Chapter 3.2, Problem 44E

(a)

To determine

To calculate and interpret the residual for the cricket who chirped $20$ times per minute when the temperature was ${88.6}^0\text{F}$ .

Answer to Problem 44E

Residual is $1.5268$ .

Explanation of Solution

It is given in the question the regression line as:

  $\widehat{y}=-0.31+0.212x$

And also the cricket chirped $20$ times per minute when the temperature was ${88.6}^0\text{F}$ . Thus,

  $\begin{array}{l}x=88.6\\ y=20\end{array}$

Thus, the predicted number of times cricket chirped is as:

  $\begin{array}{l}\widehat{y}=-0.31+0.212x\\ =-0.31+0.212(88.6)\\ =18.4732\end{array}$

Thus, the residual will be calculated as:

  $\begin{array}{l}\text{Residual}=y-\widehat{y}\\ =20-18.4732\\ =1.5268\end{array}$

This then implies that we underestimated the number of chirps per minute by $1.5268$ chirps per minute, when making a prediction using the regression line for the cricket.

(b)

To determine

To explain about how many additional chirps per minute do you expect a cricket to make if the temperature increases by ${10}^0\text{F}$ .

Answer to Problem 44E

  $2.12$ chirps per minute we expect a cricket to make.

Explanation of Solution

It is given in the question the regression line as:

  $\widehat{y}=-0.31+0.212x$

As we know that the slope is coefficient of x in the least square regression equation and represents the average increase or decrease of y per unit of x . Thus,

  $b_1=0.212$

This then implies on average the number of chirps per minute increase by $0.212$ chirps per degree Fahrenheit. In this case, we increase the temperature by ${10}^0\text{F}$ and thus we then expect the number of chirps per minute to increase by:

  $\begin{array}{l}10\times 0.212\\ =2.12\text{ chirps per minute}\end{array}$