Chapter 3.2, Problem 61E

(a)

To determine

To calculate and interpret the residual for the day where the average temperature was ${42}^0\text{F}$ and the average wind speed was $2.2$ mph.

Answer to Problem 61E

Residual is $-7.972528$ .

Explanation of Solution

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

The general equation of the least square regression line is:

  $\widehat{y}=b_0+b_{1}x$

Thus, the estimate of the constant $b_0$ is given in the row “Intercept” and in the column “Estimate” of the computer output as:

  $b_0=11.897762$

And the estimate of the slope $b_1$ is given in the row “Avg temp” and in the column “Estimate” of the computer output as:

  $b_1=-0.041077$

Thus, putting the values in the general equation we will get,

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \Rightarrow \widehat{y}=11.897762-0.041077x\end{array}$

Thus, the average wind speed where the average temperature was ${42}^0\text{F}$ is:

  $\begin{array}{l}\widehat{y}=11.897762-0.041077x\\ =11.897762-0.041077(42)\\ =10.172528\end{array}$

Thus the residual will be calculated as:

  $\begin{array}{l}\text{Residual}=y-\widehat{y}\\ =2.2-10.172528\\ =-7.972528\end{array}$

This then implies that we overestimated the average wind speed on the day with average temperature ${42}^0\text{F}$ by $7.972528$ mph when making a prediction using the regression line.

(b)

To determine

To interpret the slope.

Explanation of Solution

The relationship of average temperature and the average wind speed in given in the question. And the regression line is as:

  $\widehat{y}=11.897762-0.041077x$

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

  $b_1=-0.041077$

Thus this implies that on average the average wind speed decreases by $0.041077$ mph per degree Fahrenheit.

(c)

To determine

To find out by how much do the actual average wind speeds typically vary from the values predicted by the least square regression line with x as average temperature.

Answer to Problem 61E

The predicted average wind speed using the equation of least square regression line deviated on average by $3.655950$ mph from the actual average wind speed.

Explanation of Solution

The relationship of average temperature and the average wind speed in given in the question. And the regression line is as:

  $\widehat{y}=11.897762-0.041077x$

As, it is given that the standard error of the estimate s is given in the computer output after “Root Mean square error” as:

  $s=3.655950$

As we know that the standard error of the estimate s represents the average error of predictions thus the average deviation between actual and the predicted values. Thus the predicted average wind speed using the equation of least square regression line deviated on average by $3.655950$ mph from the actual average wind speed.

(c)

To determine

To find out what percent of the variability in average wind speed is accounted for by the least squares regression line with x as average temperature.

Answer to Problem 61E

  $4.7874\%$ of the variation in the average wind speed is explained by the least square regression line using the average temperature as explanatory variable.

Explanation of Solution

The relationship of average temperature and the average wind speed in given in the question. And the regression line is as:

  $\widehat{y}=11.897762-0.041077x$

As it is given that the coefficient of determination is given in the computer output after “RSquare” as:

  $\begin{array}{l}{r}^2=0.047874\\ =4.7874\%\end{array}$

As we know that the coefficient of determination measures the proportion of variation in the responses y variable that is explained by the least square regression model using the explanatory x variable. Thus, we can say that $4.7874\%$ of the variation in the average wind speed is explained by the least square regression line using the average temperature as explanatory variable.