Chapter 12.1, Problem 18E

To determine

(a)

To Calculate:

The correlation coefficient, $r$ from given table.

Answer to Problem 18E

Solution:

The correlation coefficient is $r=0.8453$.

Explanation of Solution

Given:

Number of Absences and class Average
Absences	2	2	3	10	3	7	9	1	12	9	1	1	13	1	10	3
Class Average	86	83	81	53	92	71	68	79	53	78	77	85	62	97	54	79

Formula used:

Pearson Correlation Coefficient:

The Pearson Correlation Coefficient for paired data from a sample:

$r=\frac{n{\displaystyle \sum x_i{y}_i-\left({\displaystyle \sum x_i}\right)\left({\displaystyle \sum y_i}\right)}}{\sqrt{n{{\displaystyle \sum x_i}}^2-{\left({\displaystyle \sum x_i}\right)}^2}\sqrt{n{{\displaystyle \sum y_i}}^2-{\left({\displaystyle \sum y_i}\right)}^2}}$

Where $n$ is the number of data pairs,

$x_i$ is the $i^{th}$ value of the explanatory variable,

$y_i$ is the $i^{th}$ value of the response variable.

While $r$ takes the value between $-1$ and $1$; (i.e.) $-1\le r\le 1$.

Calculation:

Let $x$ be the number of absences and $y$ be the Class Average.

Here, $n=16$

$x_i$	$y_i$	$x_i\cdot y_i$	$x_i{}^2$	$y_i{}^2$
2	86	172	4	7396
2	83	166	4	6889
3	81	243	9	6561
10	53	530	100	2809
3	92	276	9	8464
7	71	497	49	5041
9	68	612	81	4624
1	79	79	1	6241
12	53	636	144	2809
9	78	702	81	6084
1	77	77	1	5929
1	85	85	1	7225
13	62	806	169	3844
1	97	97	1	9409
10	54	540	100	2916
3	79	237	9	6241
${\displaystyle \sum x_i}=87$	${\displaystyle \sum y_i}=1198$	${\displaystyle \sum x_i{y}_i}=5755$	${{\displaystyle \sum x_i}}^2=763$	${{\displaystyle \sum y_i}}^2=92482$

To find Pearson Correlation Coefficient $r$,

$\begin{array}{rll}r& =& \frac{n{\displaystyle \sum x_i{y}_i-\left({\displaystyle \sum x_i}\right)\left({\displaystyle \sum y_i}\right)}}{\sqrt{n{{\displaystyle \sum x_i}}^2-{\left({\displaystyle \sum x_i}\right)}^2}\sqrt{n{{\displaystyle \sum y_i}}^2-{\left({\displaystyle \sum y_i}\right)}^2}}\\ & =& \frac{16\left(5755\right)-\left(87\right)\left(1198\right)}{\sqrt{16\left(763\right)-{\left(87\right)}^2}\sqrt{16\left(92482\right)-{\left(1198\right)}^2}}\\ & =& \frac{92080-104226}{\sqrt{12208-7569}\sqrt{1479712-1435204}}\\ & =& \frac{-12146}{\sqrt{4639}\sqrt{44508}}\\ & =& \frac{-12146}{\left(68.1102\right)\cdot \left(210.9692\right)}\\ & =& \frac{-12146}{14369.1544}\\ & =& 0.8453\end{array}$

Therefore, the correlation coefficient of $r=0.8453$.

To determine

(b)

The type of distribution to use for the test statistics and state the level of significance.

Answer to Problem 18E

Solution:

The correlation coefficient is statistically significant at the level of significance $\alpha =0.05$ for the sample size $n=16$.

Explanation of Solution

Procedure:

The correlation coefficient $r$ is statistically significant, if the absolute value of the correlation coefficient is greater than or equal to the critical value, $r_{\alpha }$, from the table critical values of the Pearson correlation coefficient.

Where $\alpha$ level of significance and $n$ is the sample size.

On other hand,

A sample correlation coefficient, $r$, is statistically significant if $\left|r\right|\ge r_{\alpha }$.

Given:

Correlation Coefficient is $r=0.8453$; $n=16\text{and}\alpha =0.05$

To find the critical value $r_{\alpha }$:

The critical value $r_{\alpha }$ for $\alpha =0.05$ with $n=16$ from the Pearson correlation coefficient table is $r_{\alpha }=0.497$.

Comparing this critical value to the absolute value of the correlation coefficient. We get the relation,

$0.845>0.497$

Thus, $\left|r\right|>r_{\alpha }$.

Therefore the correlation coefficient is statistically significant at the level of significance $\alpha =0.05$ for the sample size $n=16$.

To determine

(c)

To Find:

The coefficient of determination, $r^2$.

Answer to Problem 18E

Solution:

The coefficient of determination $r^2=0.7145$.

Explanation of Solution

Definition:

The coefficient of determination, $r^2$, is a measure of the proportion of the variation in the response variable $(y)$ that can be associated with the variation in the explanatory variable $(x)$.

Calculation:

Correlation Coefficient is $r=0.8453$.

Coefficient of determination is the square of the Correlation Coefficient.

Then,

$\begin{array}{rll}{r}^2& =& {\left(0.8453\right)}^2\\ & =& 0.7145\end{array}$

Thus, the Coefficient of determination $r^2=0.7145$.

To determine

(d)

To Interpret:

The Coefficient of determination $r^2$ for the given set of data.

Answer to Problem 18E

Solution:

$71.5\%$ of the variation in their class average that can be associated with the variation in number of absences a student had for the semester.

Explanation of Solution

The correlation coefficient for the relationship between the number of Absences a student had for the semester and their class average is $r=0.8453$

The coefficient of determination, $r^2$, is a measure of the proportion of the variation in the response variable (class average) that can be associated with the variation in the explanatory variable (Absences).

Coefficient of determination is, $r^2=0.7145$.

Interpretation:

Thus, approximately $71.5\%$ of the variation in their class average that can be associated with the variation in number of absences a student had for the semester.