•
With the growth of internet service providers, a researcher decides to examine whether there is a
correlation between cost of internet service per month (rounded to the nearest dollar) and degree
of customer satisfaction (on a scale of 1 - 10 with a 1 being not at all satisfied and a 10 being
extremely satisfied). The researcher only includes programs with comparable types of services.
Determine if customers should be happy about paying more.
Dollars (SD = 6.13)
11
18
17
15
9
5
12
19
22
25
Satisfaction (SD = 2.87)
6
8
10
4
9
6
3
5
2
10
Correlations
VAR00001
VAR00002
Pears on Correlation
Sig. (2-tailed)
N
Pears on Correlation
Sig. (2-tailed)
N
VAR00001
1
.
10
.076
.834
10
VAR00002
.076
.834
10
1
.
10
Practice
• Situation 1
• Based on a sample of 100 subjects you find the
correlation between extraversion is happiness is
r=.15. Determine if this value is significantly
different than zero.
• Situation 2
• Based on a sample of 600 subjects you find the
correlation between extraversion is happiness is
r=.15. Determine if this value is significantly
different than zero.
Step 1
• Situation 1
• H1: r is not equal to 0
– The two variables are related to each other
• H0: r is equal to zero
– The two variables are not related to each other
• Situation 2
• H1: r is not equal to 0
– The two variables are related to each other
• H0: r is equal to zero
– The two variables are not related to each other
Step 2
• Situation 1
• df = 98
• t crit = +1.985 and -1.984
• Situation 2
• df = 598
• t crit = +1.96 and -1.96
Step 3
• Situation 1
• r = .15
• Situation 2
• r = .15
Step 4
• Situation 1
1.5 
.15 100  2
100  (.15) 2
• Situation 2
3.71 
.15 600  2
1  .15 2
Step 5
• Situation 1
• If tobs falls in the critical region:
– Reject H0, and accept H1
• If tobs does not fall in the critical region:
– Fail to reject H0
• Situation 2
• If tobs falls in the critical region:
– Reject H0, and accept H1
• If tobs does not fall in the critical region:
– Fail to reject H0
Step 6
• Situation 1
• Based on a sample of 100 subjects you find the
correlation between extraversion is happiness is r=.15.
Determine if this value is significantly different than zero.
• There is not a significant relationship between
extraversion and happiness
• Situation 2
• Based on a sample of 600 subjects you find the
correlation between extraversion is happiness is r=.15.
Determine if this value is significantly different than zero.
• There is a significant relationship between
extraversion and happiness.
Practice
• You collect data from 53 females and find the
correlation between candy and depression is
-.40. Determine if this value is significantly
different than zero.
• You collect data from 53 males and find the
correlation between candy and depression is
-.50. Determine if this value is significantly
different than zero.
Practice
• You collect data from 53 females and find the
correlation between candy and depression is .40.
– t obs = 3.12
– t crit = 2.00
• You collect data from 53 males and find the
correlation between candy and depression is .50.
– t obs = 4.12
– t crit = 2.00
Practice
• You collect data from 53 females and find
the correlation between candy and
depression is -.40.
• You collect data from 53 males and find
the correlation between candy and
depression is -.50.
• Is the effect of candy significantly different
for males and females?
Hypothesis
• H1: the two correlations are different
• H0: the two correlations are not different
Testing Differences Between
Correlations
• Must be independent
for this to work
Z
r1  r2
1
1

N1  3 N 2  3
r
When the population value of r is not zero the
distribution of r values gets skewed
Easy to fix!
Use Fisher’s r transformation
Page 746
Testing Differences Between
Correlations
• Must be independent
for this to work
Z
r1  r2
1
1

N1  3 N 2  3
Testing Differences Between
Correlations
Z
 .549  (.424)
1
1

N1  3 N 2  3
Testing Differences Between
Correlations
Z
 .549  (.424)
1
1

53  3 53  3
Testing Differences Between
Correlations
 .625 
 .549  (.424)
1
1

53  3 53  3
Testing Differences Between
Correlations
 .625 
 .549  (.424)
1
1

53  3 53  3
Note: what would the z value be if there was no difference
between these two values (i.e., Ho was true)
Testing Differences
• Z = -.625
• What is the probability of obtaining a Z score of this size
or greater, if the difference between these two r values
was zero?
• p = .267
• If p is < .025 reject Ho and accept H1
• If p is = or > .025 fail to reject Ho
• The two correlations are not significantly different than
each other!
Remember this:
Statistics Needed
• Need to find the best place to draw the
regression line on a scatter plot
• Need to quantify the cluster of scores
around this regression line (i.e., the
correlation coefficient)
Regression allows us to predict!
.
12
10
Smile
8
6
4
2
.
.
.
.
0
1
2
3
Talk
4
5
Straight Line
Y = mX + b
Where:
Y and X are variables representing scores
m = slope of the line (constant)
b = intercept of the line with the Y axis
(constant)
Excel Example
That’s nice but. . . .
• How do you figure out the best values to
use for m and b ?
• First lets move into the language of
regression
Straight Line
Y = mX + b
Where:
Y and X are variables representing scores
m = slope of the line (constant)
b = intercept of the line with the Y axis
(constant)
Regression Equation
Y = a + bX
Where:
Y = value predicted from a particular X value
a = point at which the regression line intersects
the Y axis
b = slope of the regression line
X = X value for which you wish to predict a Y
value
Practice
• Y = -7 + 2X
• What is the slope and the Y-intercept?
• Determine the value of Y for each X:
• X = 1, X = 3, X = 5, X = 10
Practice
• Y = -7 + 2X
• What is the slope and the Y-intercept?
• Determine the value of Y for each X:
• X = 1, X = 3, X = 5, X = 10
• Y = -5, Y = -1, Y = 3, Y = 13
Finding a and b
• Uses the least squares method
• Minimizes Error
Error = Y - Y
 (Y - Y)2 is minimized
.
12
10
Smile
8
6
4
2
.
.
.
.
0
1
2
3
Talk
4
5
Error = Y - Y
 (Y - Y)2 is minimized
12
10
Error = 1
Smile
8
6
4
2
.
Error = .5
.
Error = 0
.
Error = -.5
.
.
Error = -1
0
1
2
3
Talk
4
5
Finding a and b
• Ingredients
• COVxy
• Sx2
• Mean of Y and X
Regression
a  Y  bX
COV XY
b
2
SX
Jerry
Smile
(Y)
9
Talk
(X)
5
Elan
2
1
George
5
3
Newman
4
4
Kramer
3
2
SY =2.70 SX =1.58
2
M = 4.6 SX = 2.50
M=3
XY
Regression
a  Y  bX
Ingredients
Mean Y =4.6
Mean X = 3
Covxy = 3.75
S2X = 2.50
COV XY
b
2
SX
Regression
a  Y  bX
Ingredients
Mean Y =4.6
Mean X = 3
Covxy = 3.75
S2x = 2.50
3.75
1.50 
2.50
Regression
a  .10  4.6  (1.50)3
3.75
1.50 
2.50
Ingredients
Mean Y =4.6
Mean X = 3
Covxy = 3.75
S2x = 2.50
Regression Equation
Y = a + bx
Equation for predicting smiling from talking
Y = .10+ 1.50(x)
Coefficientsa
Model
1
(Constant)
TALK
Unstandardized
Coefficients
B
Std. Error
1.000E-01
1.567
1.500
.473
a. Dependent Variable: SMILE
Standardi
zed
Coefficien
ts
Beta
.878
t
.064
3.174
Sig.
.953
.050
Regression Equation
Y = .10+ 1.50(x)
How many times would a person likely smile
if they talked 15 times?
Regression Equation
Y = .10+ 1.50(x)
How many times would a person likely smile
if they talked 15 times?
22.6 = .10+ 1.50(15)
Y = 0.1 + (1.5)X
.
12
10
Smile
8
6
4
2
.
.
.
.
0
1
2
3
Talk
4
5
Y = 0.1 + (1.5)X
X = 1; Y = 1.6
.
12
10
Smile
8
6
4
2
..
.
.
.
0
1
2
3
Talk
4
5
Y = 0.1 + (1.5)X
X = 5; Y = 7.60
.
.
12
10
Smile
8
6
4
2
..
.
.
.
0
1
2
3
Talk
4
5
Y = 0.1 + (1.5)X
.
.
12
10
Smile
8
6
4
2
..
.
.
.
0
1
2
3
Talk
4
5
Mean Y = 14.50; Sy = 4.43
Mean X = 6.00; Sx= 2.16
Aggression Happiness
Y
X
Mr. Blond
10
9
Mr. Blue
20
4
Mr. Brown
12
5
Mr. Pink
16
6
Quantify the relationship with a correlation and draw a regression line
that predicts aggression.
COV XY 
∑XY = 326
∑Y = 58
∑X = 24
N=4
X Y

 XY 
N 1
N
24(58)
326 
4
 7.33 
4 1
∑XY = 326
∑Y = 58
∑X = 24
N=4
COV XY
r 
S X SY
• COV = -7.33
• Sy = 4.43
Sx= 2.16
 7.33
 .77 
2.16(4.43)
• COV = -7.33
• Sy = 4.43
Sx= 2.16
Regression
a  Y  bX
Ingredients
Mean Y =14.5
Mean X = 6
Covxy = -7.33
S2X = 4.67
COV XY
b
2
SX
Regression
a  23.92  14.5  (1.57)6
b  1.57 
Ingredients
Mean Y =14.5
Mean X = 6
Covxy = -7.33
S2X = 4.67
 7.33
4.67
Regression Equation
Y = a + bX
Y = 23.92 + (-1.57)X
Y = 23.92 + (-1.57)X
.
22
12
20
Aggression
10
.
18
8
16
6
.
14
4
12
2
.
10
0
1
2
3
4
5
6
Happiness
7
8
9
10
Y = 23.92 + (-1.57)X
22
12
20
Aggression
10
.
.
.
18
8
16
6
.
14
4
12
2
.
10
0
1
2
3
4
5
6
Happiness
7
8
9
10
Y = 23.92 + (-1.57)X
22
12
20
Aggression
10
.
.
.
18
8
16
6
.
14
4
12
2
..
10
0
1
2
3
4
5
6
Happiness
7
8
9
10
Y = 23.92 + (-1.57)X
22
12
20
Aggression
10
.
.
.
18
8
16
6
.
14
4
12
2
..
10
0
1
2
3
4
5
6
Happiness
7
8
9
10
Hypothesis Testing
• Have learned
– How to calculate r as an estimate of
relationship between two variables
– How to calculate b as a measure of the rate
of change of Y as a function of X
• Next determine if these values are
significantly different than 0
Testing b
• The significance test for r and b are
equivalent
• If X and Y are related (r), then it must be
true that Y varies with X (b).
• Important to learn b significance tests for
multiple regression
Steps for testing b value
•
•
•
•
•
•
1) State the hypothesis
2) Find t-critical
3) Calculate b value
4) Calculate t-observed
5) Decision
6) Put answer into words
Practice
• You are interested in if candy consumption
significantly alters a persons depression.
• Create a graph showing the relationship
between candy consumption and
depression
• (note: you must figure out which is X and
which is Y)
Practice
Candy
Depression
Charlie
5
55
Augustus
7
43
Veruca
4
59
Mike
3
108
Violet
4
65
Step 1
• H1: b is not equal to 0
• H0: b is equal to zero
Step 2
• Calculate df = N - 2
– df = 3
• Page 747
– First Column are df
– Look at an alpha of .05 with two-tails
– t crit = 3.182 and -3.182
Step 3
Candy
Depression
Charlie
5
55
Augustus
7
43
Veruca
4
59
Mike
3
108
Violet
4
65
COV = -30.5
N=5
r = -.81
Sy = 24.82
Sx = 1.52
Step 3
Y = 127 + -13.26(X)
b = -13.26
COV = -30.5
r = -.81
Sx = 1.52
Sy = 24.82
N=5
Step 4
• Calculate t-observed
b
t
Sb
b = Slope
Sb = Standard error of slope
Step 4
SY . X
Sb 
S X N 1
Syx = Standard error of estimate
Sx = Standard Deviation of X
Step 4
SY . X  SY
N 1
(1  r )
N 2
2
Sy = Standard Deviation of y
r = correlation between x and y
Note
SY . X 
2
ˆ
 (Y  Y )
N 2
Error = Y - Y
 (Y - Y)2 is minimized
12
10
Error = 1
Smile
8
6
4
2
.
Error = .5
.
Error = 0
.
Error = -.5
.
.
Error = -1
0
1
2
3
Talk
4
5
Step 4
16.80  24.82
5 1
(1  (.81) )
52
Sy = Standard Deviation of y
r = correlation between x and y
2
Step 4
SY . X
Sb 
S X N 1
Syx = Standard error of estimate
Sx = Standard Deviation of X
Step 4
16.80
5.53 
1.52 5  1
Syx = Standard error of estimate
Sx = Standard Deviation of X
Step 4
• Calculate t-observed
b
t
Sb
b = Slope
Sb = Standard error of slope
Step 4
• Calculate t-observed
 13.26
 2.39 
5.53
b = Slope
Sb = Standard error of slope
Step 4
• Note: same value at t-observed for r
 2.39 
 .81 5  2
1  (.81)
2
Step 5
• If tobs falls in the critical region:
– Reject H0, and accept H1
• If tobs does not fall in the critical region:
– Fail to reject H0
t distribution
tcrit = -3.182
tcrit = 3.182
0
t distribution
tcrit = -3.182
tcrit = 3.182
0
-2.39
Step 5
• If tobs falls in the critical region:
– Reject H0, and accept H1
• If tobs does not fall in the critical region:
– Fail to reject H0
Coefficientsa
Model
1
(Constant)
CANDY
Unstandardized
Coefficients
B
Std. Error
127.000
26.555
-13.261
5.537
a. Dependent Variable: DEPRESSI
Standardi
zed
Coefficien
ts
Beta
-.810
t
4.783
-2.395
Sig.
.017
.096
Practice
Practice
• Page 290
• 9.23
9.23
• r = .68
• r = .51
r1 = .829
r1 = .563
• Z = .797
• p = .2119
• Correlations are not different from each other
SPSS Problem
Due March 27th
• 9.31