that is indeed the case. To find the intercept of the linear regression, we make use of one other property of the best fit line: in order for it to minimize the SSE, it must pass through the point (X,Y). Again, I will not prove this, but note that the point of the two mean values is the central point of the "cloud" of points in the scatterplot, and if the line missed that central point, the deviations will be larger. Assuming that is the case, we have the following equation for the line: Y=aX+b, which we can solve for the intercept b: 5.2.3 Execises b=Ÿ- Cov(X,Y)X Var (X) Standard Caviar o(X) --4-² (5.5)
that is indeed the case. To find the intercept of the linear regression, we make use of one other property of the best fit line: in order for it to minimize the SSE, it must pass through the point (X,Y). Again, I will not prove this, but note that the point of the two mean values is the central point of the "cloud" of points in the scatterplot, and if the line missed that central point, the deviations will be larger. Assuming that is the case, we have the following equation for the line: Y=aX+b, which we can solve for the intercept b: 5.2.3 Execises b=Ÿ- Cov(X,Y)X Var (X) Standard Caviar o(X) --4-² (5.5)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
I need help with number 5 please!!
![m of squared errors, but
that is indeed the case. To find the intercept of the linear regression, we make use of one other
property of the best fit line: in order for it to minimize the SSE, it must pass through the
point (X,Y). Again, I will not prove this, but note that the point of the two mean values is
the central point of the "cloud" of points in the scatterplot, and if the line missed that central
point, the deviations will be larger. Assuming that is the case, we have the following equation
for the line: Y = aX+b, which we can solve for the intercept b:
5.2.3 Execises
b=Y
Mean (B)=7+5+36 +3,3 + 24 +2₁1
6
Mean (B) = 3.9) ming
7.0
-5.0
3.6
Mean (H) = 0,103+0 091+0,014 + 0.024 +0.031 +0,008.3
2014+
2.4
Mean (H)= 0.044 Comin
2.1
Cov(X, Y)X
Var (X)
Standard
my I
o(X)
Table 5.1: Body leanness (B) and heat loss rate (H) in boys; partial data set from [?]
B(m²/kg) H(°C/min) Standard Dewatton (B) =
L
Claviano
A
13/
92
1₁)²
(5.5)
(3,9-7)+(39-5)²+(3,9-36)+(3.9-3.3) 39-24) (3.9-21)
= 18308467.99 1.83
5
0.103
0.091
0.014
0.024
Standard deviation (H) =
0.031) (0.044-0.103)² + (0,044-0,091) + (0,044-0.014) +-(0,044-0.024)
+(0.044 -0,031) + (0,044-0.006)
0.006
S
Use the data set in table 5.2.3 to answer the following questions:
1. Compute the means and standard deviations of each variable.
Mean (B) = 3.9 m²/ling mean (H)= 0,044 |C/min Standard deviation (B) = 1.83
J
2. Compute the covariance between the two variables.
10.0687 -coverianc
3. Calculate the slope and intercept of the linear regression for the data with B as the
explanatory variable.
-0.0336
intercept
Slope 0.0201
(1-variable)
4. Make a scatterplot of the data set with B as the explanatory variable and sketch the
linear regression line with the parameters you computed.
-0.0950357 824 0.095
5. Calculate the slope and intercept of the linear regression the data with H as the explana-
tory variable.
6. Make a scatterplot of the data set, with H as the explanatory variable and sketch the
linear regression line with the parameters you computed.
Standard deviation (A)= 0.093
Covariance
0
(3.94-7) (0.044-0.103)+(3.9-5) (0.044 -0.091) + (3.9-36) (0.044 -0.014) + (3.9-3.5) (0.044-0m2y)+(3.9-24) (0.044-00x)
+(3.9 -2.1) (0044-0.006)
Covariance = 0.0687
+](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4bb26d64-bb49-4ec5-afc7-ea9f4d31dfb0%2Fa575d9fb-d555-458c-8528-8d38b7441d65%2F8iio5z5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:m of squared errors, but
that is indeed the case. To find the intercept of the linear regression, we make use of one other
property of the best fit line: in order for it to minimize the SSE, it must pass through the
point (X,Y). Again, I will not prove this, but note that the point of the two mean values is
the central point of the "cloud" of points in the scatterplot, and if the line missed that central
point, the deviations will be larger. Assuming that is the case, we have the following equation
for the line: Y = aX+b, which we can solve for the intercept b:
5.2.3 Execises
b=Y
Mean (B)=7+5+36 +3,3 + 24 +2₁1
6
Mean (B) = 3.9) ming
7.0
-5.0
3.6
Mean (H) = 0,103+0 091+0,014 + 0.024 +0.031 +0,008.3
2014+
2.4
Mean (H)= 0.044 Comin
2.1
Cov(X, Y)X
Var (X)
Standard
my I
o(X)
Table 5.1: Body leanness (B) and heat loss rate (H) in boys; partial data set from [?]
B(m²/kg) H(°C/min) Standard Dewatton (B) =
L
Claviano
A
13/
92
1₁)²
(5.5)
(3,9-7)+(39-5)²+(3,9-36)+(3.9-3.3) 39-24) (3.9-21)
= 18308467.99 1.83
5
0.103
0.091
0.014
0.024
Standard deviation (H) =
0.031) (0.044-0.103)² + (0,044-0,091) + (0,044-0.014) +-(0,044-0.024)
+(0.044 -0,031) + (0,044-0.006)
0.006
S
Use the data set in table 5.2.3 to answer the following questions:
1. Compute the means and standard deviations of each variable.
Mean (B) = 3.9 m²/ling mean (H)= 0,044 |C/min Standard deviation (B) = 1.83
J
2. Compute the covariance between the two variables.
10.0687 -coverianc
3. Calculate the slope and intercept of the linear regression for the data with B as the
explanatory variable.
-0.0336
intercept
Slope 0.0201
(1-variable)
4. Make a scatterplot of the data set with B as the explanatory variable and sketch the
linear regression line with the parameters you computed.
-0.0950357 824 0.095
5. Calculate the slope and intercept of the linear regression the data with H as the explana-
tory variable.
6. Make a scatterplot of the data set, with H as the explanatory variable and sketch the
linear regression line with the parameters you computed.
Standard deviation (A)= 0.093
Covariance
0
(3.94-7) (0.044-0.103)+(3.9-5) (0.044 -0.091) + (3.9-36) (0.044 -0.014) + (3.9-3.5) (0.044-0m2y)+(3.9-24) (0.044-00x)
+(3.9 -2.1) (0044-0.006)
Covariance = 0.0687
+
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