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. (Some 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) Based on the scatterplot below, is a linear model appropriate to describe the relationship between seed count and seed weight? Explain

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

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 $3700$.

(d) Interpret the value of$r^2$ for your model.

Short answer

(a) No, the linear model is appropriate to describe the relationship between seed count and seed weight .

(b) The answer to this part is given below.

(c) The seed weight if the seed count is $3700$is $19.7760mg$

(d) The value of $r^2$for model is$86.3\%$.

Step-by-step solution

Part (a) step 1: Given Information

We need to find the linear model appropriate to describe the relationship between seed count and seed weight or not.

Part (a) step 2: Explanation

No, The scatterplot shows a strong curved pattern.

Part (b) step 1: Given Information

We need to graphs and computer output from a least-squares regression analysis on the transformed data are shown.

Part (b) step 2:Explanation

Its scatterplot shows a more linear pattern and its residual plot shows no visible pattern.

Part (c) step 1: Given Information

We need to find the seed weight if the seed count is$3700$.

Part (c) step 2: Explanation

On the scatterplot, we that the variable on the horizontal axis is "ln(count)" and the variable on the vertical axis is "ln(weight)", so the variable is " while the -variable is " . The general regression equation is then:

$\ln \left(weight\right)=a+b\ln \left(count\right)$

The constant is given in the row "Constant" and in the column "Coef" of the computer output of model $B$:

$a=15.491$

The slope is given in the row "ln(count)" and in the column "Coef" of the computer output of model B:

$b=-1.5222$

Replacing with $15.491$and with $-1.5222$in the general regression equation, we have:

$\ln \left(weight\right)=15.491-1.5222\ln \left(count\right)$

Replace "count" by $3700$and evaluate:

$\ln \left(weight\right)=15.491-1.5222\ln \left(3700\right)\approx 2.9845$

Take the exponential of each side:

$weight=e^{\ln weight}=e^{2.9845}\approx 19.7760$

Thus the predicted weight is$19.7760mg$.

Part (d) step 1: Given Information

We need to find the value of$r^2$for in model.

Part (d) step 2: Explanation

About $86.3\%$of the variation in $In$(seed weight) is accounted for by the linear modal relating $In$(seed weight ) to$In$(seed count.)