91Ó°ÊÓ

Given table is

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

STEP 1: the null and alternative hypothesis are:

$H_0:{\mathrm{β}}_1=0$

$H_{\mathrm{Î\pm }}:{\mathrm{β}}_1≠0$

STEP 2: Determine the $\mathrm{Î\pm }$significance level.

The hypothesis test should be run at a significance level of $5$percent, or $\mathrm{Î\pm }=0.05$

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

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

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

SST is given by

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

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

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

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

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

STEP-4

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

We discovered that critical values are$Â\pm t_{\mathrm{Î\pm }/2}=Â\pm t_{0.025}=Â\pm 2.262$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$=0.319982>\mathrm{Î\pm }=0.05$, 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.