Temperature and wind The average temperature (in degrees Fahrenheit) and average wind speed (in miles per hour) were recorded for 365 consecutive days at Chicago’s O’Hare International Airport. Here is the computer output for regression of y = average wind speed on x = average temperature:

a. Calculate and interpret the residual for the day where the average temperature was 42°F and the average wind speed was 2.2 mph.

b. Interpret the slope.

c. By how much do the actual average wind speeds typically vary from the values predicted by the least-squares regression line with x = average temperature?

d. What percent of the variability in average wind speed is accounted for by the least-squares regression line with x = average temperature?

The relationship of average temperature and the average wind speed in given in the question.

The formula used:$Residual=y-\stackrel{⏜}{y}$

The general equation of the least square regression line is:

$\stackrel{⏜}{y}=b_0+b_{1}x$

As a result, the estimate of the constant $b_0$ is given in the computer output's row "Intercept" and column "Estimate" as:

$b_0=11.897762$

In the computer output, the estimate of the slope $b_1$ is given in the row "Avg temp" and the column "Estimate" as:

$b_1=-0.041077$

As a result, when we plug the values into the general equation, we get:

$$\begin{gathered} \stackrel{⏜}{y}=b_0+b_{1}x \\ \Rightarrow \stackrel{⏜}{y}=11.897762-0.041077x \end{gathered}$$

As a result, the average wind speed at $42$degrees Fahrenheit is:

$$\begin{gathered} \stackrel{⏜}{y}=11.897762-0.041077x \\ =11.897762-0.041077\left(42\right) \\ =10.172528 \end{gathered}$$

Thus the residual will be calculated as:.

$$\begin{gathered} Residual=y-\stackrel{⏜}{y} \\ =2.2-10.172528 \\ =-7.972528 \end{gathered}$$

This means that while using the regression line to make a prediction, we underestimated the average wind speed on the day with an average temperature of $40F$by$7.972528$ mph.

The question specifies the link between average temperature and average wind speed. The regression line is as follows:

$\stackrel{⏜}{y}=11.897762-0.041077x$

The slope is the coefficient of $x$ in the least-squares regression equation, and it reflects the average rise or decrease of y per unit of $x$ as we all know. Thus,

$b_0=-10.041077$

As a result, the average wind speed reduces by$0.041077$ miles per hour per degree Fahrenheit.

The question specifies the link between average temperature and average wind speed. The regression line is as follows:

$\stackrel{⏜}{y}=11.897762-0.041077x$

The standard error of the estimate s is calculated as follows in the computer output following "root mean square error":

$s=3.655950$

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. As a result, using the equation of the least square regression line, the predicted average wind speed differed by $3.655950$ mph on average from the actual average wind speed.

The question specifies the link between average temperature and average wind speed. The regression line is as follows:

$\stackrel{⏜}{y}=11.897762-0.041077x$

The coefficient of determination is given as follows in the computer output after "RSquare":

$$\begin{gathered} r^2=0.0478742 \\ =4.7874\% \end{gathered}$$

The coefficient of determination, as we all know, represents the proportion of variance in the answers $y$ variable that can be explained by a least square regression model using the explanatory $x$ variable. As a result, we can state that the least square regression line employing average temperature as an explanatory variable explains $4.7874$ percent of the variation in average wind speed.