With the growth of internet service providers, a researcher decides to examine whether there is a correlation between cost of internet service per month

• 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 satisfaction11 618 817 1015 49 95 6

12 319 522 225 10

Correlations

1 .076

. .834

10 10

.076 1

.834 .

10 10

Pearson Correlation

Sig. (2-tailed)

N

Pearson Correlation

Sig. (2-tailed)

N

VAR00001

VAR00002

VAR00001 VAR00002

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.



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

• Situation 2

2)15(.100

210015.5.1

215.1

260015.71.3

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:


• If tobs does not fall in the critical region:– Fail to reject H0

Step 6



• There is not a significant relationship between extraversion and happiness



• 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

31

31

21

21

NN

rrZ

rWhen 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


• Must be independent for this to work

31

31

21

21

NN

rrZ


31

31

)424.(549.

21

NN

Z


3531

3531

)424.(549.

Z


3531

3531

)424.(549.625.


3531

3531

)424.(549.625.

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!

0

2

4

6

8

10

12

1 2 3 4 5

Talk

Smile

.

.. ..

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

0

2

4

6

8

10

12

1 2 3 4 5

Talk

Smile

.

.. ..

0

2

4

6

8

10

12

1 2 3 4 5

Talk

Smile

.

.. ..

Error = 1

Error = -1Error = .5

Error = -.5Error = 0

Error = Y - Y


Finding a and b

• Ingredients

• COVxy

• Sx2

• Mean of Y and X

Regression

XbYa

2X

XY

S

COVb

Smile (Y)

Talk (X)

XY

Jerry 9 5

Elan 2 1

George 5 3

Newman 4 4

Kramer 3 2

SY =2.70 M = 4.6

SX =1.58 SX

2 = 2.50 M = 3

Regression

XbYa

2X

XY

S

COVb

Ingredients

Mean Y =4.6

Mean X = 3

Covxy = 3.75

S2X = 2.50

Regression

XbYa

50.2

75.350.1

Ingredients

Mean Y =4.6

Mean X = 3

Covxy = 3.75

S2x = 2.50

Regression

3)50.1(6.410. a

Ingredients

Mean Y =4.6

Mean X = 3

Covxy = 3.75

S2x = 2.50

50.2

75.350.1

Regression Equation

Y = a + bx

Equation for predicting smiling from talking

Y = .10+ 1.50(x)

1.000E-01 1.567 .064 .953

1.500 .473 .878 3.174 .050

(Constant)

TALK

Model1

B Std. Error

UnstandardizedCoefficients

Beta

Standardized

Coefficients

t Sig.

Coefficientsa

Dependent Variable: SMILEa.

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)

0

2

4

6

8

10

12

1 2 3 4 5

Talk

Smile

Y = 0.1 + (1.5)X

.

.. ..

0

2

4

6

8

10

12

1 2 3 4 5

Talk

Smile

Y = 0.1 + (1.5)XX = 1; Y = 1.6

.

.

.. ..

0

2

4

6

8

10

12

1 2 3 4 5

Talk

Smile

Y = 0.1 + (1.5)XX = 5; Y = 7.60

.

.

.

.. ..

0

2

4

6

8

10

12

1 2 3 4 5

Talk

Smile

Y = 0.1 + (1.5)X

.

.

.

.. ..

Aggression Y

Happiness X

Mr. Blond 10 9

Mr. Blue 20 4

Mr. Brown 12 5

Mr. Pink 16 6

Mean Y = 14.50; Sy = 4.43Mean X = 6.00; Sx= 2.16

Quantify the relationship with a correlation and draw a regression line that predicts aggression.

∑XY = 326

∑Y = 58

∑X = 24

N = 4

1

NN

YXXY

COVXY

∑XY = 326

∑Y = 58

∑X = 24

N = 4

144

)58(24326

33.7

• COV = -7.33

• Sy = 4.43Sx= 2.16

YX

XY

SS

COVr

• COV = -7.33

• Sy = 4.43Sx= 2.16

)43.4(16.2

33.777.

Regression

XbYa

2X

XY

S

COVb

Ingredients

Mean Y =14.5

Mean X = 6

Covxy = -7.33

S2X = 4.67

Regression

6)57.1(5.1492.23 a

67.4

33.757.1

b

Ingredients

Mean Y =14.5

Mean X = 6

Covxy = -7.33

S2X = 4.67

Regression Equation

Y = a + bX

Y = 23.92 + (-1.57)X

0

2

4

6

8

10

12

1 2 3 4 5 6 7 8 9 10

Happiness

Aggression

.

.

..

10

12

14

16

18

20

22

Y = 23.92 + (-1.57)X

0

2

4

6

8

10

12

1 2 3 4 5 6 7 8 9 10

Happiness

Aggression

.

.

..

10

12

14

16

18

20

22

Y = 23.92 + (-1.57)X

.

0

2

4

6

8

10

12

1 2 3 4 5 6 7 8 9 10

Happiness

Aggression

.

.

..

10

12

14

16

18

20

22

Y = 23.92 + (-1.57)X

.

.

0

2

4

6

8

10

12

1 2 3 4 5 6 7 8 9 10

Happiness

Aggression

.

.

..

10

12

14

16

18

20

22

Y = 23.92 + (-1.57)X

.

.

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

COV = -30.5 N = 5

r = -.81

Sx = 1.52

Sy = 24.82

Y = 127 + -13.26(X)

b = -13.26

Step 4

• Calculate t-observed

bS

bt

b = Slope

Sb = Standard error of slope

Step 4

1.

NS

SS

X

XYb

Syx = Standard error of estimate

Sx = Standard Deviation of X

Step 4

2

1)1( 2

.

N

NrSS YXY

Sy = Standard Deviation of y

r = correlation between x and y

Note

2

)ˆ( 2

.

N

YYS XY

0

2

4

6

8

10

12

1 2 3 4 5

Talk

Smile

.

.. ..

Error = 1

Error = -1Error = .5

Error = -.5Error = 0

Error = Y - Y


Step 4

25

15))81.(1(82.2480.16 2

Sy = Standard Deviation of y

r = correlation between x and y

Step 4

1.

NS

SS

X

XYb



Step 4

1552.1

80.1653.5



Step 4


bS

bt

b = Slope


Step 4


53.5

26.1339.2

b = Slope


Step 4

• Note: same value at t-observed for r

2)81.(1

2581.39.2

Step 5

• If tobs falls in the critical region:


• If tobs does not fall in the critical region:

– Fail to reject H0

t distribution

tcrit = 3.182tcrit = -3.182

0

t distribution

tcrit = 3.182tcrit = -3.182

0

-2.39

Step 5

• If tobs falls in the critical region:


• If tIf tobsobs does not fall in the critical region: does not fall in the critical region:

– Fail to reject HFail to reject H00

127.000 26.555 4.783 .017

-13.261 5.537 -.810 -2.395 .096

(Constant)

CANDY

Model1

B Std. Error

UnstandardizedCoefficients

Beta

Standardized

Coefficients

t Sig.

Coefficientsa

Dependent Variable: DEPRESSIa.

Practice

Practice

• Page 288

• 9.18

9.18

• The regression equation for faculty shows that the best estimate of starting salary for faculty is $15,000 (intercept). For every additional year the salary increases on average by $900 (slope). For administrative staff the best estimate of starting salary is $10,000 (slope), for every additional year the salary increases on average by $1500 (slope). They will be equal at 8.33 years of service.

Practice

• Page 290

• 9.23

9.23

• r = .68 r1 = .829• r = .51 r1 = .563

• Z = .797

• p = .2119

• Correlations are not different from each other

SPSS Problem #3Due March 14th

• Page 287– 9.2– 9.3

– 9.10 and create a graph by hand

Documents

With the growth of internet service providers, a researcher decides to examine whether there is a correlation between cost of internet service per month