for it.

Answer to Problem 23P

Solution: Yes, the data for average has a higher correlation coefficient than the data for individual measurement because, according to the central limit theorem, the standard deviation of averages will be smaller than the standard deviation of individual values.

Explanation of Solution

Given: The values of correlation coefficient from part (a) and part (b) are 0.991 and 0.794, respectively.

It can be seen that $0.991>0.794$. The data for average has a higher correlation coefficient than the data for individual measurement. This is because the standard deviation for the average is smaller than the standard deviation for individual measurements.

According to the central limit theorem, the standard deviation is smaller for the $\overline{x}$ distribution than the corresponding x distribution.