Question

The following table gives data on the mean number of seeds produced in a year by several common tree species and the mean weight (in milligrams) of the seeds produced. Two species appear twice because their seeds were counted in two locations. We might expect that trees with heavy seeds produce fewer of them, but what mathematical model best describes the relationship?

(a) Describe the association between seed count and seed weight shown in the scatterplot.

(b) Two alternative models based on transforming the original data are proposed to predict the seed weight from the seed count. Here are graphs and computer output from a least-squares regression analysis of the transformed data.

Model A:

Model B:

Which model, A or B, is more appropriate for predicting seed weight from seed count? Justify your answer.

(c) Using the model you chose in part (b), predict the seed weight if the seed count is 3700.

Short answer

(a) The scatterplot's unusual features: There appears to be one outlier because the scatterplot's rightmost point is far from the other points.

(b) Model B is suitable.

(c) The estimated seed weight is $19.7766$mg.

Step-by-step solution

Part (a) Step 1: Given information

To describe the scatterplot's association between seed count and weight.

Explanation

The data and scatterplot on the mean number of seeds produced in a year by several common tree species, as well as the mean weight of the seeds produced, are provided. We can tell from the scatterplot that the scatterplot's direction is negative because the pattern in the scatterplot slopes downward.

And the scatterplot's shape is curved because the scatterplot has a strong curvature. The scatterplot's strength is also high because the points in the scatterplot do not deviate significantly from the general pattern of the points. The scatterplot's unusual features: There appears to be one outlier because the scatterplot's rightmost point is far from the other points.

Part (b) Step 1: Given information

To determine whether model A or B is better suited for predicting seed weight from seed count.

Explanation

The data and scatterplot on the mean number of seeds produced in a year by several common tree species, as well as the mean weight of the seeds produced, are provided. To predict the seed weight from the seed count, two alternative models are proposed. As a result, the scatterplot of model A has strong curvature, as does the residual plot of model A, indicating that model A is not appropriate.

The scatterplot B, on the other hand, has no strong curvature, and neither does the residual plot of model B. Furthermore, the residuals in the residual plot appear to be randomly distributed about the horizontal line at zero, implying that model B is appropriate for predicting seed weight from seed count.

Part (c) Step 1: Given information

To forecast the seed weight if the seed count is 3700, use the model you selected in part (b).

Explanation

The data and scatterplot on the mean number of seeds produced in a year by several common tree species, as well as the mean weight of the seeds produced, are provided. To predict the seed weight from the seed count, two alternative models are proposed. Part (b) reveals that model B is more appropriate. Then we'll apply model B. As a result, the general equation of the least square regression line is as follows:

$\hat{y}=b_0+b_{1}x$

Thus, the estimate of the constant is given in the row "Constant" and column "Coef" of the computer output as:

$b_0=15.491$

The slope b1 is given in the computer output's row "Mentos" and column "Coef" as:

$b_1=-1.5222$

Now, in place of the values in the equation,

$\hat{y}=b_0+b_{1}x=15.491-1.5222x$

Take the logarithm in the equation and solve it as follows:

$ln\hat{y}=15.491-1.5222x$

Replace $x$by $3700$,

$$\begin{gathered} ln\hat{y}=15.491-1.5222x \\ =15.491-1.5222\left(3700\right)=2.9845 \end{gathered}$$

Taking the exponential on both sides, we have

As a result, the estimated seed weight is $19.7766$mg.