Crickets keep chirping In Exercise 44, we summarized the relationship between x = temperature in degrees Fahrenheit and y = chirps per minute for the striped ground cricket, with the regression line y^=–0.31+0.212x. For this model, technology gives s = 0.97 and r2 = 0.697.

a. Interpret the value of s.

b. Interpret the value of r2.

$\stackrel{⏜}{y}=-0.31+0.212x$

In the question, the link between the temperature in degrees Fahrenheit and the number of chirps per minute for the striped ground cricket is given. The regression line is as follows:

$\stackrel{⏜}{y}=0.31+0.212x$

$s=0.97$ is the value of $s$ The standard error of the estimations, as we all know, is the average error of forecasts, and thus the average difference between actual and predicted values. Thus, using the equation of least squares regression line, the projected chirps per minute for the striped ground cricket differed on average by $0.97$ chirps from the actual chirps per minute for the striped ground cricket.

The relationship between the temperature in degrees Fahrenheit and chirps per minute for the striped ground cricket is given in the question. And the regression line is as:

$\stackrel{⏜}{y}=0.31+0.212x$

And the value of $r^2$is,

$$\begin{gathered} r^2=0.697 \\ =69.7\% \end{gathered}$$

The coefficient of determination, as we know, is a measurement of how much variation in the answers $y$ variable is explained by the least square regression model with the explanatory variable. As a result, the least square regression line employing temperature as an explanatory variable can explain $69.7\%$ of the variation in chirps per minute for the striped ground cricket.