91Ó°ÊÓ

The given data is

Regression $t-$interval is

For a $95\%$confidence interval,

$\mathrm{Î\pm }=0.05$

$n=11$,

$df=n-2$

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

From the technology

$t_{\mathrm{Î\pm }/2}=t_{0.05/2}$

$$\begin{gathered} =t_{0.025} \\ =2.262 \end{gathered}$$

The formula for computing the confidence interval endpoints for ${\mathrm{β}}_1$is

$b_1Â\pm t_{\mathrm{Î\pm }/2}\frac{s_e}{\sqrt{S_{xx}}}$

Computation table

$S_{xy}=∑x_i{y}_i-\left(∑x_i\right)\left(∑y_i\right)/n$

$=10486-\left(723\right)(156.5)/11$

$=10486-113149.5/11$

$$\begin{gathered} =10486-10286.31818 \\ =199.6818182 \end{gathered}$$

$S_{xx}=∑x_i^2-{\left(∑x_i\right)}^2/n$

$=48747-(723{)}^2/11$

$=48747-522729/11$

$$\begin{gathered} =48747-47520.81818 \\ =1226.181818 \end{gathered}$$

The total sum of square SST is

$S_{yy}=∑y_i^2-{\left(∑y_i\right)}^2/n$

$=2523.25-(156.5{)}^2/11$

$=2523.25-24492.25/11$

$$\begin{gathered} =2523.25-2226.568182 \\ =296.6818182 \end{gathered}$$

The regression sum of squares SSR is

$SSR=\frac{S_{xy}^2}{S_{xx}}$

$=\frac{(199.6818{)}^2}{1226.182}$

$=\frac{39872.82851}{1226.182}$

$=32.51787616$

$SSE=SST-SSR$

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

The slope of the regression line is

$b_1=\frac{S_{xy}}{S_{xx}}$

$=\frac{199.6818182}{126.181818}$

$=0.162848458$

The standard error of the estimate is

$s_e=\sqrt{\frac{SSE}{n-2}}$

$=\sqrt{\frac{264.163942}{11-2}}$

$$\begin{gathered} =5.417706998 \\ ≈5.42 \end{gathered}$$

Substituting the values

$0.162848458Â\pm 2.262×\frac{5.42}{\sqrt{1226.18}}$

$0.162848458Â\pm 0.35011801,Or-0.187to0.512$

We may be $95\%$positive that the change in the mean quantity of volatile emission per 1-gram increase in weight for the potato plant is between $19$and $51$Nanograms.