91影视

It is given the data and the scatterplot on the mean number of seeds produced in a year by several common tree species and the mean weight of the seeds produced. By looking at the scatterplot we can say that,

The direction of the scatterplot is negative because the pattern in the scatterplot slopes downward. And the form of the scatterplot is curved because there is a strong curvature present in the scatterplot. Also the strength of the scatterplot is strong because the points in the scatterplot do not deviate much from the general pattern in the points. The unusual features in the scatterplot: There appear to be one outlier because the right most point in the scatterplot lies far from the other points in the scatterplot.

It is given the data and the scatterplot on the mean number of seeds produced in a year by several common tree species and the mean weight of the seeds produced. The two alternative models are proposed to predict the seed weight from the seed count. Thus, the scatterplot of model A contains strong curvature and the residual plot of the model A contains strong curvature as well which indicates that model A is not appropriate. Whereas, the scatterplot B contains no strong curvature and the residual plot of model B contains no strong curvature as well. Moreover, the residuals in the residual plot appear to be randomly scattered about the horizontal line at zero and thus model B appears to be appropriate for predicting seed weight from the seed count.

It is given the data and the scatterplot on the mean number of seeds produced in a year by several common tree species and the mean weight of the seeds produced. The two alternative models are proposed to predict the seed weight from the seed count. In part (b), we find out the model B is more appropriate. Then we will use the model B. Thus, the general equation of the least square regression line is:

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

Thus, form the computer output, we have that the estimate of the constant is given in the row 鈥淐onstant鈥� and in the column 鈥淐oef鈥� as:

$b_0=15.491$

The slope $b_1$is given in the row 鈥淢entos鈥� and in the column 鈥淐oef鈥� of the given computer output as:

$b_1=-1.5222$

Now replacing the values in the equation we have,

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

Now, take the logarithm in the equation and solve it as:

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

Replace $x$by $3700,$

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

Now taking exponential on both sides we have,

$$\begin{gathered} \hat{y}=e^{\ln \hat{y}} \\ =e^{2.9845} \\ =19.7766 \end{gathered}$$

Thus, the predicted seed weight is$19.7766mg$.