Chapter 5.3, Problem 43E

a.

To determine

Find the value of r, for which, the value of $s_e$ is as large as $s_y$.

Identify the least squares line in this situation.

Answer to Problem 43E

The value of r, for which, the value of $s_e$ is as large as $s_y$ is 0.

The least squares line in this situation is $\underset{\_}{\widehat{y}=\overline{y}}$.

Explanation of Solution

Calculation:

It is given that the residual sum of squares can be expressed as $\text{SSRedid}=\left(1-r^2\right){\displaystyle \sum \left(y-\overline{y}\right)}$, which implies that $s_e=\sqrt{\frac{n-1}{n-2}}\sqrt{1-r^2}{s}_y$. If n is not too small, $s_e\approx \sqrt{1-r^2}{s}_y$.

If the value of $s_e$ is as large as $s_y$, then the following calculation can be shown:

$\begin{array}{c}{s}_e\approx \sqrt{1-r^2}{s}_y\\ \sqrt{1-r^2}\approx \frac{s_e}{s_y}\\ \approx 1.\\ 1-r^2={\left(1\right)}^2\end{array}$

       $\begin{array}{c}=1\\ r^2=1-1\\ =0\\ r=0.\end{array}$

Thus, the value of r, for which, the value of $s_e$ is as large as $s_y$ is 0.

Alternate form of the regression equation:

A regression equation involving one response variable, y and one predictor, x can be expressed as: $\widehat{y}=\overline{y}+r\left(\frac{s_y}{s_x}\right)\left(x-\overline{x}\right)$, where $\overline{x}$ and $\overline{y}$ are the means of x and y respectively; $s_x$ and $s_y$ are the standard deviations of x and y respectively.

Substitute $r=0$ in the equation:

$\begin{array}{c}\widehat{y}=\overline{y}+r\left(\frac{s_y}{s_x}\right)\left(x-\overline{x}\right)\\ =\overline{y}+0\cdot \left(\frac{s_y}{s_x}\right)\left(x-\overline{x}\right)\\ =\overline{y}.\end{array}$

Thus, the least squares line in this situation is $\underset{\_}{\widehat{y}=\overline{y}}$.

b.

To determine

Find the value of r, for which, the value of $s_e$ is much smaller than $s_y$.

Answer to Problem 43E

The value of r, for which, the value of $s_e$ is much smaller than $s_y$ is $\underset{\_}{\pm 1}$.

Explanation of Solution

Calculation:

If the value of $s_e$ is much smaller than $s_y$, then it can be said that $s_e$ is negligible with respect to $s_y$. In other words, the ratio, $\frac{s_e}{s_y}$ becomes approximately 0.  In this case, the following calculation can be shown:

$\begin{array}{c}{s}_e\approx \sqrt{1-r^2}{s}_y\\ \sqrt{1-r^2}\approx \frac{s_e}{s_y}\\ \approx 0.\\ 1-r^2={\left(0\right)}^2\end{array}$

       $\begin{array}{c}=0\\ r^2=1-0\\ =1\\ r=\pm 1.\end{array}$

Thus, the value of r, for which, the value of $s_e$ is much smaller than $s_y$ is $\underset{\_}{\pm 1}$.

c.

To determine

Find the value of $s_e$ for the regression line to predict the height at 18 years, from height in 6 years.

Answer to Problem 43E

The value of $s_e$ for the regression line to predict the height at 18 years, from height in 6 years is 1.5 inches.

Explanation of Solution

Calculation:

For the given institute, the value of r is approximately 0.80; average height at 6 years is 46 inches with standard deviation 1.7 inches; average height at 18 years is 70 inches with standard deviation 2.5 inches.

Consider the given relation $s_e\approx \sqrt{1-r^2}{s}_y$.

Here, $r=0.8$, $s_y=2.5$. Substitute these values in the given relation:

$\begin{array}{c}{s}_e\approx \sqrt{1-r^2}{s}_y\\ =\sqrt{1-{\left(0.8\right)}^2}\times 2.5\\ =\sqrt{1-0.64}\times 2.5\\ =\sqrt{0.36}\times 2.5\\ =0.6\times 2.5\\ =\mathrm{1.5.}\end{array}$

Thus, the value of $s_e$ for the regression line to predict the height at 18 years, from height in 6 years is 1.5 inches.

d.

To determine

Find the least-squares equation to predict the height at 6 years, using the height at 18 years.

Find the corresponding value of $s_e$.

Answer to Problem 43E

The least-squares equation to predict the height at 6 years, using the height at 18 years is $\underset{\_}{\widehat{y}=7.92+0.544x}$.

The corresponding value of $s_e$ is 1.02 inches.

Explanation of Solution

Calculation:

Consider the form of the least-squares equation given in Part a: $\widehat{y}=\overline{y}+r\left(\frac{s_y}{s_x}\right)\left(x-\overline{x}\right)$. While attempting to predict the height at 6 years, using the height at 18 years, the response variable, y is the height at 6 years and the predictor variable, x is the height at 18 years.

Under this consideration, $r=0.8$, $\overline{x}=70$, $\overline{y}=46$, $s_x=2.5$ and $s_y=1.7$. Substitute these values in the given form of the least-squares equation:

$\begin{array}{c}\widehat{y}=\overline{y}+r\left(\frac{s_y}{s_x}\right)\left(x-\overline{x}\right)\\ =46+\left(0.8\right)\left(\frac{1.7}{2.5}\right)\left(x-70\right)\\ =46+0.544\left(x-70\right)\\ =46+0.544x-\left(0.544\times 70\right)\\ =46+0.544x-38.08\\ =7.92+0.544x.\end{array}$

Thus, the least-squares equation to predict the height at 6 years, using the height at 18 years is $\underset{\_}{\widehat{y}=7.92+0.544x}$.

Substitute the above mentioned values in the relation $s_e\approx \sqrt{1-r^2}{s}_y$:

$\begin{array}{c}{s}_e\approx \sqrt{1-r^2}{s}_y\\ =\sqrt{1-{\left(0.8\right)}^2}\times 1.7\\ =\sqrt{1-0.64}\times 1.7\\ =\sqrt{0.36}\times 1.7\\ =0.6\times 1.7\\ =\mathrm{1.02.}\end{array}$

Thus, the corresponding value of $s_e$ is 1.02 inches.