91Ó°ÊÓ

The researcher looked at the weeds among the crop yields in the query. As a result, the scatterplot for the variables utilized is also included in the question. As a result, we can deduce from the scatterplot that

Because the scatterplot slopes downhill, the direction is negative.

Because the points appear to nearly lie along a line, the form is linear.

Because the points are so wide apart, the strength is weak.

Minitab output from a linear regression on these data is shown below $\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 166.483& 2.725& 61.11& 0.000\\ \text{Weeds permeter}& -1.0987& 0.5712& -1.92& 0.075\end{array}$

$$\begin{gathered} \mathrm{S}=7.97665 \\ \mathrm{R}-\mathrm{Sq}=20.9\% \\ \mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=15.3\% \end{gathered}$$

Now, we know that the researcher evaluated weeds among corn yields in the inquiry, and the computer output of this data is provided. Also, the generic regression equation is as follows:

$\hat{y}=a+bx$

In the "Coef" column of the computer output, the estimates $a$and $b$are given:

$$\begin{gathered} \hat{y}=a+bx \\ =166.483-1.0987x \end{gathered}$$

$\hat{y}$represents the expected corn yield, and $x$represents the number of weeds per meter.

Minitab output from a linear regression on these data is shown below.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 166.483& 2.725& 61.11& 0.000\\ \text{Weeds permeter}& -1.0987& 0.5712& -1.92& 0.075\end{array}$

$$\begin{gathered} \mathrm{S}=7.97665 \\ \mathrm{R}-\mathrm{Sq}=20.9\% \\ \mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=15.3\% \end{gathered}$$

Now, we know that the researcher evaluated weeds among corn yields in the inquiry, and the computer output of this data is provided. As we all know, the general regression equation is as follows:

$\hat{y}=a+bx$

In the "Coef" column of the computer output, the estimates $a$and $b$are given:

$$\begin{gathered} \hat{y}=a+bx \\ =166.483-1.0987x \end{gathered}$$

$\hat{y}$represents the expected corn yield, and $x$represents the number of weeds per meter.

The coefficient of $x$is thus $-1.0987$, and the slope $b$is thus$-1.0987$.

And the $y$-intercept an is the regression equation's constant, resulting in$166.483$.

Minitab output from a linear regression on these data is shown below.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 166.483& 2.725& 61.11& 0.000\\ \text{Weeds permeter}& -1.0987& 0.5712& -1.92& 0.075\end{array}$

$$\begin{gathered} \mathrm{S}=7.97665 \\ \mathrm{R}-\mathrm{Sq}=20.9\% \\ \mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=15.3\% \end{gathered}$$

The following is taken from the computer output:

$$\begin{gathered} n=16 \\ b=-1.0987 \\ {\mathrm{SE}}_b=0.5712 \end{gathered}$$

As a result, we define the hypothesis as follows:

$H_0:\mathrm{β}=0$

$H_a:\mathrm{β}<0$

As a result, the test statistics have the following value:

$$\begin{gathered} t=\frac{b-\mathrm{β}}{{\mathrm{SE}}_b} \\ =\frac{-1.0987-0}{0.5712} \\ =-1.923 \end{gathered}$$

The degrees of freedom are as follows:

$$\begin{gathered} df=n-2 \\ =16-2 \\ =14 \end{gathered}$$

As a result, the $p$-value is:

$0.025<P<0.05$

Alternatively, the $p$-value can be calculated using technology as follows:

$P=0.03753$

The null hypothesis is rejected if the $p$-value is less than or equal to the significance level, as follows:

$P<0.05⇒\text{Reject}{H}_0$