Please explain step by step The scatter plot and table below show 8 data points that were collected to compare battery life to thetemperature of the battery's environment.60y82585587578255555558565455505154535350 5560 65 70 75 80 85 90Тетрerature (F)The sums of x , y , x , y“ , and xy are shown below:.2xy522443357322454929031The estimates for the slope and intercept of the regression line are0.075 hours per degreea50.481 hoursBattery Lifetime (Hrs) a) Find the 95% prediction interval for the slope.Threshold value: tcritVariance of the response: SyStandard deviation of the slope: StSlope confidence interval is fromup to(Round answers to 3 decimal place as needed)

Answered: a) Find the 95% prediction interval for…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Please explain step by step

The scatter plot and table below show 8 data points that were collected to compare battery life to the
temperature of the battery's environment.
60
y
82
58
55
87
57
82
55
55
55
58
56
54
55
50
51
54
53
53
50 55
60 65 70 75 80 85 90
Тетрerature (F)
The sums of x , y , x , y“ , and xy are shown below:
.2
xy
522
443
35732
24549
29031
The estimates for the slope and intercept of the regression line are
0.075 hours per degree
a
50.481 hours
Battery Lifetime (Hrs)

a) Find the 95% prediction interval for the slope.
Threshold value: tcrit
Variance of the response: Sy
Standard deviation of the slope: St
Slope confidence interval is from
up to
(Round answers to 3 decimal place as needed)

Expert Solution

Step 1

The regression equation is given by :

$y = a + b x a = i n t e r c e p t b = s l o p e$

The formula for the confidence interval for slope coefficient is given by :

$\begin{array}{rcl} C . I & = & \hat{b} \pm t_{α / 2, n - 2} S . E (\hat{b}) \\ S . E (\hat{b}) & = & \frac{s}{\sqrt{S S_{x x}}} \\ s & = & \sqrt{\frac{\sum_{i} {(Y_{i} - \hat{Y})}^{2}}{n - 2}} \\ = & \sqrt{\frac{8.49}{6}} \\ = & 1.1895 \\ S S_{x x} & = & \sum_{i} {(x_{i} - x)}^{2} = \sum_{i} {(x_{i})}^{2} - n x^{2} \\ = & 35732 - 8 \times {(522 / 8)}^{2} = 1671.5 \end{array}$

	X	Y	$\hat{y}$	${(y - \hat{y})}^{2}$
	82	58	56.636	1.860
	87	57	57.011	0.000
	82	55	56.636	2.676
	55	55	54.611	0.151
	58	56	54.836	1.355
	54	55	54.536	0.215
	51	54	54.311	0.097
	53	53	54.461	2.135
Sum	522	443		8.49

$\begin{array}{rcl} S . E (\hat{b}) & = & \frac{1.1895}{\sqrt{1671.5}} \\ = & 0.0290945 \end{array}$

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