Chapter 8, Problem 31E

(a)

To determine

To write the equation of the regression line.

Answer to Problem 31E

  $\text{<mover accent="true"><mo>N</mo><mo>^</mo></mover> icotine}=0.15403+0.065052\text{Tar}$ .

Explanation of Solution

In the previous question, it is given the researcher wants to find out is the nicotine content of a cigarette related to the “tars”. Thus, a collection of data in milligrams on $29$ cigarettes produced the scatterplot, residual plot and the regression analysis shown in the question. And also it is given that,

  $R^2=92.4\%$

  $\begin{array}{l}\text{Constant}=0.154030\\ \text{Tar}=0.065052\end{array}$

And the dependent variable is nicotine. Thus, the regression line will be:

  $\text{<mover accent="true"><mo>N</mo><mo>^</mo></mover> icotine}=0.15403+0.065052\text{Tar}$

(b)

To determine

To estimate the nicotine content of cigarettes with $4$ milligrams of tar.

Answer to Problem 31E

  $0.414$ milligrams.

Explanation of Solution

In the previous question, it is given the researcher wants to find out is the nicotine content of a cigarette related to the “tars”. Thus, a collection of data in milligrams on $29$ cigarettes produced the scatterplot, residual plot and the regression analysis shown in the question. And also it is given that,

  $R^2=92.4\%$

  $\begin{array}{l}\text{Constant}=0.154030\\ \text{Tar}=0.065052\end{array}$

And the dependent variable is nicotine. And, the regression line is:

  $\text{<mover accent="true"><mo>N</mo><mo>^</mo></mover> icotine}=0.15403+0.065052\text{Tar}$

Thus, the nicotine content of cigarettes with $4$ milligrams of tar is calculated as:

  $\begin{array}{l}\text{<mover accent="true"><mo>N</mo><mo>^</mo></mover> icotine}=0.15403+0.065052\text{Tar}\\ =0.15403+0.065052\times 4\\ =0.414\end{array}$

The nicotine will be $0.414$ milligrams.

(c)

To determine

To interpret the meaning of the slope of the regression line in this context.

Explanation of Solution

In the previous question, it is given the researcher wants to find out is the nicotine content of a cigarette related to the “tars”. Thus, a collection of data in milligrams on $29$ cigarettes produced the scatterplot, residual plot and the regression analysis shown in the question. And also it is given that,

  $R^2=92.4\%$

  $\begin{array}{l}\text{Constant}=0.154030\\ \text{Tar}=0.065052\end{array}$

And the dependent variable is nicotine. And, the regression line is:

  $\text{<mover accent="true"><mo>N</mo><mo>^</mo></mover> icotine}=0.15403+0.065052\text{Tar}$

Thus, the meaning of the slope of the regression line in this context is that predicted nicotine content increases by $0.065$ milligram of nicotine per additional milligram of tar.

(d)

To determine

To explain what does the y -intercept mean.

Answer to Problem 31E

Explanation of Solution

In the previous question, it is given the researcher wants to find out is the nicotine content of a cigarette related to the “tars”. Thus, a collection of data in milligrams on $29$ cigarettes produced the scatterplot, residual plot and the regression analysis shown in the question. And also it is given that,

  $R^2=92.4\%$

  $\begin{array}{l}\text{Constant}=0.154030\\ \text{Tar}=0.065052\end{array}$

And the dependent variable is nicotine. And, the regression line is:

  $\text{<mover accent="true"><mo>N</mo><mo>^</mo></mover> icotine}=0.15403+0.065052\text{Tar}$

Thus the y -intercept means that we had expect a cigarette with no tar to have $0.154$ milligrams of nicotine.

(e)

To determine

To find out what is the nicotine content.

Answer to Problem 31E

  $0.1094$ mg of nicotine.

Explanation of Solution

In the previous question, it is given the researcher wants to find out is the nicotine content of a cigarette related to the “tars”. Thus, a collection of data in milligrams on $29$ cigarettes produced the scatterplot, residual plot and the regression analysis shown in the question. And also it is given that,

  $R^2=92.4\%$

  $\begin{array}{l}\text{Constant}=0.154030\\ \text{Tar}=0.065052\end{array}$

And the dependent variable is nicotine. And, the regression line is:

  $\text{<mover accent="true"><mo>N</mo><mo>^</mo></mover> icotine}=0.15403+0.065052\text{Tar}$

It is given if a new brand of cigarette contains $7$ milligrams of tar and a nicotine level whose residual is $-0.5$ mg then the nicotine content will be as:

  $\begin{array}{l}\text{<mover accent="true"><mo>N</mo><mo>^</mo></mover> icotine}=0.15403+0.065052\text{Tar}\\ =0.15403+0.065052\times 7\\ =0.609\\ \text{Residual}=\text{Actual}-\text{predicted}\\ \Rightarrow -0.5=\text{Actual}-0.609\\ \Rightarrow \text{Actual}=0.609-0.5\\ =0.1094\end{array}$