91影视

The regression line is given as follows in the question:

y= 鈭�0.31 + 0.212x

Also, when the temperature was $88.6$degrees Fahrenheit, the insect chirped$20$ times per minute.

Thus,

$$\begin{gathered} x=88.6 \\ y=20 \end{gathered}$$

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

$$\begin{gathered} \stackrel{铃�}{y}=鈭�0.31+0.212x \\ =鈭�0.31+0.212(88.6) \\ =18.4732 \end{gathered}$$

Thus, the residual will be calculated as:

$$\begin{gathered} Residual=y鈭�\stackrel{铃�}{y} \\ =20鈭�18.4732 \\ =1.5268 \end{gathered}$$

This means that while using the regression line to make a prediction for the cricket, we underestimated the amount of chirps per minute by $1.5268$chirps per minute.

The regression line is given as follows in the question:

$\stackrel{铃�}{y}=鈭�0.31+0.212x$

The slope in the least square regression equation is the coefficient of $x$and it indicates the average rise or decrease of $y$per unit of $x$Thus,

$b_1=0.212$

As a result, the average number of chirps per minute increases by $0.212$ per degree Fahrenheit. In this scenario, we raise the temperature by, and we anticipate an increase in the number of chirps per minute of:

$$\begin{gathered} 10脳0.212 \\ =2.12chirpsperminute \end{gathered}$$