Chapter 12.1, Problem 12E

(a)

To determine

To find out what height would you predict for a student with an arm span of $76$ inches.

Answer to Problem 12E

We predict that the student will be $75.41892$ inches tall with an arm span of $76$ inches.

Explanation of Solution

Now, as we know that in the previous exercise the computer output of this data is given. And also as we know the general regression equation be as:

  $\widehat{y}=a+bx$

And the estimate a and b are given in the column of the computer output “Coef”:

  $\begin{array}{l}\widehat{y}=a+bx\\ \Rightarrow \widehat{y}=11.547+0.84042x\end{array}$

Thus, we will evaluate it for finding the height for a student with an arm span of $76$ inches as:

  $\begin{array}{l}\widehat{y}=a+bx\\ \Rightarrow \widehat{y}=11.547+0.84042x\\ =11.547+0.84042(76)\\ =75.41892\end{array}$

Thus, this means that we predict that the student will be $75.41892$ inches tall with an arm span of $76$ inches.

(b)

To determine

To explain about how far do you expect the prediction in part (a) to be from the student’s actual height.

Answer to Problem 12E

The prediction will be off by $1.613$ on average.

Explanation of Solution

It is given in the part (a) that:

We predict that the student will be $75.41892$ inches tall with an arm span of $76$ inches.

Then as it is given in the computer output that,

  $s=1.613$

Thus, the expected error made when making predictions is given by $\sigma$ which is estimated by s . thus, we expect that the prediction will be off by $1.613$ on average.