Chapter 12.2, Problem 49E

(a)

To determine

To construct: an appropriate scatter plot predicting heart weight from the length.

Answer to Problem 49E

Positive curve strong correlation

Explanation of Solution

Given:

Mammal	Length of Cavity of left ventricle (cm)	Heart weight
Mouse	0.55	0.13
Rat	1	0.64
Rabbit	2.2	5.8
Dog	4	102
Sheep	6.5	210
Ox	12	2030
Horse	16	3900

Graph:

Scatter plot

  

Form: curve, the reason is that the points are not lying on the straight line

Direction: Positive, the reason is that the scatter plot slope is going to upwards.

Strength: strong, the reason is that all points lie very near together in the same pattern.

(b)

To determine

To Explain: that this relationship follows an exponential model or a power model.

Answer to Problem 49E

Power Model

Explanation of Solution

Given:

Mammal	Length of Cavity of left ventricle (cm)	Heart weight
Mouse	0.55	0.13
Rat	1	0.64
Rabbit	2.2	5.8
Dog	4	102
Sheep	6.5	210
Ox	12	2030
Horse	16	3900

Graph:

  

  

The model for the relation between the variables is the model with the most linear pattern in the associate scatter plot” Power Model

(c)

To determine

To Explain: the equation of the regression line.

Answer to Problem 49E

  $\ln y=-0.314+3.139\ln x$

Explanation of Solution

Given:

Mammal	Length of Cavity of left ventricle (cm)	Heart weight
Mouse	0.55	0.13
Rat	1	0.64
Rabbit	2.2	5.8
Dog	4	102
Sheep	6.5	210
Ox	12	2030
Horse	16	3900

Formula used:

  $\begin{array}{l}a=\frac{{\displaystyle \sum y}.{\displaystyle \sum x^2}-{\displaystyle \sum x}{\displaystyle \sum xy}}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ b=\frac{n{\displaystyle \sum xy}-{\displaystyle \sum x}.{\displaystyle \sum y}}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ y=a+b.x\end{array}$

Calculation:

X is the logarithm of length and Y is the logarithm of heart weight

Length of Cavity of left ventricle X	Heart weight Y	XY	$X^2$
-0.597837001	-2.040220829	1.219719501	0.357409
0	-0.446287103	0	0
0.78845736	1.757857918	1.385996014	0.621665
1.386294361	4.624972813	6.411573731	1.921812
1.871802177	5.347107531	10.00872752	3.503643
2.48490665	7.615791072	18.92452988	6.174761
2.772588722	8.268731832	22.92579262	7.687248
${\displaystyle \sum x}$ =8.70621227	${\displaystyle \sum y}$ =25.12795323	${\displaystyle \sum xy}$ =60.87633927	${\displaystyle \sum x^2}$ =20.26654

Using from the table find the value of a and b

  $a=\frac{{\displaystyle \sum y}.{\displaystyle \sum x^2}-{\displaystyle \sum x}{\displaystyle \sum xy}}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}$ =-0.314

  $b=\frac{n{\displaystyle \sum xy}-{\displaystyle \sum x}.{\displaystyle \sum y}}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}$ =3.139

Substituting the value of a and b in the regression equation formula

  $\begin{array}{l}y=a+b.x\\ y=-0.314+3.139x\end{array}$

The least squares regression equation

  $\ln y=-0.314+3.139\ln x$

Where x is the length and y the heart weight

(d)

To determine

To Explain: the prediction from the part (c) of heart weight of a human who has a left ventricle 6.8 cm long.

Answer to Problem 49E

299.825 grams

Explanation of Solution

Given:

Mammal	Length of Cavity of left ventricle (cm)	Heart weight
Mouse	0.55	0.13
Rat	1	0.64
Rabbit	2.2	5.8
Dog	4	102
Sheep	6.5	210
Ox	12	2030
Horse	16	3900

Calculation:

Using the part (c)

  $\ln y=-0.314+3.139\ln x$

Putting the value of x

  $\begin{array}{c}\ln y=-0.314+3.139\ln x\\ \ln y=-0.314+3.139\ln \left(6.8\right)\\ =5.7032\end{array}$

Taking the exponential

  $\begin{array}{c}y=e^{\ln y}=e^{5.7032}\\ =299.825\end{array}$

Therefore the expected heart weight is 299.825 grams.