Chapter 4.2, Problem 35E

Least-squares regression line for z-scores: The following table presents math and verbal SAT scores for

1. Compute the correlation coefficient between math and verbal SAT score.
2. Compute the mean and the standard deviation $s_x$ for the verbal scores.
3. Compute the mean and the standard deviation $s_y$ for the math scores.
4. Compute the least- squares regression he for predicting math score from verbal score.
5. Compute the z-score for each x-value: $z_x=\frac{x-\bar{x}}{s_x}$
6. Compute the z-score for each y-value: $z_y=\frac{y-\bar{y}}{s_y}$
7. Compute the correlation coefficient r between $z_x$ and $z_y$. Is it the same as the correlation between math and verbal SAT scores?
8. Compute the least-squares regression line for predicting $z_y$ from $z_x$. Explain why the equation of the line is $z_y=r{z}_x.$

To determine

To compute:The correlation coefficient between math and verbal SAT score.

Answer to Problem 35E

The correlation coefficient is 0.750.

Explanation of Solution

Given information:The following table presents the math and verbal SAT scores for six freshmen.

Verbal, x	428	386	653	316	438	323
Math, y	373	571	686	319	607	440

Formula used:The correlation coefficient is given by

  $r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}$ where $\overline{x}$ and $\overline{y}$ are the means of variable x and y and $s_x$ and $s_y$ are the standard deviations of x and y respectively.

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ \overline{y}=\frac{{\displaystyle \sum y}}{n}\\ s_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\\ s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n-1}}\end{array}$

Calculation:

From the table, the number of variables is $n=6$

The means of x and y are given by

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{428+386+653+316+438+323}{6}\\ =424\\ \overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{373+571+686+319+607+440}{6}\\ =499.33\end{array}$

The standard deviations of x and y are

Verbal, x	Math, y	$x-\overline{x}$	$y-\overline{y}$
428	373	4	-126.33
386	571	-38	71.67
653	686	229	186.61
316	319	-108	-180.33
438	607	14	107.67
323	440	-101	-59.33
${\displaystyle \sum x}=2544$	${\displaystyle \sum x}=2996$	${\displaystyle \sum \left(x-\overline{x}\right)=0}$	${\displaystyle \sum \left(y-\overline{y}\right)=252.62}$

${\left(x-\overline{x}\right)}^2$	${\left(y-\overline{y}\right)}^2$
16	15959.2689
1444	5136.5889
52441	34823.2921
11664	32518.9089
196	11592.8289
10,201	3520.0489
${\displaystyle \sum {\left(x-\overline{x}\right)}^2}=75962$	${\displaystyle \sum {\left(y-\overline{y}\right)}^2}=103550.9366$

  $\begin{array}{c}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{75962}{6-1}}\\ =\sqrt{15192.4}\\ =123.257\end{array}$

  $\begin{array}{c}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n-1}}\\ =\sqrt{\frac{103550.9366}{6-1}}\\ =\sqrt{20,710.18732}\\ =143.91\end{array}$

$\frac{x-\overline{x}}{s_x}$	$\frac{y-\overline{y}}{s_y}$	$\left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)$
0.0325	-0.8778	-0.0285
-0.3083	0.4980	0.1535
1.8579	1.2967	2.4091
-0.8762	-1.2531	1.098
0.1136	0.7482	0.085
-0.8194	-0.4123	0.3378
${\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)=0.0001}$	${\displaystyle \sum \left(\frac{y-\overline{y}}{s_y}\right)=-0.0003}$	${\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)=3.7512}$

The correlation coefficient is $r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}$

  $\begin{array}{c}r=\frac{1}{6-1}\left(3.7512\right)\\ =\frac{3.75}{5}\\ =0.750\end{array}$

Hence, the correlation coefficient is 0.750.

To determine

To compute:The mean $\overline{x}$ and the standard deviation $s_x$ for the verbal scores.

Answer to Problem 35E

The mean and standard deviation of x are $\overline{x}=424$ and $s_x=123.257$ .

Explanation of Solution

Given information:The following table presents the math and verbal SAT scores for six freshmen.

Verbal, x	428	386	653	316	438	323
Math, y	373	571	686	319	607	440

Formula used:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ s_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\end{array}$

Calculation:

The mean of verbal, x is given by

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{428+386+653+316+438+323}{6}\\ =424\end{array}$

The standard deviations of x is

  $\begin{array}{c}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{75962}{6-1}}\\ =\sqrt{15192.4}\\ =123.257\end{array}$

Hence, the mean and standard deviation of x are $\overline{x}=424$ and $s_x=123.257$ .

To determine

To compute: The mean $\overline{y}$ and the standard deviation $s_y$ for the verbal scores.

Answer to Problem 35E

The mean and standard deviation of y are $\overline{y}=499.33$ and $s_y=143.91$ .

Explanation of Solution

Given information:The following table presents the math and verbal SAT scores for six freshmen.

Verbal, x	428	386	653	316	438	323
Math, y	373	571	686	319	607	440

Formula used:

  $\begin{array}{l}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n-1}}\end{array}$

Calculation:

  $\begin{array}{c}\overline{y}=\frac{{\displaystyle \sum y}}{n}\\ =\frac{373+571+686+319+607+440}{6}\\ =499.33\end{array}$

The standard deviations of y is

  $\begin{array}{c}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n-1}}\\ =\sqrt{\frac{103550.9366}{6-1}}\\ =\sqrt{20,710.18732}\\ =143.91\end{array}$

Hence, the mean and standard deviation of y are $\overline{y}=499.33$ and $s_y=143.91$ .

To determine

To compute: The least-square regression line for predicting math from verbal scores.

Answer to Problem 35E

The least-square regression line for predicting math from verbal scores is $\widehat{y}=128.0332+0.8757x$ .

Explanation of Solution

Given information:The following table presents the math and verbal SAT scores for six freshmen.

Verbal, x	428	386	653	316	438	323
Math, y	373	571	686	319	607	440

Formula used:The least-square regression line for predicting y from x is

  $\widehat{y}=b_0+b_{1}x$

Where $b_1=r\frac{s_y}{s_x}$ and $b_0=\overline{y}-b_1\overline{x}$ .

Calculation:

We have

  $\overline{x}=424$ and $\overline{y}=499.33$

  $s_x=123.257$ and $s_y=143.91$

  $r=0.750$

Then $b_1=r\frac{s_y}{s_x}$

  $\begin{array}{c}{b}_1=\left(0.750\right)\frac{143.91}{123.257}\\ =0.8757\end{array}$

  $\begin{array}{c}{b}_0=\overline{y}-b_1\overline{x}\\ =499.33-\left(0.8757\right)\left(424\right)\\ =128.0332\end{array}$

Therefore, the least-square regression line for predicting math from verbal scores is $\widehat{y}=b_0+b_{1}x$

  $\Rightarrow \widehat{y}=128.0332+0.8757x$

Hence, the least-square regression line for predicting math from verbal scores is $\widehat{y}=128.0332+0.8757x$ .

To determine

To compute: The z-score for each x-value: $z_x=\frac{x-\overline{x}}{s_x}$

Answer to Problem 35E

The values of $z_x=\frac{x-\overline{x}}{s_x}$ are given by $0.0325,-0.3083,1.8579,-0.8762,0.1136 and -\mathrm{0.8194.}$

Explanation of Solution

Given information:The following table presents the math and verbal SAT scores for six freshmen.

Verbal, x	428	386	653	316	438	323
Math, y	373	571	686	319	607	440

Calculation:

The values of $z_x=\frac{x-\overline{x}}{s_x}$ are given in the following table:

Verbal, x	Math, y	$z_x=\frac{x-\overline{x}}{s_x}$
428	373	0.0325
386	571	-0.3083
653	686	1.8579
316	319	-0.8762
438	607	0.1136
323	440	-0.8194

Hence, the values of $z_x=\frac{x-\overline{x}}{s_x}$ are given by $0.0325,-0.3083,1.8579,-0.8762,0.1136 and -\mathrm{0.8194.}$

To determine

To compute: The z-score for each y-value: $z_y=\frac{y-\overline{y}}{s_y}$

Answer to Problem 35E

The values of $z_y=\frac{y-\overline{y}}{s_y}$ are given by $-0.8778,0.4980,1.2967,-1.2531,0.7482 and -\mathrm{0.4123.}$

Explanation of Solution

Given information:The following table presents the math and verbal SAT scores for six freshmen.

Verbal, x	428	386	653	316	438	323
Math, y	373	571	686	319	607	440

Calculation:

The values of $z_y=\frac{y-\overline{y}}{s_y}$ are given in the following table:

Verbal, x	Math, y	$z_y=\frac{y-\overline{y}}{s_y}$
428	373	-0.8778
386	571	0.4980
653	686	1.2967
316	319	-1.2531
438	607	0.7482
323	440	-0.4123

Hence, the values of $z_y=\frac{y-\overline{y}}{s_y}$ are given by $-0.8778,0.4980,1.2967,-1.2531,0.7482 and -\mathrm{0.4123.}$

To determine

To compute: The correlation coefficient r between $z_x$ and $z_y$ , to find whether it is same as the correlation between math and verbal SAT score.

Answer to Problem 35E

The correlation coefficient r between $z_x$ and $z_y$ is 0.75 which is same as the correlation between math and verbal SAT score.

Explanation of Solution

Given information:The following table presents the math and verbal SAT scores for six freshmen.

Verbal, x	428	386	653	316	438	323
Math, y	373	571	686	319	607	440

Formula used:

The correlation coefficient is given by

  $r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}$ where $\overline{x}$ and $\overline{y}$ are the means of variable x and y and $s_x$ and $s_y$ are the standard deviations of x and y respectively.

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ \overline{y}=\frac{{\displaystyle \sum y}}{n}\\ s_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\\ s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n-1}}\end{array}$

Calculation:

$z_x=\frac{x-\overline{x}}{s_x}$	$z_y=\frac{y-\overline{y}}{s_y}$
0.0325	-0.8778
-0.3083	0.4980
1.8579	1.2967
-0.8762	-1.2531
0.1136	0.7482
-0.8194	-0.4123
$\begin{array}{l}{\displaystyle \sum z_x}=0.0001\\ \overline{z_x}=0\end{array}$	$\begin{array}{l}{\displaystyle \sum z_y}=-0.0003\\ \overline{z_y}=0\end{array}$

  ${{\displaystyle \sum \left(z_x-\overline{z_x}\right)}}^2=4$

  ${{\displaystyle \sum \left(z_y-\overline{z_y}\right)}}^2=5$

The standard deviation of $z_x$ and $z_y$ are given by

  $\begin{array}{c}{s}_{z_x}=\sqrt{\frac{{\displaystyle \sum {\left(z_x-\overline{z_x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{5}{5}}\\ =1\\ s_{z_y}=\sqrt{\frac{{\displaystyle \sum {\left(z_y-\overline{z_y}\right)}^2}}{n-1}}\\ =\sqrt{\frac{5}{5}}\\ =1\end{array}$

$\frac{z_x-\overline{z_x}}{s_{z_x}}$	$\frac{z_y-\overline{z_y}}{s_{z_y}}$	$\left(\frac{z_x-\overline{z_x}}{s_{z_x}}\right)\left(\frac{z_y-\overline{z_y}}{s_{z_y}}\right)$
0.0325	-0.8778	-0.02853
-0.3083	0.4980	-0.1535
1.8579	1.2967	2.4091
-0.8762	-1.2531	1.098
0.1136	0.7482	0.085
-0.8194	-0.4123	0.3378
$\begin{array}{l}{\displaystyle \sum z_x}=0.0001\\ \overline{z_x}=0\end{array}$	$\begin{array}{l}{\displaystyle \sum z_y}=-0.0003\\ \overline{z_y}=0\end{array}$	${\displaystyle \sum \left(\frac{z_x-\overline{z_x}}{s_{z_x}}\right)\left(\frac{z_y-\overline{z_y}}{s_{z_y}}\right)}=3.75$

The correlation coefficient is given by

  $\begin{array}{c}r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{z_x-\overline{z_x}}{s_{z_x}}\right)\left(\frac{z_y-\overline{z_y}}{s_{z_y}}\right)}\\ =\frac{1}{6-1}\left(3.75\right)\\ =0.750\end{array}$

Hence, the correlation coefficient r between $z_x$ and $z_y$ is 0.75 which is same as the correlation between math and verbal SAT score.

To determine

To compute: The least-square regression line for predicting $z_x$ and $z_y$ and to explain why the equation is $z_y=r{z}_x$ .

Explanation of Solution

Given information:The following table presents the math and verbal SAT scores for six freshmen.

Verbal, x	428	386	653	316	438	323
Math, y	373	571	686	319	607	440

Formula used: The least-square regression line for predicting y from x is

  $\widehat{y}=b_0+b_{1}x$

Where $b_1=r\frac{s_y}{s_x}$ and $b_0=\overline{y}-b_1\overline{x}$ .

Calculation:

The least-square regression line for predicting $z_y$ from $z_x$ is

  $\widehat{z_y}=b_0+b_1{z}_x$

Where $b_1=r\frac{s_{z_y}}{s_{z_x}}$ and $b_0=\overline{z_y}-b_1\overline{z_x}$ .

We have $s_{z_x}=s_{z_y}=1$ and $\overline{z_x}=\overline{z_y}=0$

Then,

  $\begin{array}{c}{b}_1=r\frac{s_{z_y}}{s_{z_x}}\\ =r\frac{1}{1}\\ =r\end{array}$

  $\begin{array}{c}{b}_0=\overline{z_y}-b_1\overline{z_x}\\ =0-0\\ =0\end{array}$

Then the least-square regression line for predicting $z_y$ from $z_x$ is $\widehat{z_y}=r{z}_x$ .

The least-square regression line for predicting $z_y$ from $z_x$ is $\widehat{z_y}=r{z}_x$ because the means of $\overline{z_x}and\overline{z_y}$ are both zero, the least-square regression line goes through the point of average

, the y-intercept is zero, the slope of the least-square regression line is $b_1=r\frac{s_{z_y}}{s_{z_x}}=r$ because $s_{z_x}=s_{z_y}=1$ .