Question

In each of Exercises 14.64-14.69, apply Procedure 14.2 on page 567 to find and interpret a confidence interval, at the specified confidence level, for the slope of the population regression line that relates the response variable to the predictor variable.
14.67 Plant Emissions. Refer to Exercise 14.61: 95%.

Short answer

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 gram, according to $95\%$ confidence.

Step-by-step solution

Given information

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

Weight	57	85	57	65	52	67	62	80	77	53	68
Emission	8	22	10.5	22.5	12	11.5	7.5	13	16.5	21	12

Explanation

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

$=t_{0.025}$

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

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

Explanation

Table of calculations as follows:

x	y	$xy$	$x^2$	$y^2$
57	8	456	3249	64
85	22	1870	7225	484
57	10.5	598.5	3249	110.25
65	22.5	1462.5	4225	506.25
52	12	624	2704	144
67	11.5	770.5	4489	132.25
62	7.5	465	3844	56.25
80	13	1040	6400	169
77	16.5	1270.5	5929	272.25
53	21	1113	2809	441
68	12	816	4624	144
$\sum x_i=723$	$\sum y_i=156.5$	$\sum x_i{y}_i=10486$	$\sum {x^2}_i=48747$	$\sum {y^2}_i=2523.25$

Explanation

Let,

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

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

Explanation

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:

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

$=\frac{199.6818182}{126.181818}$

$=0.162848458$

Explanation

The standard error of the estimate is calculated using the formula:
$s_e=\sqrt{\frac{SSE}{n-2}}$
$=\sqrt{\frac{264.163942}{11-2}}$
$=5.417706998$
$\approx 5.42$
Since,$b_1=0.162848458$
$s_e=5.42$
$S_{xx}=1226.181818$
Hence, $0.162848458\pm 2.262\times \frac{5.42}{\sqrt{1226.18}}$
Also, $0.162848458\pm 0.35011801$, 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.