Subject: NUMERICAL METHODSDirections: In example no. 2, the problem and solution are both given. Please answer letter a.Thank you in advance tutors! Example 2:Use least-squares regression to fit a straight line to x and y values given below.Along with the slope and intercept, compute the standard error of the estimateand the correlation coefficient. Plot the data and the regression line.4.11121517199 8 7y57101212nEx¡yt-Ex; £ yiη Σx2-Σ x)210(911)-(95)(82)%3Dn = 10Σy82a, == 0.352469959910(1277)-(95)zEx¡yi = 91195= 9.510ao = ỹ – ajã = 8.2 – 0.3524699599(9.5) = 4.851535381Σχ 1277== 8.2ΣΧ 95Therefore, the least square fit is:y = 4.851535381 + 0.3524699599xLeast-Squares Fit of a Straight Line14Standard error of the estimate12SrVn-2Evi-ao-a;x;)²%3Dп-2109.0739652878Sy == 1.065009710-26Correlation coefficient4EGi-y)²-E(y;-ao-a,x;)?r2 = St-Sr -St55.60-9.073965287r2 =55.60101520r2 = 0.8367991855r = 0.9147672849a.)Repeat the problem Example 2, but regress x versus y - that is, switch thevariables. Interpret your results. Plot the data and the regression line.Interpret your results.49.1112151719y67698 7101212

Answered: Image /qna-images/answer/ff8e6a07-45f2-4c8f-aa32-01bef4a9ef8f.jpg

Example 2: Use least-squares regression to fit a straight line to x and y values given below. Along with the slope and intercept, compute the standard error of the estimate and the correlation coefficient. Plot the data and the regression line. 15 17 4. 11 12 19 9 8 7 6. 7 6 10 12 12 Σy 82 nExiy-ExEy a = nEx1²-Ex1)² n = 10 10(911)-(95)(82) = 0.3524699599 10(1277)-(95)z Ex;yi = 911 95 = 9.5 10 ao = ỹ – aqã = 8.2 – 0.3524699599(9.5) = 4.851535381 Ex;? = 1277 y =- = 8.2 10 Σχ 95 Therefore, the least square fit is: y = 4.851535381 + 0.3524699599x Least-Squares Fit of a Straight Line 14 Standard error of the estimate 12 Sr Evi-ao-a‚x¡)² Vn-2 n-2 10 Sy 9.073965287 %3D = 1.0650097 10-2 6 Correlation coefficient r2 = St-Sr - EGi-y)²-E(yi-ao-a,x;)² St 2 55.60-9.073965287 r2 = 55,60 10 15 20 r2 = 0.8367991855 r = 0.9147672849 a.) Repeat the problem Example 2, but regress x versus y variables. Interpret your results. Plot the data and the regression line. Interpret your results. that is, switch the 4. 6. 11 12 15 17 19 y 7 8 7 10 12 12

Example 2: Use least-squares regression to fit a straight line to x and y values given below. Along with the slope and intercept, compute the standard error of the estimate and the correlation coefficient. Plot the data and the regression line. 15 17 4. 11 12 19 9 8 7 6. 7 6 10 12 12 Σy 82 nExiy-ExEy a = nEx1²-Ex1)² n = 10 10(911)-(95)(82) = 0.3524699599 10(1277)-(95)z Ex;yi = 911 95 = 9.5 10 ao = ỹ – aqã = 8.2 – 0.3524699599(9.5) = 4.851535381 Ex;? = 1277 y =- = 8.2 10 Σχ 95 Therefore, the least square fit is: y = 4.851535381 + 0.3524699599x Least-Squares Fit of a Straight Line 14 Standard error of the estimate 12 Sr Evi-ao-a‚x¡)² Vn-2 n-2 10 Sy 9.073965287 %3D = 1.0650097 10-2 6 Correlation coefficient r2 = St-Sr - EGi-y)²-E(yi-ao-a,x;)² St 2 55.60-9.073965287 r2 = 55,60 10 15 20 r2 = 0.8367991855 r = 0.9147672849 a.) Repeat the problem Example 2, but regress x versus y variables. Interpret your results. Plot the data and the regression line. Interpret your results. that is, switch the 4. 6. 11 12 15 17 19 y 7 8 7 10 12 12

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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Question

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Subject: NUMERICAL METHODS

Directions: In example no. 2, the problem and solution are both given. Please answer letter a.

Thank you in advance tutors!

Example 2:
Use least-squares regression to fit a straight line to x and y values given below.
Along with the slope and intercept, compute the standard error of the estimate
and the correlation coefficient. Plot the data and the regression line.
4.
11
12
15
17
19
9 8 7
y
5
7
10
12
12
nEx¡yt-Ex; £ yi
η Σx2-Σ x)2
10(911)-(95)(82)
%3D
n = 10
Σy82
a, =
= 0.3524699599
10(1277)-(95)z
Ex¡yi = 911
95
= 9.5
10
ao = ỹ – ajã = 8.2 – 0.3524699599(9.5) = 4.851535381
Σχ 1277
=
= 8.2
ΣΧ 95
Therefore, the least square fit is:
y = 4.851535381 + 0.3524699599x
Least-Squares Fit of a Straight Line
14
Standard error of the estimate
12
Sr
Vn-2
Evi-ao-a;x;)²
%3D
п-2
10
9.073965287
8
Sy =
= 1.0650097
10-2
6
Correlation coefficient
4
EGi-y)²-E(y;-ao-a,x;)?
r2 = St-Sr -
St
55.60-9.073965287
r2 =
55.60
10
15
20
r2 = 0.8367991855
r = 0.9147672849
a.)
Repeat the problem Example 2, but regress x versus y - that is, switch the
variables. Interpret your results. Plot the data and the regression line.
Interpret your results.
4
9.
11
12
15
17
19
y
6
7
6
9
8 7
10
12
12

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