Chapter 8, Problem 49E

(a)

To determine

To create a model to predict %body fat from weight.

Answer to Problem 49E

  $\text{\%<mover accent="true"><mo>B</mo><mo>^</mo></mover> ody fat}=-27.4+0.25(\text{Weight)}$ .

Explanation of Solution

In the question, the %body fat and weight of the people is given as researcher wants to find out ways to make a good estimate to calculate the body fat. Thus, the table is given in which the body fat and their weights are given. So, the table is as:

Weight	Body
175	6
181	21
200	15
159	6
196	22
192	31
205	32
173	21
187	25
188	30
188	10
240	20
175	22
168	9
246	38
160	10
215	27
159	12
146	10
219	28

Now we will create the model to predict %body fat from weight by using the excel function as:

We will first select the data given in the table and then go to the insert tab. In the tab we will use the scatterplot option from the charts options and then the scatterplot will appear on the screen. Now, we will go to the design tab from the chart tools. We will then select the quick layout option from it. Then in it we will select the layout $9$ from it and the scatterplot with the model and regression line will appear as:

  

Thus, from this we can say that the regression line to predict %body fat from weight is as:

  $\begin{array}{l}\text{\%<mover accent="true"><mo>B</mo><mo>^</mo></mover> ody fat}=\alpha +\beta (\text{Weight)}\\ =-27.4+0.25(\text{Weight)}\end{array}$

(b)

To determine

To explain do you think a linear model is appropriate.

Answer to Problem 49E

Yes, the linear model is appropriate.

Explanation of Solution

In the question, the %body fat and weight of the people is given as researcher wants to find out ways to make a good estimate to calculate the body fat. Thus, the table is given in which the body fat and their weights are given.

And the regression line is as:

  $\text{\%<mover accent="true"><mo>B</mo><mo>^</mo></mover> ody fat}=-27.4+0.25(\text{Weight)}$

We can see that from the residual plot that the residual look randomly scattered around zero, so the conditions are satisfied and thus, the linear model is appropriate. The residual plot is as:

  

(c)

To determine

To interpret the slope of your model.

Explanation of Solution

In the question, the %body fat and weight of the people is given as researcher wants to find out ways to make a good estimate to calculate the body fat. Thus, the table is given in which the body fat and their weights are given.

And the regression line is as:

  $\text{\%<mover accent="true"><mo>B</mo><mo>^</mo></mover> ody fat}=-27.4+0.25(\text{Weight)}$

Thus, the slope of the regression line interprets that %body fat increases on average by $0.25\%$ per pound of weight.

(d)

To determine

To explain is your model likely to make reliable estimates.

Answer to Problem 49E

It is relatively reliable.

Explanation of Solution

In the question, the %body fat and weight of the people is given as researcher wants to find out ways to make a good estimate to calculate the body fat. Thus, the table is given in which the body fat and their weights are given.

And the regression line is as:

  $\text{\%<mover accent="true"><mo>B</mo><mo>^</mo></mover> ody fat}=-27.4+0.25(\text{Weight)}$

Thus, we have from part (a) that the coefficient of determination is $R^2=48.5\%$ . So, we know that $48.5\%$ of the variation is showed by our model. So we can say that our model is likely to be relatively reliable and also because residual have a standard deviation of $7\%$ so the variation around line is large.

(e)

To determine

To find out what is the residual for a person who weighs $190$ pounds and has $21\%$ body weight.

Answer to Problem 49E

The residual is $0.9\%$ .

Explanation of Solution

In the question, the %body fat and weight of the people is given as researcher wants to find out ways to make a good estimate to calculate the body fat. Thus, the table is given in which the body fat and their weights are given.

And the regression line is as:

  $\text{\%<mover accent="true"><mo>B</mo><mo>^</mo></mover> ody fat}=-27.4+0.25(\text{Weight)}$

The body weight of the person who weighs $190$ pounds is calculated as:

  $\begin{array}{l}\text{\%<mover accent="true"><mo>B</mo><mo>^</mo></mover> ody fat}=-27.4+0.25(\text{Weight)}\\ =-27.4+0.25\times 190\\ =20.1\end{array}$

Thus, the residual is as:

  $\begin{array}{l}\text{Residual}=\text{Actual}-\text{Predicted}\\ =21-20.1\\ =0.9\end{array}$

The residual is $0.9\%$ .