Chapter 4.2, Problem 23E

Pass the ball: The following table lists the heights (inches) and weights (pounds) of 14 National Football League quarterbacks in the 2016 season.

1. Compute the least-squares regression line for predicting weight from height.
2. Is it possible to interpret the y-intercept? Explain.
3. If two quarterbacks differ in height by two inches: by how much would you predict their weights to differ?
4. Predict the of a quarterback who is 74.5 inches tall.
5. Kirk Cousins is 75 inches tall and weighs 214 pounds. Does he weigh more or less than the weight predicted by the least-squares regression line?

To determine

To find: The least-square regression line for the given data set.

Answer to Problem 23E

The least square regression line of the given data set is,

  $\widehat{y}=-146.766+4.9439x$

Explanation of Solution

Following table with the heights and the weights of the national football quarterbacks in the season $2016$ has been given.

Name	Height	Weight
Aaron Rogers	77	230
Cam Newton	77	244
Russell Wilson	71	206
Andrew Luck	76	234
Drew Brees	72	209
Blake Bortles	77	232
Ben Roethlisberger	77	241
Philip Rivers	77	228
Eli Manning	76	225
Tyrod Taylor	73	217
Jamies Winston	76	231
Carson Palmer	77	235
Kirk Cousins	75	214
Tom Brady	76	225

Calculation:

The least-square regression is given by the formula,

  $\widehat{y}=b_0+b_{1}x$

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

  $r$ is the correlation coefficient.

  $s_x$ is the standard deviation of $x$ .

  $s_y$ is the standard deviation of $y$ .

The correlation coefficient is given by the formula,

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

Let $x$ be the sales price and $y$ be the monthly rental. Here MINITAB is used.

  

The correlation coefficient can be obtained by the following table.

  $\begin{array}{l}\begin{array}{ccccc}x& y& \frac{x-\overline{x}}{s_x}& \frac{y-\overline{y}}{s_y}& \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)\\ 77& 230& 0.7394& 0.3072& 0.2271\\ 77& 244& 0.7394& 1.5360& 1.1357\\ 71& 206& -2.2182& -1.7993& 3.9913\\ 76& 234& 0.2465& 0.6583& 0.1622\\ 72& 209& -1.7253& -1.5360& 2.6500\\ 77& 232& 0.7394& 0.4827& 0.3569\\ 77& 241& 0.7394& 1.2727& 0.9410\\ 77& 228& 0.7394& 0.1317& 0.0973\\ 76& 225& 0.2465& -0.1317& -0.0325\\ 73& 217& -1.2324& -0.8338& 1.0276\\ 76& 231& 0.2465& 0.3950& 0.0973\\ 77& 235& 0.7394& 0.7461& 0.5516\\ 75& 214& -0.2465& -1.0971& 0.2704\\ 76& 225& 0.2465& -0.1317& -0.0325\end{array}\\ {\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}=11.4438\end{array}$

Hence, the correlation coefficient is,

  $\begin{array}{c}r=\frac{1}{14-1}\times 11.4438\\ =\frac{11.4438}{13}\\ r=0.8803\end{array}$

Then, the coefficient $b_1$ should be,

  $\begin{array}{c}{b}_1=r\frac{s_y}{s_x}\\ =0.8803\times \frac{11.3933}{2.0286}\\ b_1=4.9439\end{array}$

Therefore,

  $\begin{array}{c}{b}_0=\overline{y}-b_1\overline{x}\\ =226.5+4.9439\times 75.5\\ =226.5+725.8420\\ b_0=-146.766\end{array}$

Conclusion:

The least square regression line is found to be,

  $\widehat{y}=-146.766+4.9439x$

To determine

To explain:The possibility to interpret the $y$ intercept of the least-square regression line.

Answer to Problem 23E

The $y$ intercept cannot be interpreted for the weight of the players.

Explanation of Solution

The least-square regression line has been computed as $\widehat{y}=-146.766+4.9439x$ in the part (a).

Considering the $x$ as the height and $y$ as the weight of these $14$ players, the relationship is said to be $\widehat{y}=-146.766+4.9439x$ . Here the $y$ intercept has been obtained as $-146.766$ . It is clear that this is a negative value.

The $y$ intercept is defined as the $y$ value when $x$ is zero. Regarding this case, that $y$ intercept is negative which means when the height of a player is zero, his weight is a negative value.

Conclusion:

There is no chance to a weight be a negative value. Therefore, this $y$ intercept cannot be interpreted for this height-to-weight relationship.

To determine

To calculate:The difference in the weight of two players when their heights differ by two inches.

Answer to Problem 23E

The difference of weight is found to be $9.8878$ pounds.

Explanation of Solution

The least-square regression line has been computed as $\widehat{y}=-146.766+4.9439x$ in the part (a).

Calculation:

Let the height of the shorter payer be $x_0$ . Then, the taller one’s height should be $x_0+2$ . The corresponding weight of the shorter player can be obtained as,

  ${\widehat{y}}_{x_0}=-146.766+4.9439x_0$

Also, the weight of the taller player should be,

  ${\widehat{y}}_{x_0+2}=-146.766+4.9439\left(x_0+2\right)$

Simplifying the obtained weight,

  $\begin{array}{c}{\widehat{y}}_{x_0+2}=-146.766+4.9439\left(x_0+2\right)\\ =\underset{{\widehat{y}}_{x_0}}{\underbrace{-146.766+4.9439\times x_0}}+4.9439\times 2\\ {\widehat{y}}_{x_0+2}={\widehat{y}}_{x_0}+9.8878\end{array}$

Therefore, the difference of two weights should be,

  ${\widehat{y}}_{x_0+2}-{\widehat{y}}_{x_0}=9.8878$

Interpretation:

As the calculation above, the weight difference for two inches change in the height is found to be $9.8878$ pounds.

To determine

To find:The predicted weight of a player whose height is $74.5$ inches.

Answer to Problem 23E

The weight of a player whose height is $74.5$ inches is found to be $221.5546$ pounds.

Explanation of Solution

Calculation:

By the computed least-square regression line, we can calculate the predictions for the weights of the players by substituting their heights into the formula.

Here, the height is said to be $74.5$ inches. Hence, $x$ should be $74.5$ .

  $\begin{array}{c}\widehat{y}=-146.766+4.9439\times 74.5\\ =-146.766+368.3206\\ \widehat{y}=221.5546\end{array}$

Conclusion:

The weight of a player whose height is $74.5$ inches is found to be $221.5546$ pounds.

To determine

To find:Whether theactual weight of this player is greater than the predicted weight or not.

Answer to Problem 23E

The player weighs less than the predicted weight.

Explanation of Solution

The player is $214$ pounds in weight where his height is $75$ inches.

Calculation:

When the height is said to be $75$ inches, the variable $x$ should be equal to $75$ . Then, by substituting $x=75$ into the formula of least-square regression line, we can obtain the predicted weight of the player.

  $\begin{array}{c}\widehat{y}=-146.766+4.9439\times 75\\ =-146.766+368.3206\\ \widehat{y}=224.0265\end{array}$

The predicted weight is found to be $224.0265$ pounds.

Conclusion:

Since $224.0265>214$ the predicted weight of the player is greater than the actual weight.