91Ó°ÊÓ

To find the confidence interval at $95\%$ that refer to Exercise 14.61. The given data at Exercise 14.61 as follows:

The regression $t$-interval as follows:
Since, $n=11$for a $95\%$confidence interval, $\mathrm{Î\pm }=0.05$.
$df=n-2$
$=11-2$
$=9$
According to calculation:
$t_{\mathrm{Î\pm }/2}=t_{0.05/2}$

$=t_{0.025}$

$=2.262$
For ${\mathrm{β}}_1$, the formula for finding the end points of the confidence interval is as follows:

Table of calculations as follows:

Let,

$S_{xy}=∑x_i{y}_i-\left(∑x_i\right)\left(∑y_i\right)/n$
$=10486-\left(723\right)(156.5)/11$
$=10486-113149.5/11$
$=10486-10286.31818$
$=199.6818182$
Then,

$S_{xx}=∑x_i-{\left(∑x_i\right)}^2/n$
$=48747-(723{)}^2/11$
$=48747-522729/11$
$=48747-47520.81818$
$=1226.181818$

The total sum of square SST is calculated as follows:
$SSR=\frac{S_{xy}^2}{S_{xx}}$
$=\frac{(199.6818{)}^2}{1226.182}$
$=\frac{39872.82851}{1226.182}$
$=32.51787616$

Since,

$SSE=SST-SSR$

$=296.6818182-32.51787616$

$=264.163942$

The slope of the regression line can be calculated using the following formula:

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

$=\frac{199.6818182}{126.181818}$

$=0.162848458$

The standard error of the estimate is calculated using the formula:
$s_e=\sqrt{\frac{SSE}{n-2}}$
$=\sqrt{\frac{264.163942}{11-2}}$
$=5.417706998$
$≈5.42$
Since,$b_1=0.162848458$
$s_e=5.42$
$S_{xx}=1226.181818$
Hence, $0.162848458Â\pm 2.262×\frac{5.42}{\sqrt{1226.18}}$
Also, $0.162848458Â\pm 0.35011801$, Or $-0.187$ to $0.512$
Asa result, the change in the mean quantity of volatile emission per one-gram increase in weight for the potato plant is somewhere between $19$ and $51$ Nano grams, according to $95\%$ confidence.