Chapter 12.2, Problem 44E

(a)

To determine

To build: a scatterplot for estimating and describing heart weight from the length.

Answer to Problem 44E

True positive curved correlation

Explanation of Solution

Given:

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

Graph:

Length is in the horizontal dimension, and weight of the heart is in the vertical axis.

  

Direction: Positive, as the scatterplot swings upwards.

Form: Curved, since there is no straight line to the ends.

Strength: Solid, since all points in the same pattern lay very close together.

(b)

To determine

To Explain: The heart weight-to-length relationship tends to follow an exponential formula, or a power model.

Answer to Problem 44E

Power Model

Explanation of Solution

Given:

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

Graph:

Exponential Model

  

Power Model

  

In Exponential Model the Length is in the horizontal dimension, and the Heart weight logarithm is in the y axis and in Power Model the Length logarithm is in the horizontal axis, and the Heart weight logarithm is in the vertical axis.

Power Model is the model for the interaction between the variables, with the corresponding scatterplot having the more linear form.

(c)

To determine

To perform: minimal square regression on the transformed data, and give equation to the regression line to define some variables.

Answer to Problem 44E

  $\ln \widehat{y}=-0.314+3.139\ln x$

Explanation of Solution

Given:

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

Formula used:

  $\begin{array}{l}a=\frac{{\displaystyle \sum Y}.{\displaystyle \sum X^2}-{\displaystyle \sum X}.{\displaystyle \sum XY}}{n{\displaystyle \sum X^2}-{\left({\displaystyle \sum X}\right)}^2}\\ b=\frac{n{\displaystyle \sum XY}-{\displaystyle \sum X}.{\displaystyle \sum Y}}{n{\displaystyle \sum X^2}-{\left({\displaystyle \sum X}\right)}^2}\end{array}$

Calculation:

Value of Length of left ventricle cavity (X) and Heart weight(grams) is taken as logarithmic

Length of left ventricle cavity (X)	Heart weight(grams) (Y)	X.X	XY
-0.597837001	-2.040220829	0.357409	1.21972
0	-0.446287103	0	0
0.78845736	1.757857918	0.621665	1.385996
1.386294361	4.624972813	1.921812	6.411574
1.871802177	5.347107531	3.503643	10.00873
2.48490665	7.615791072	6.174761	18.92453
2.772588722	8.268731832	7.687248	22.92579
${\displaystyle \sum X}$=8.70621227	${\displaystyle \sum Y}$=25.12795323	=20.26654	${\displaystyle \sum X^2}$=60.87634

  $\begin{array}{l}a=\frac{{\displaystyle \sum Y}.{\displaystyle \sum X^2}-{\displaystyle \sum X}.{\displaystyle \sum XY}}{n{\displaystyle \sum X^2}-{\left({\displaystyle \sum X}\right)}^2}=-0.314\\ b=\frac{n{\displaystyle \sum XY}-{\displaystyle \sum X}.{\displaystyle \sum Y}}{n{\displaystyle \sum X^2}-{\left({\displaystyle \sum X}\right)}^2}=3.139\end{array}$

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

Therefore, the equation is

  $\ln \widehat{y}=-0.314+3.139\ln x$

Here $x$ is the Length of left ventricle cavity and $y$ Heart weight(grams).

(d)

To determine

To Predict: the heart weight of a person left 6.8 centimetres long ventricle using the part (c).

Answer to Problem 44E

  $\widehat{y}=1242.41$

Explanation of Solution

Given:

From part (c)

  $\ln \widehat{y}=-0.314+3.139\ln x$

Calculation:

Putting the value of $x$ =6.8 in the equation of part (c)

  $\begin{array}{c}\ln \widehat{y}=-0.314+3.139\ln x\\ =-0.314+3.139\ln \left(6.8\right)\\ =5.70322\end{array}$

Taking the exponential

  $\begin{array}{c}\widehat{y}=e^{5.70322}\\ =299.831\end{array}$