Chapter 8, Problem 45E

(a)

To determine

To create a scatterplot of calories versus fat.

Explanation of Solution

In the question we are examining the association between the fat and calories in fast-food hamburgers. Thus, provided the data on the fat and calories, the scatterplot of fat versus calories is given as:

  

In the above scatterplot we can see that the fat is on the horizontal axis and the calories is on the vertical axis. From the points in the scatterplot we can say that the points are in the upward direction thus the relationship must be positive in nature. And the association between the fat and the calories must be strong and linear in nature.

(b)

To determine

To interpret the value of $R^2$ in this context.

Answer to Problem 45E

The value of $R^2$ is $92.3\%$ .

Explanation of Solution

In the question we are examining the association between the fat and calories in fast-food hamburgers. Thus, the data is as below:

Fat	Calories
19	410
31	580
34	590
35	570
39	640
39	680
43	660

Now for calculating the value of $R^2$ , we have to find the value of the correlation. The value of correlation is calculated by using the excel function. The function used for calculating the correlation is as:

  $\text{CORREL(array1,array2)}$

Thus, the calculation is as follows:

Correlation =

=CORREL(D64:D70,E64:E70)

And the result will be as follows:

Correlation =

0.9606

Now, the value of $R^2$ can be calculated as:

  $\begin{array}{l}{R}^2=r^2\\ $ \Rightarrow R^2={(0.9606)}^2$=0.923\\ =92.3\%\end{array}$

Thus, we can interpret that $92.3\%$ of the variation in calories can be explained by the fat content in the hamburgers. As the value of determination examine the variation among the variables.

(c)

To determine

To write the equation of the line of regression.

Answer to Problem 45E

  $\text{<mover accent="true"><mo>C</mo><mo>^</mo></mover> alories}=211+11.056(\text{Fat)}$ .

Explanation of Solution

In the question we are examining the association between the fat and calories in fast-food hamburgers. Thus, the data is as below:

Fat	Calories
19	410
31	580
34	590
35	570
39	640
39	680
43	660

Now, the equation of the regression can be calculated by using the excel. Thus, we have,

Formula used:

The formula for mean is as:

  $\text{AVERAGE}\left(\text{number1,}\left[\text{number2}\right]\text{, }\mathrm{...}\right)$

And the formula for the standard deviation is as:

  $\text{STDEV}\left(\text{number1,}\left[\text{number2}\right]\text{,}\mathrm{...}\right]\text{)}$

Calculation:

The mean, standard deviation and correlation is calculated as:

For the mean,

	Fat	Calories
Average =	=AVERAGE(D64:D70)	=AVERAGE(E64:E70)

For the standard deviation,

	Fat	Calories
Standard deviation =	=STDEV(D64:D70)	=STDEV(E64:E70)

Thus, the result is as:

	Fat	Calories
Average =	34	590
	Fat	Calories
Standard deviation =	7.80	89.81

So, we have,

  $\begin{array}{l}r=0.9606\\ {\mu }_1=590\\ {\mu }_2=34\\ {\sigma }_1=89.81\\ {\sigma }_2=7.80\end{array}$

The value of correlation is taken from the part (b), thus the regression equation can be calculated as:

  $\begin{array}{l}\beta =r\times \frac{{\sigma }_1}{{\sigma }_2}\\ =0.9606\times \frac{89.81}{7.80}\\ =11.056\\ \alpha ={\mu }_1-\beta \times {\mu }_2\\ =590-11.056\times 34\\ =211\\ \text{<mover accent="true"><mo>C</mo><mo>^</mo></mover> alories}=\alpha +\beta (\text{Fat)}\\ =211+11.056(\text{Fat)}\end{array}$

(d)

To determine

To use the residual plot to explain whether your linear model is appropriate.

Answer to Problem 45E

Yes, linear model is appropriate.

Explanation of Solution

In the question we are examining the association between the fat and calories in fast-food hamburgers. Thus, the data is as below:

Fat	Calories
19	410
31	580
34	590
35	570
39	640
39	680
43	660

Also, it is given that,

  $\begin{array}{l}r=0.9606\\ {\mu }_1=590\\ {\mu }_2=34\\ {\sigma }_1=89.81\\ {\sigma }_2=7.80\end{array}$

And the regression line is:

  $\text{<mover accent="true"><mo>C</mo><mo>^</mo></mover> alories}=211+11.056(\text{Fat)}$

Thus, the residual plot is as follows:

  

Thus, from the residual plot we can see that the residual shows no clear points as the points are in different direction so, we can conclude that the model seems appropriate by concluding from the residual plot.

(e)

To determine

To explain the meaning of the y -intercept of the line.

Explanation of Solution

In the question we are examining the association between the fat and calories in fast-food hamburgers. Thus, the data is as below:

Fat	Calories
19	410
31	580
34	590
35	570
39	640
39	680
43	660

Also, it is given that,

  $\begin{array}{l}r=0.9606\\ {\mu }_1=590\\ {\mu }_2=34\\ {\sigma }_1=89.81\\ {\sigma }_2=7.80\end{array}$

And the regression line is:

  $\text{<mover accent="true"><mo>C</mo><mo>^</mo></mover> alories}=211+11.056(\text{Fat)}$

Thus, the y -intercept interprets that we can say that a fat-free burger still has $211$ calories, nut this is extrapolation as no data is close to zero.

(f)

To determine

To explain the meaning of the slope of the line.

Explanation of Solution

In the question we are examining the association between the fat and calories in fast-food hamburgers. Thus, the data is as below:

Fat	Calories
19	410
31	580
34	590
35	570
39	640
39	680
43	660

Also, it is given that,

  $\begin{array}{l}r=0.9606\\ {\mu }_1=590\\ {\mu }_2=34\\ {\sigma }_1=89.81\\ {\sigma }_2=7.80\end{array}$

And the regression line is:

  $\text{<mover accent="true"><mo>C</mo><mo>^</mo></mover> alories}=211+11.056(\text{Fat)}$

Thus, the slope of the line explains that every gram of fat adds to $11.056$ calories on average. As we know that the slope of the line explains the variation of the one variable by the variation of the other variable.

(g)

To determine

To find out how many calories does the burger have.

Answer to Problem 45E

The new burger has $553.5$ grams of calories.

Explanation of Solution

In the question we are examining the association between the fat and calories in fast-food hamburgers. Thus, the data is as below:

Fat	Calories
19	410
31	580
34	590
35	570
39	640
39	680
43	660

Also, it is given that,

  $\begin{array}{l}r=0.9606\\ {\mu }_1=590\\ {\mu }_2=34\\ {\sigma }_1=89.81\\ {\sigma }_2=7.80\end{array}$

And the regression line is:

  $\text{<mover accent="true"><mo>C</mo><mo>^</mo></mover> alories}=211+11.056(\text{Fat)}$

It is given that a new burger containing $\text{28}$ grams of fat is introduced and its residual for calories is $\text{+33}$ . Thus, the predicted calories that the new burger is containing:

  $\begin{array}{l}\text{<mover accent="true"><mo>C</mo><mo>^</mo></mover> alories}=211+11.056(\text{Fat)}\\ =211+11.056\times 28\\ =520.5\end{array}$

Thus, the actual calories that the burger have is:

  $\begin{array}{l}\text{Residual}=\text{Actual}-\text{Predicted}\\ \Rightarrow 33=\text{Actual}-520.5\\ \Rightarrow \text{Actual}=553.5\end{array}$

Thus, the new burger has $553.5$ grams of calories.