Question

Here are some data on the hearts of various mammals

$\begin{array}{lcc}\text{Mammal}& \text{Heart weight (grams)}& \begin{array}{c}\text{Length of cavity of left}\\ \text{ventricle (centimeters)}\end{array}\\ \text{Mouse}& 0.13& 0.55\\ \text{Rat}& 0.64& 1.0\\ \text{Rabbit}& 5.8& 2.2\\ \text{Dog}& 102& 4.0\\ \text{Sheep}& 210& 6.5\\ \text{Ox}& 2030& 12.0\\ \text{Horse}& 3900& 16.0\\ & & \end{array}$

(a) Make an appropriate scatterplot for predicting heart weight from length. Describe what you see.

(b) Use transformations to linearize the relationship. Does the relationship between heart weight and length seem to follow an exponential model or a power model? Justify your answer.

(c) Perform least-squares regression on the transformed data. Give the equation of your regression line. Define any variables you use.

(d) Use your model from part (c) to predict the heart weight of a human who has a left ventricle $6.8$centimeters long. Show your work.

Short answer

(a) The scatter plot diagram is Positive, Curved, and Solid.

(b) The relationship between heart weight and length seems to follow a power model.

(c) The regression line equation is $\ln \hat{y}=-0.314+3.139\ln x$.

(d) The heart weight of a human who has a left ventricle $6.8$centimeters long is$\hat{y}=299.831$.

Step-by-step solution

Part(a) Step 1: Given Information

$\begin{array}{|lll|}\hline \text{Mammal}& \text{Heart weight(grams)}& \text{Length of left ventricle cavity}\\ \text{Mouse}& 0.13& 0.55\\ \text{Rat}& 0.64& 1\\ \text{Rabbit}& 5.8& 2.2\\ \text{Dog}& 102& 4\\ \text{Sheep}& 210& 6.5\\ \text{Ox}& 2030& 12\\ \text{Horse}& 3900& 16\\ \hline\end{array}$

Part(a) Step 2: Explanation

The horizontal dimension is length, and the vertical axis represents the weight of the heart.

The scatterplot is swinging higher, hence the direction is positive.

Because there is no straight line connecting the ends, the form is curved.

Because all spots in the same pattern are so close together, the strength is solid.

Part(b) Step 1: Given Information

$\begin{array}{|lll|}\hline \text{Mammal}& \text{Heart weight(grams)}& \text{Length of left ventricle cavity}\\ \text{Mouse}& 0.13& 0.55\\ \text{Rat}& 0.64& 1\\ \text{Rabbit}& 5.8& 2.2\\ \text{Dog}& 102& 4\\ \text{Sheep}& 210& 6.5\\ \text{Ox}& 2030& 12\\ \text{Horse}& 3900& 16\\ \hline\end{array}$

Part(b) Step 2: Explanation

Exponential Model

Power Model

The length logarithm is in the horizontal axis in the Exponential Model, and the heart weight logarithm is in the y axis in the Power Model, and the length logarithm is in the horizontal axis in the Power Model.

The Power Model is a model for the interaction of variables, and the related scatterplot has a more linear structure.

Part(c) Step 1: Given Information

$\begin{array}{|lll|}\hline \text{Mammal}& \text{Heart weight(grams)}& \text{Length of left ventricle cavity}\\ \text{Mouse}& 0.13& 0.55\\ \text{Rat}& 0.64& 1\\ \text{Rabbit}& 5.8& 2.2\\ \text{Dog}& 102& 4\\ \text{Sheep}& 210& 6.5\\ \text{Ox}& 2030& 12\\ \text{Horse}& 3900& 16\\ \hline\end{array}$

Part(c) Step 2: Explanation

The length of the left ventricular cavity (X) and the weight of the heart (grams) are taken as logarithmic values.

$\begin{array}{|rrll|}\hline \text{Length of left ventricle cavity (X)}& \text{Heart weight(grams) (Y)}& \backslash \mathrm{multicolumn}1\left|l\right|\text{X.X}& \backslash \mathrm{multicolumn}1l|\text{XY}\\ -0.597837001& -2.040220829& 0.357409& 1.21972\\ 0.78845736& -0.446287103& 0& 0\\ 1.386294361& 1.757857918& 0.621665& 1.385996\\ 1.871802177& 4.624972813& 1.921812& 6.411574\\ 2.48490665& 5.347107531& 3.503643& 10.00873\\ 2.772588722& 7.615791072& 6.174761& 18.92453\\ X=8.70621227& \sum Y=25.12795323& =20.26654& \sum X^2=60.87634\\ \hline\end{array}$

$a=\frac{\sum Y·\sum X^2-\sum X·\sum XY}{n\sum X^2-{\left(\sum X\right)}^2}=-0.314$

$b=\frac{n\sum XY-\sum X·\sum Y}{n\sum X^2-{\left(\sum X\right)}^2}=3.139$

role="math" localid="1652862916129" $$\begin{gathered} y=a+bx \\ =-0.314+3.139x \end{gathered}$$

Part(d) Step 1: Given Information

From part(c)

$\ln \hat{y}=-0.314+3.139\ln x$

Part(d) Step 2: Explanation

Using the value of $x=6.8$in the component equation (c)

$$\begin{gathered} \ln \hat{y}=-0.314+3.139\ln x \\ =-0.314+3.139\ln (6.8) \\ =5.70322 \end{gathered}$$

We take the Exponential of above equation

$$\begin{gathered} \hat{y}=e^{5.70322} \\ =299.831 \end{gathered}$$