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Topic 7:
Regression Analysis 1
This topic will cover:
◦ Simple linear regression
◦ Pearson’s (product moment) correlation coefficient
◦ Spearman’s (rank order) correlation coefficient
By the end of this topic students will be able
to:
◦ Understand a straight line fit to bivariate data
◦ Calculate and interpret Pearson’s correlation
coefficient
◦ Calculate and interpret Spearman’s correlation
coefficient
◦ The search for, and strength of, predictors
 What is a good predictor of future job
performance?
 What is this product’s price-demand curve?
 Which process factors affect production
yield?
 How does a particular share price move with
the market average?
income
saving
s
x
y
xy x
y
y = mx +
c
•Assuming interval or ratio
data.
◦ yi = mxi + c + ei
◦ The least squared line
is the line that
minimizes the sum of
square errors
 e1
2
+ e2
2
+ ⋯ + en
2
◦ y = mxi + c
x
y
e5
e1
e2
e3
e4
yi = mxi + c
+ ei
◦ For the least SSE
straight line, y = mxi +
c
◦ m =
xi − x yi − y
xi − x
2
◦ c = y − mx
x
y
e5
e1
e2
e3
e4
yi = mxi + c
+ ei
◦ For the least SSE
straight line, y = mxi +
c
◦ m =
xi − x yi − y
xi − x
2
◦ c = y − mx
• For the least SSE straight
line, y = mxi + c
• m =
n xiyi − xi yi
n xi
2 − xi
2
• c = y − mx
A company has been carrying out experiments with the
position of a button on its website.
Page
Position
(% vertical)
Click
Throughs
(%)
25 3.07
50 5.64
75 9.63
100 10.26
Independent
Variable
(X axis)
Dependent
Variable
(Y axis)
X
Y
◦ A company has been carrying out experiments with the
position of a button on its website.
x y xy x2
25 3.07
50 5.64
75 9.63
100 10.26
total
mean
m = {(4 × 2107) + (250 ×28.6)}
{(4 × 18750) -(250 ×250)}
m = 0.10224,c = y − mx
c = 7.15 – (0.10224 ×62.5) = 0.76
m =
n xy − x y
n x2− x 2
x y xy x2
25 3.07 76.75 625
50 5.64 282.00 2500
75 9.63 722.25 5625
100 10.26 1026.00 10000
total 250 28.6 2107 18750
mean 62.5 7.15
◦ A company has been carrying out experiments with the
position of a button on its website.
Page
Position
(% vertical)
Click
Throughs
(%)
25 3.07
50 5.64
75 9.63
100 10.26
◦ A company has been carrying out experiments with the
position of a button on its website.
x y xy x2
25 1.44
50 5.58
75 14.64
100 6.94
total
mean
m =
n xy − x y
n x2− x 2
c = y − mx
x y xy x2
25 1.44 36 625
50 5.58 279 2500
75 14.64 1098 5625
100 6.94 694 10000
total 250 28.6 2107 18750
mean 62.5 7.15
= 0.10224
= 0.76
x y
25 1.44
50 5.58
75 14.64
100 6.94
x y
25 3.07
50 5.64
75 9.63
100 10.26
explained
unexplained
y
• Total variation in two parts
y − y 2 = y
x
◦ R2 or r2 is called the coefficient of determination
◦ 0 ≤r2 ≤ 1
◦ r is called the Pearson correlation coefficient
◦ -1 ≤ r ≤ 1
◦ Following rearrangement
R = r =
n xiyi − xi yi
n xi
2
− xi
2
n yi
2
− yi
2
9.425
31.810
92.737
105.268
239.239
R = r =
n xiyi − xi yi
n xi
2
− xi
2
n yi
2
− yi
2
x y xy x2 y2
25 3.07 76.75 625
50 5.64 282.00 2500
75 9.63 722.25 5625
100 10.26
1026.0
0
1000
0
250 28.6 2107 18750
x y
25 1.44
50 5.58
75
14.6
4
100 6.94
x y
25 3.07
50 5.64
75 9.63
100
10.2
6
◦ Sign
 r > 0 positive linear relationship
 r = 0 no linear relationship
 r < 0 negative linear relationship
◦ Strength
 Physical sciences / engineering R2 > 0.6 often found
 Social sciences / policy R2 > 0.25 sometimes useful
 Business and management – includes science &
social science!
 But you will see language like R2 > 0.8 strong, R2 >
0.5 moderate, R2 > 0.25 weak relationship
 Be careful; context and numbers often more
informative than descriptive word, but words help to
communicate.
What is r?
A B C
D E F
r = 1 r = 0.9 r = 0.7
r = 0 r = -0.7 r = -0.9
◦ Interpolation - Estimates between values already
known
◦ Extrapolation - Estimates outside known values
◦ Plot scatter graph to intuit whether straight line is
reasonable
◦ Look at r2 = R2 for strength of relationship
◦ Look at sign of r for direction
 Check agrees with graph
◦ Look at m for gradient of relationship
◦ Use straight line equation to interpolate (with care)
◦ Use straight line equation to extrapolate (with caution)
◦ Sometimes we only have ordinal data
 two interviewers rank candidates
◦ Can we still define a correlation function? Yes
◦ 𝑟𝑠 = 1 −
6 𝑑2
n n2
−1
 where d is the difference between ranking.
◦ Two interviewers individually rank prospective job
candidates. What is the Spearman correlation
coefficient?
Candidat
e
Interviewer
1
Interviewer 2
Hidayat 3 E
Elisa 2 A
Nouman 1 B
Bernie 4 C
Li Ren 5 D
Ahere 6 F
◦ Two interviewers individually rank prospective job
candidates. What is the Spearman correlation coefficient?
◦ 𝑟𝑠 = 1 −
6 𝑑2
n n2
−1
= 1 −
8
35
= 0.771
Candidat
e
Interviewer
1
Interviewer 2 d d2
Hidayat 3 E 5 -2 4
Elisa 2 A 1 1 1
Nouman 1 B 2 -1 1
Bernie 4 C 3 1 1
Li Ren 5 D 4 1 1
Ahere 6 F 6 0 0
◦ For tied ranks use mean rank, then
◦ Use formula for Pearson correlation
Candidat
e
Interviewer
1
Interviewer 2
Hidayat 3
Elisa 2
Nouman 1
Bernie 4
Li Ren 5
Ahere 6
4
1
2
4
4
6
OK
Excellent
Good
OK
OK
Poor
By the end of this topic students will be able
to:
◦ Understand a straight line fit to bivariate data
◦ Calculate and interpret Pearson’s correlation
coefficient
◦ Calculate and interpret Spearman’s correlation
coefficient
◦ Burton, G., Carrol, G. and Wall, S. Quantitative
Methods for Business and Economics. Longman.
◦ Buglear, J. Quantitative Methods for Business.
Elsevier Butterworth Heinemann
◦ Hinton, PR. Statistics Explained. Routledge
Any Questions?

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Lecture 07 Regression Analysis Part 1

  • 2. This topic will cover: ◦ Simple linear regression ◦ Pearson’s (product moment) correlation coefficient ◦ Spearman’s (rank order) correlation coefficient
  • 3. By the end of this topic students will be able to: ◦ Understand a straight line fit to bivariate data ◦ Calculate and interpret Pearson’s correlation coefficient ◦ Calculate and interpret Spearman’s correlation coefficient
  • 4. ◦ The search for, and strength of, predictors  What is a good predictor of future job performance?  What is this product’s price-demand curve?  Which process factors affect production yield?  How does a particular share price move with the market average?
  • 5. income saving s x y xy x y y = mx + c •Assuming interval or ratio data.
  • 6. ◦ yi = mxi + c + ei ◦ The least squared line is the line that minimizes the sum of square errors  e1 2 + e2 2 + ⋯ + en 2 ◦ y = mxi + c x y e5 e1 e2 e3 e4 yi = mxi + c + ei
  • 7. ◦ For the least SSE straight line, y = mxi + c ◦ m = xi − x yi − y xi − x 2 ◦ c = y − mx x y e5 e1 e2 e3 e4 yi = mxi + c + ei
  • 8. ◦ For the least SSE straight line, y = mxi + c ◦ m = xi − x yi − y xi − x 2 ◦ c = y − mx • For the least SSE straight line, y = mxi + c • m = n xiyi − xi yi n xi 2 − xi 2 • c = y − mx
  • 9. A company has been carrying out experiments with the position of a button on its website. Page Position (% vertical) Click Throughs (%) 25 3.07 50 5.64 75 9.63 100 10.26 Independent Variable (X axis) Dependent Variable (Y axis) X Y
  • 10. ◦ A company has been carrying out experiments with the position of a button on its website. x y xy x2 25 3.07 50 5.64 75 9.63 100 10.26 total mean m = {(4 × 2107) + (250 ×28.6)} {(4 × 18750) -(250 ×250)} m = 0.10224,c = y − mx c = 7.15 – (0.10224 ×62.5) = 0.76 m = n xy − x y n x2− x 2 x y xy x2 25 3.07 76.75 625 50 5.64 282.00 2500 75 9.63 722.25 5625 100 10.26 1026.00 10000 total 250 28.6 2107 18750 mean 62.5 7.15
  • 11. ◦ A company has been carrying out experiments with the position of a button on its website. Page Position (% vertical) Click Throughs (%) 25 3.07 50 5.64 75 9.63 100 10.26
  • 12. ◦ A company has been carrying out experiments with the position of a button on its website. x y xy x2 25 1.44 50 5.58 75 14.64 100 6.94 total mean m = n xy − x y n x2− x 2 c = y − mx x y xy x2 25 1.44 36 625 50 5.58 279 2500 75 14.64 1098 5625 100 6.94 694 10000 total 250 28.6 2107 18750 mean 62.5 7.15 = 0.10224 = 0.76
  • 13. x y 25 1.44 50 5.58 75 14.64 100 6.94 x y 25 3.07 50 5.64 75 9.63 100 10.26
  • 14. explained unexplained y • Total variation in two parts y − y 2 = y x
  • 15. ◦ R2 or r2 is called the coefficient of determination ◦ 0 ≤r2 ≤ 1 ◦ r is called the Pearson correlation coefficient ◦ -1 ≤ r ≤ 1 ◦ Following rearrangement R = r = n xiyi − xi yi n xi 2 − xi 2 n yi 2 − yi 2
  • 16. 9.425 31.810 92.737 105.268 239.239 R = r = n xiyi − xi yi n xi 2 − xi 2 n yi 2 − yi 2 x y xy x2 y2 25 3.07 76.75 625 50 5.64 282.00 2500 75 9.63 722.25 5625 100 10.26 1026.0 0 1000 0 250 28.6 2107 18750
  • 17. x y 25 1.44 50 5.58 75 14.6 4 100 6.94 x y 25 3.07 50 5.64 75 9.63 100 10.2 6
  • 18. ◦ Sign  r > 0 positive linear relationship  r = 0 no linear relationship  r < 0 negative linear relationship
  • 19. ◦ Strength  Physical sciences / engineering R2 > 0.6 often found  Social sciences / policy R2 > 0.25 sometimes useful  Business and management – includes science & social science!  But you will see language like R2 > 0.8 strong, R2 > 0.5 moderate, R2 > 0.25 weak relationship  Be careful; context and numbers often more informative than descriptive word, but words help to communicate.
  • 20. What is r? A B C D E F r = 1 r = 0.9 r = 0.7 r = 0 r = -0.7 r = -0.9
  • 21. ◦ Interpolation - Estimates between values already known ◦ Extrapolation - Estimates outside known values
  • 22. ◦ Plot scatter graph to intuit whether straight line is reasonable ◦ Look at r2 = R2 for strength of relationship ◦ Look at sign of r for direction  Check agrees with graph ◦ Look at m for gradient of relationship ◦ Use straight line equation to interpolate (with care) ◦ Use straight line equation to extrapolate (with caution)
  • 23. ◦ Sometimes we only have ordinal data  two interviewers rank candidates ◦ Can we still define a correlation function? Yes ◦ 𝑟𝑠 = 1 − 6 𝑑2 n n2 −1  where d is the difference between ranking.
  • 24. ◦ Two interviewers individually rank prospective job candidates. What is the Spearman correlation coefficient? Candidat e Interviewer 1 Interviewer 2 Hidayat 3 E Elisa 2 A Nouman 1 B Bernie 4 C Li Ren 5 D Ahere 6 F
  • 25. ◦ Two interviewers individually rank prospective job candidates. What is the Spearman correlation coefficient? ◦ 𝑟𝑠 = 1 − 6 𝑑2 n n2 −1 = 1 − 8 35 = 0.771 Candidat e Interviewer 1 Interviewer 2 d d2 Hidayat 3 E 5 -2 4 Elisa 2 A 1 1 1 Nouman 1 B 2 -1 1 Bernie 4 C 3 1 1 Li Ren 5 D 4 1 1 Ahere 6 F 6 0 0
  • 26. ◦ For tied ranks use mean rank, then ◦ Use formula for Pearson correlation Candidat e Interviewer 1 Interviewer 2 Hidayat 3 Elisa 2 Nouman 1 Bernie 4 Li Ren 5 Ahere 6 4 1 2 4 4 6 OK Excellent Good OK OK Poor
  • 27. By the end of this topic students will be able to: ◦ Understand a straight line fit to bivariate data ◦ Calculate and interpret Pearson’s correlation coefficient ◦ Calculate and interpret Spearman’s correlation coefficient
  • 28. ◦ Burton, G., Carrol, G. and Wall, S. Quantitative Methods for Business and Economics. Longman. ◦ Buglear, J. Quantitative Methods for Business. Elsevier Butterworth Heinemann ◦ Hinton, PR. Statistics Explained. Routledge