Long-term records from the Serengeti National Park in Tanzania show interesting ecological relationships. When wildebeest are more abundant, they graze the grass more heavily, so there are fewer fires and more trees grow. Lions feed more successfully when there are more trees, so the lion population increases. Researchers collected data on one part of this cycle, wildebeest abundance (in thousands of animals) and the percent of the grass area burned in the same year. The results of a least-squares regression on the data are shown here.

a. Is a linear model appropriate for describing the relationship between wildebeest abundance and percent of grass area burned? Explain.

b. Give the equation of the least-squares regression line. Be sure to define any variables you use.

c. Interpret the slope. Does the value of the y-intercept have meaning in this context? If so, interpret the y-intercept. If not, explain why.

d. Interpret the standard deviation of the residuals and r2.

The regression line is:

$$\begin{gathered} \stackrel{⏜}{y}=92.29-0.05762x \\ s=15.9880 \\ {r}^2=64.6\% \end{gathered}$$

Yes, given there is no strong curvature in the scatterplot and no strong curvature in the residual plot, the linear model is adequate. As a result, we can conclude that a linear model is suitable for capturing the link between wildebeest abundance and the fraction of grass area burned.

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

$$\begin{gathered} b_0=92.29 \\ {b}_1=-0.05762 \end{gathered}$$

Thus, the regression line is:

$$\begin{gathered} y=b_0+b_{1}x \\ \Rightarrow \stackrel{⏜}{y}=92.29-0.05762x \end{gathered}$$

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

The regression line is:

$$\begin{gathered} \stackrel{⏜}{y}=92.29-0.05762x \\ {b}_0=92.29 \\ {b}_1=0.05762 \end{gathered}$$

The slope in a regression equation is the coefficient of $x$and it reflects the average rise or decrease in $y$per unit of $x$This means that per thousand wildebeest, the proportion burned lowers by $0.05762$percent on average.

When x is 0, the y-intercept is the regression line's constant and represents the average y-value. When there are wildebeest, this means that 92.29 percent of the grass area is burned on average. However, having zero wildebeest is not acceptable, as the value of the $y-$intercept then has no relevance in this situation.

The standard error indicates that the regression line's projected percent of grass area burned differs by $15.9880$on average from the actual percent of grass area burned.

The coefficient of determination indicates that the least-squares regression line using the number of wildebeest as the explanatory variable explains $64.6$ percent of the variation in the proportion of grass burned.