Chapter 3, Problem 13PT

(a)

To determine

To give: the equation of the least-squares regression line.

Answer to Problem 13PT

The equation of the least-squares regression line is $92.29–0.05762$ (Wildebeest)

Explanation of Solution

Given:

  

  

Calculation:

We think that “wildebeest” will help explain “percent burned”. So, “wildebeest” is the explanatory variable and “percent burned” is the response variable. From the out put we see that,

Y- Intercept = 92.29

Slope= -0.05762

Equation of least squares regression line is:

Percent burned $=92.29–0.05762$ (Wildebeest)

Conclusion:

Hence,the equation of the least-squares regression line is $92.29–0.05762$ (Wildebeest)

(b)

To determine

To explain: the slope of the regression line

Answer to Problem 13PT

Slope is negative, we can say that every wildebeest abundance (in 1000s) increases percent of the grass burned decreases by 0.05762.

Explanation of Solution

The slope b of the regression line $\widehat{Y}=a+bX$ is the rate at which the predicted response changes along the line as explanatory variable X changes.

Specifically, b is the predicted change in Y when X increases by 1 unit.

(c)

To determine

To find: The correlation

Answer to Problem 13PT

The value of Correlation (r) is $-0.8037$

Explanation of Solution

Calculation:

From the output we have,

  $\begin{array}{l}R-sq=64.6\%\\ r^2=0.646\\ r=\pm 0.8037\end{array}$

The graph 1 shows a clear direction: The overall pattern moves from upper left to lower right.

That is, if wildebeest abundance increases, percent of the grass burned decreases. We call this a negative association between the two variables.

Conclusion:

Hence, the value of Correlation (r) is $-0.8037$

This means a strong relationship between variables.

(d)

To determine

To find: whether a linear model is appropriatefor describing the relationship

Answer to Problem 13PT

A line does seem to be an appropriate model for these data.

Explanation of Solution

A line does seem to be an appropriate model for these data. The residual plot shows a fairly random scatter of points about the horizontal line representing error = 0. The residuals are between +30 to -30 except one point.

On average, predictions of percent of the grass burned would be off by about$s=15.9880$ for the data.

That may be too much prediction error.