Question

To find and interpret a confidence interval, at the specified confidence level $95\%$for the slope of the population regression line that relates the response variables to the predictor variable.

Short answer

It 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.

Step-by-step solution

Given information

The given data is

Explanation

Regression $t-$interval is

For a $95\%$confidence interval,

$\alpha =0.05$

$n=11$,

$df=n-2$

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

From the technology

$t_{\alpha /2}=t_{0.05/2}$

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

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

$b_1\pm t_{\alpha /2}\frac{s_e}{\sqrt{S_{xx}}}$

Computation table

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

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

$=10486-113149.5/11$

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

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

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

$=48747-522729/11$

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

The total sum of square SST is

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

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

$=2523.25-24492.25/11$

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

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

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

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

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

Substituting the values

$0.162848458\pm 2.262\times \frac{5.42}{\sqrt{1226.18}}$

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

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.