Lamb’s-quarter is a common weed that interferes with the growth of corn. An agriculture researcher planted corn at the same rate in $16$small plots of ground and then weeded the plots by hand to allow a fixed number of lamb’s-quarter plants to grow in each meter of the cornrow. The decision of how many of these plants to leave in each plot was made at random. No other weeds were allowed to grow. Here are the yields of corn (bushels per acre) in each of the plots.

$\begin{array}{ccccc}\begin{array}{c}\text{Weeds}\\ \text{per meter}\end{array}& \text{Corn yield}& & \begin{array}{c}\text{Weeds}\\ \text{per meter}\end{array}& \text{Corn yield}\\ 0& 166.7& & 3& 158.6\\ 0& 172.2& & 3& 176.4\\ 0& 165.0& & 3& 153.1\\ 0& 176.9& & 3& 156.0\\ 1& 166.2& & 9& 162.8\\ 1& 157.3& & 9& 142.4\\ 1& 166.7& & 9& 162.8\\ 1& 161.1& & 9& 162.4\\ & & & & \end{array}$

(a) A scatterplot of the data with the least-squares line added is shown below. Describe what this graph tells you about the relationship between these two variables.

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

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

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

(b) What is the equation of the least-squares regression line for predicting corn yield from the number of lamb’s quarter plants per meter? Define any variables you use.

(c) Interpret the slope and y-intercept of the regression line in context.

(d) Do these data provide convincing evidence that more weeds reduce corn yield? Carry out an appropriate test at the A $\alpha =0.05$level to help answer this question.

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:

$H_0:\beta =0$

$H_a:\beta <0$

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

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

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:

$P<0.05\Rightarrow \text{Reject}{H}_0$