Chapter 1.2, Problem 23E

a.

To determine

To make: a scatter plot and check whether a linear model is appropriate.

Answer to Problem 23E

Yes, a linear model is appropriate in the scatter plot.

Explanation of Solution

Given:

The given table shows the winning heights for the Olympic pole vault competitions up to the year 2020.

Year	Height (m)	Year	Height (m)
$1896$	1 $3.30$	3 $1956$	$4.56$
$1900$	$3.30$	$1960$	$4.70$
$1904$	$3.50$	$1964$	$5.10$
$1908$	$3.71$	$1968$	$5.40$
$1912$	$3.95$	$1972$	$5.64$
$1920$	$4.09$	$1976$	$5.64$
$1924$	$3.95$	$1980$	$5.78$
$1928$	$4.20$	$1984$	$5.75$
$1932$	$4.31$	$1988$	$5.90$
$1936$	$4.35$	$1992$	$5.87$
$1948$	$4.30$	$1996$	$5.92$
$1952$	$4.55$	$2000$	$5.90$

Calculation:

The scatter plot of the data is shown below:

  

The points in the scatter plots follows a linear trend, so it is appropriate to use a linear model to describe this situation.

b.

To determine

To find: the graph and regression line.

Answer to Problem 23E

  $H=0.026455172\left(Y\right)-46.8758896552$

Explanation of Solution

Given:

The given table shows the winning heights for the Olympic pole vault competitions up to the year 2020.

Year	Height (m)	Year	Height (m)
$1896$	$3.30$	$1956$	$4.56$
$1900$	$3.30$	$1960$	$4.70$
$1904$	$3.50$	$1964$	$5.10$
$1908$	$3.71$	$1968$	$5.40$
$1912$	$3.95$	$1972$	$5.64$
$1920$	$4.09$	$1976$	$5.64$
$1924$	$3.95$	$1980$	$5.78$
$1928$	$4.20$	$1984$	$5.75$
$1932$	$4.31$	$1988$	$5.90$
$1936$	$4.35$	$1992$	$5.87$
$1948$	$4.30$	$1996$	$5.92$
$1952$	$4.55$	$2000$	$5.90$

Calculation:

According to the table, to make a linear regression line and obtained the equation:

  $H=0.026455172\left(Y\right)-46.8758896552$

Now, graph of the regression line is shown below

  

Hence, a linear regression line model is $H=0.026455172\left(Y\right)-46.8758896552$ .

c.

To determine

To predict: the height of the winning pole vault at 2004 Olympics and compare with actual winning height of 5.95 meters.

Answer to Problem 23E

  $6.27m$

Explanation of Solution

From part (b) $H=0.026455172\left(Y\right)-46.8758896552$

Chirping rate is ${100}^{\circ }F$

Calculation:

The regression line of the equation is $H=0.026455172\left(Y\right)-46.8758896552$ .

The objective is to estimate the height for 2008 Olympics,

Plug $Y=2008$ in the equation, then

  $\begin{array}{l}H=0.026455172\left(2008\right)-46.8758896552\\ H=6.266868\end{array}$

According to the linear model the winning height in 2008 Olympics should be close $6.27$ meters. But the actual winning height was 5.96 meter which is below the estimated height with the model.

It seems the model overvalued the winning height in 2008.

d.

To determine

To check: that it is reasonable to apply the model to predict the winning height at the 2010 Olympics.

Answer to Problem 23E

It is not reasonable.

Explanation of Solution

Given:

The given model is $H=0.026455172\left(Y\right)-46.8758896552$

Calculation:

Plug the value of $H=2100$ in the linear model found in part (b).

  $\begin{array}{l}H=0.026455172\left(2100\right)-46.8758896552\\ H=8.70169605\end{array}$

It is hard to believe these athletes will be able to jump outstanding height of 8 meter.