Chapter 3, Problem 3.30SE

a.

To determine

Draw a scatterplot for the given points.

Answer to Problem 3.30SE

The relation is linear and positive.

Explanation of Solution

Given:

The table shows therelation between the number of passes completed and the total number of passing yards.

Calculation:

The scatterplot for the given points are.

The horizontal axis x represents the completion and the vertical axis y represents the total yards.

b.

To determine

Find the relationship between pretest and posttest scores.

Answer to Problem 3.30SE

The relationship is positive, moderately strong and linear and the outlier is $\left(7,46\right)$ .

Explanation of Solution

Given:

The table shows the relation between the number of passes completed and the total number of passing yards.

Calculation:

The scatterplot for the given points are.

The horizontal axis x represents the completion and the vertical axis y represents the total yards.

In the scatterplot,

The pattern position is upward $\to$ The direction is positive.

Strength $\to$ moderately strong, because points far apart from general pattern.

Formed $\to$ graph shows the linear pattern, because there is no curvature.

Outliers $\to$ the outliers in the plot is $\left(7,46\right)$ which lie left corner of the scatterplot.

Hence the relationship is positive, moderately strong and linear and the outlier is $\left(7,46\right)$ .

c.

To determine

Find the correlation coefficient using the given points.

Answer to Problem 3.30SE

The correlation coefficient $r\approx 0.796$ .

Explanation of Solution

Given:

The table shows the relation between the number of passes completed and the total number of passing yards.

Calculation:

First determine ${\displaystyle \sum x_i}{y}_i$ , ${\displaystyle \sum x_i}$ and ${\displaystyle \sum y_i}$ .

  $x_i\to$ first coordinates of the ordered pairs.

  $y_i\to$ second coordinates of the ordered pairs.

  $\begin{array}{l}{\displaystyle \sum x_i}{y}_i=19\cdot 188+19\cdot 255+34\cdot 316+12\cdot 181+27\cdot 293+18\cdot 313+21\cdot 295+15\cdot 170+27\cdot 289+22\cdot 301+26\cdot 344+21\cdot 295+21\cdot 298+7\cdot 46+25\cdot 404+19\cdot 229=88023\\ {\displaystyle \sum x_i}=19+19+34+12+27+18+21+15+27+22+26+21+7+25+19=312\\ {{\displaystyle \sum x_i}}^2={19}^2+{19}^2+{34}^2+{12}^2+{27}^2+{18}^2+{21}^2+{15}^2+{27}^2+{22}^2+{26}^2+{21}^2+7^2+{25}^2+{19}^2=65993\\ {\displaystyle \sum y_i=188+255+316+181+293+313+295+170+289+301+344+295+298+46+404+229=39}22\\ {\displaystyle \sum y_i{}^2={188}^2+{255}^2+{316}^2+{181}^2+{293}^2+{313}^2+{295}^2+{170}^2+{289}^2+{301}^2+{344}^2+{295}^2+{298}^2+{46}^2+{404}^2+{229}^2=1131764}\end{array}$

Find the sample variance $s^2$ using the formula $s^2=\frac{{\displaystyle \sum x_i{}^2}-\frac{{\displaystyle \sum x_i{}^2}}{n}}{n-1}$

Where $n=$ number of ordered pairs.

  $n=15$

  $\begin{array}{l}{s}_x{}^2=\frac{7106-\frac{{312}^2}{15}}{15-1}=\frac{7106-\frac{{312}^2}{15}}{14}\approx 44.028\\ s_y{}^2=\frac{1131764-\frac{{3922}^2}{15}}{15-1}=\frac{1131764-\frac{{3922}^2}{15}}{14}\approx 7592.267\end{array}$

Find sample standard deviation.

  $\begin{array}{l}{s}_x=\sqrt{44.028}\approx 6.635\\ s_y=\sqrt{7592.267}\approx 87.134\end{array}$

For covariance $s_{xy}$ using formula $s_{xy}=\frac{{\displaystyle \sum x_i{y}_i}-\frac{{\displaystyle \sum x_i{\displaystyle \sum y_i}}}{n}}{n-1}$ .

  $\begin{array}{l}{s}_{xy}=\frac{88023-\frac{312\cdot 3922}{15}}{15-1}\\ =\frac{88023-\frac{312\cdot 3922}{15}}{14}\\ \approx 460.38\end{array}$

Find the correlation coefficient $r$ using the formula $r=\frac{s_{xy}}{s_x{s}_y}$ .

  $r=\frac{460.38}{6.64\cdot 87.134}\approx 0.796$

Hence the correlation coefficient $r\approx 0.796$

d.

To determine

Find the correlation coefficient using the given points.

Answer to Problem 3.30SE

The regression line is $y=43.97+10.456x$ .

Explanation of Solution

Given:

The table shows the relation between the number of passes completed and the total number of passing yards.

Calculation:

First determine ${\displaystyle \sum x_i}{y}_i$ , ${\displaystyle \sum x_i}$ and ${\displaystyle \sum y_i}$ .

  $x_i\to$ first coordinates of the ordered pairs.

  $y_i\to$ second coordinates of the ordered pairs.

  $\begin{array}{l}{\displaystyle \sum x_i}{y}_i=19\cdot 188+19\cdot 255+34\cdot 316+12\cdot 181+27\cdot 293+18\cdot 313+21\cdot 295+15\cdot 170+27\cdot 289+22\cdot 301+26\cdot 344+21\cdot 295+21\cdot 298+7\cdot 46+25\cdot 404+19\cdot 229=88023\\ {\displaystyle \sum x_i}=19+19+34+12+27+18+21+15+27+22+26+21+7+25+19=312\\ {{\displaystyle \sum x_i}}^2={19}^2+{19}^2+{34}^2+{12}^2+{27}^2+{18}^2+{21}^2+{15}^2+{27}^2+{22}^2+{26}^2+{21}^2+7^2+{25}^2+{19}^2=65993\\ {\displaystyle \sum y_i=188+255+316+181+293+313+295+170+289+301+344+295+298+46+404+229=39}22\\ {\displaystyle \sum y_i{}^2={188}^2+{255}^2+{316}^2+{181}^2+{293}^2+{313}^2+{295}^2+{170}^2+{289}^2+{301}^2+{344}^2+{295}^2+{298}^2+{46}^2+{404}^2+{229}^2=1131764}\end{array}$

Find the sample variance $s^2$ using the formula $s^2=\frac{{\displaystyle \sum x_i{}^2}-\frac{{\displaystyle \sum x_i{}^2}}{n}}{n-1}$

Where $n=$ number of ordered pairs.

  $n=15$

  $\begin{array}{l}{s}_x{}^2=\frac{7106-\frac{{312}^2}{15}}{15-1}=\frac{7106-\frac{{312}^2}{15}}{14}\approx 44.028\\ s_y{}^2=\frac{1131764-\frac{{3922}^2}{15}}{15-1}=\frac{1131764-\frac{{3922}^2}{15}}{14}\approx 7592.267\end{array}$

Find sample standard deviation.

  $\begin{array}{l}{s}_x=\sqrt{44.028}\approx 6.635\\ s_y=\sqrt{7592.267}\approx 87.134\end{array}$

For covariance $s_{xy}$ using formula $s_{xy}=\frac{{\displaystyle \sum x_i{y}_i}-\frac{{\displaystyle \sum x_i{\displaystyle \sum y_i}}}{n}}{n-1}$ .

  $\begin{array}{l}{s}_{xy}=\frac{88023-\frac{312\cdot 3922}{15}}{15-1}\\ =\frac{88023-\frac{312\cdot 3922}{15}}{14}\\ \approx 460.38\end{array}$

Find the correlation coefficient $r$ using the formula $r=\frac{s_{xy}}{s_x{s}_y}$ .

  $r=\frac{460.38}{6.64\cdot 87.134}\approx 0.796$

Hence the correlation coefficient $r\approx 0.796$

Find the slope $b$ using the formula $b=r\cdot \frac{s_y}{s_x}$

  $b=0.79\cdot \frac{87.13}{6.64}\approx 10.456$

Find the value of y-intercept.

  $\begin{array}{l}a=\overline{y}-b\overline{x}=\frac{{\displaystyle \sum y_i}}{n}-b\frac{{\displaystyle \sum x_i}}{n}\\ =\frac{3922}{15}-\left(10.456\right)\frac{312}{15}\approx 43.97\end{array}$

Regression line is $y=a+bx$

  $y=43.97+10.456x$

Hence the regression line is $y=43.97+10.456x$ .

e.

To determine

Find the total number of passing yards for given completion.

Answer to Problem 3.30SE

The total number of yards is $253$ .

Explanation of Solution

Given:

The given completion is $20$ .

Calculation:

The equation is $y=43.97+10.456x$ .

Substitute the value of $x=20$ in the equation.

  $y=43.97+10.456\cdot 20=253$

Hence the total number of yards is $253$ .