Question

Following are the data on plant weight and quantity of volatile emissions.

$\alpha =0.05$

presuming that the assumption for regression inference are met, decide at the specified significance level whether the data provide sufficient evidence to conclude that the predictor variable is useful for providing the response variable.

Short answer

The data do not provide sufficient evidence to conclude that the variable, weight $x$is useful as a predictor of quantity of volatile emission$y$for the potato plant.

Step-by-step solution

Given Information

Given table is

$\alpha =0.05$we have to find out whether the data provide sufficient evidence to conclude that the predictor variable is useful for providing the response variable.

Explanation

STEP 1: the null and alternative hypothesis are:

$H_0:{\beta }_1=0$

$H_{\alpha }:{\beta }_1\ne 0$

STEP 2: Determine the $\alpha$significance level.

The hypothesis test should be run at a significance level of $5$percent, or $\alpha =0.05$

Table of computation:$H_0:{\beta }_1=0$

$$\begin{gathered} S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n \\ =10486-\left(723\right)(156.5)/11 \\ =199.6818182 \end{gathered}$$

$$\begin{gathered} S_{xx}=\sum x_i^2-{\left(\sum x_i\right)}^2/n \\ =48747-(723{)}^2/11 \\ =1226.181818 \end{gathered}$$

SST is given by

$$\begin{gathered} S_{yy}=\sum y_i^2-{\left(\sum y_i\right)}^2/n \\ =2523.25-(156.5{)}^2/11 \\ =296.6818182 \end{gathered}$$

SSR is given by

$$\begin{gathered} SSR=\frac{S_{xy}^2}{S_{xx}} \\ =\frac{(199.6818{)}^2}{1226.182} \\ =32.51787616 \end{gathered}$$

$$\begin{gathered} SSE=SST-SSR \\ =296.6818182-32.51787616 \\ =264.163942 \end{gathered}$$

Slope of regression line can be calculated by

$$\begin{gathered} b_1=\frac{S_{xy}}{S_{xx}} \\ =\frac{199.6818182}{1226.181818} \\ =0.162848458 \end{gathered}$$

Standard error can be estimated by

$$\begin{gathered} s_e=\sqrt{\frac{SSE}{n-2}} \\ =\sqrt{\frac{264.163942}{11-2}} \\ \approx 5.42 \end{gathered}$$

STEP-3 The value of test statistic can be calculated as

$$\begin{gathered} t=\frac{b_1}{s_e/\sqrt{s_{xx}}} \\ =\frac{0.162848458}{5.417706998\sqrt{1226.181818}} \\ \approx 1.053 \end{gathered}$$

STEP-4

$$\begin{gathered} df=n-1 \\ =11-2 \\ =9 \end{gathered}$$

We discovered that critical values are$\pm t_{\alpha /2}=\pm t_{0.025}=\pm 2.262$using technology.

STEP 4: The test statistic's value is $t=1.053$as of Step 3. Because the test is two-sided, the P-value represents the likelihood of seeing a t value of $1.053$or larger if the null hypothesis is true. We get the$p=0.319982$through using technology.

STEP 5: Because the test statistics have a lower value than the crucial value. Our null hypothesis is not rejected. $H_0$i.e. $t=1.053<t_{0.025,9}=2.262$

or

STEP 5: Since the P-value is$=0.319982>\alpha =0.05$, this is the final step. Our null hypothesis is not rejected.$H_0$

STEP 6: The data do not give adequate evidence to establish that the variable, weight $\left(x\right)$is useful as a predictor of quantity of volatile emission ($\left(y\right)$ for the potato plant at the$5\%$significance level.