Chapter 3, Problem T3.13SPT

(a)

To determine

To explain is a linear model appropriate for describing the relationship between wildebeest abundance and percent of grass area burned.

Answer to Problem T3.13SPT

Yes, it is appropriate.

Explanation of Solution

Yes, the linear model is appropriate because there is no strong curvature present in the scatterplot and there is in strong curvature present in the residual plot. Thus, we can say that a linear model is appropriate for describing the relationship between wildebeest abundance and percent of grass area burned.

(b)

To determine

To give the equation of the least squares regression line.

Answer to Problem T3.13SPT

  $\widehat{y}=92.29-0.05762x$ .

Explanation of Solution

From the computer output we have the slope and the constant as,

  $\begin{array}{l}{b}_0=92.29\\ b_1=-0.05762\end{array}$

Thus, the regression line is:

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \Rightarrow \widehat{y}=92.29-0.05762x\end{array}$

Where x be wildebeest and y be the percent of the grass area burned.

(c)

To determine

To interpret the slope and explain does this value of the y intercept have meaning in this context.

Answer to Problem T3.13SPT

No, the value of y intercept does not have meaning in this context.

Explanation of Solution

The regression line is:

  $\widehat{y}=92.29-0.05762x$

  $\begin{array}{l}{b}_0=92.29\\ b_1=-0.05762\end{array}$

The slope is the coefficient of x in the regression equation and represents the average increase or decrease of y per unit of x . This implies that on average the percent burned decreases by $0.05762\%$ per thousand wildebeest.

And the y intercept is the constant in the regression line and represents the average y -value when x is zero. This then implies that on average, $92.29\%$ of the grass area is burned when there are no wildebeest. However, it is not reasonable to have zero wildebeest and then the value of y intercept does not have meaning in this context.

(d)

To determine

To interpret the standard deviation of the residuals and $r^2$ .

Explanation of Solution

The regression line is:

  $\widehat{y}=92.29-0.05762x$

  $\begin{array}{l}s=15.9880\\ r^2=64.6\%\end{array}$

The standard error interprets that the predicted percent of grass area burned using the regression line deviates on average by $15.9880$ from the actual perent of grass area burned.

And the coefficient of determination represents that $64.6\%$ of the variation in the percent of grass burned is explained by the least squares repression line using the number of wildebeest as explanatory variable.