1 Chapter 9. Section 9-1 and 9-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman C.M. Pascual E LEMENTARY S TATISTICS

1Chapter 9. Section 9-1 and 9-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman

C.M. PascualC.M. Pascual

ELEMENTARY

STATISTICS

Chapter 9 Correlation and Regression


Chapter 9Correlation and Regression

9-1 Overview

9-2 Correlation

9-3 Regression

9-4 Variation and Prediction Intervals

9-5 Multiple Regression

9-6 Modeling


9-1 Overview

Paired Data

is there a relationship

if so, what is the equation

use the equation for prediction


9-2

Correlation


Definition

Correlation

exists between two variables when one of them is related to the other in some way


Assumptions

1. The sample of paired data (x,y) is a random sample.

2. The pairs of (x,y) data have a bivariate normal distribution.


Definition

Scatterplot (or scatter diagram)

is a graph in which the paired (x,y) sample data are plotted with a horizontal x axis and a vertical y axis. Each individual (x,y) pair is plotted as a single point.


Scatter Diagram of Paired Data


Scatter Diagram of Paired Data


Positive Linear Correlation

x x

yy y

x

Figure 9-1 Scatter Plots

(a) Positive (b) Strong positive

(c) Perfect positive


Negative Linear Correlation

x x

yy y

x(d) Negative (e) Strong

negative(f) Perfect

negative



No Linear Correlation

x x

yy

(g) No Correlation (h) Nonlinear Correlation



DefinitionLinear Correlation Coefficient r

measures strength of the linear relationship between paired x and y values in a sample


nxy - (x)(y)

n(x2) - (x)2 n(y2) - (y)2r =


Formula 9-1



nxy - (x)(y)

n(x2) - (x)2 n(y2) - (y)2r =


Formula 9-1


Calculators can compute r

(rho) is the linear correlation coefficient for all paired data in the population.


Notation for the Linear Correlation Coefficient

n = number of pairs of data presented

denotes the addition of the items indicated.

x denotes the sum of all x values.

x2 indicates that each x score should be squared and then those squares added.

(x)2 indicates that the x scores should be added and the total then squared.

xy indicates that each x score should be first multiplied by its

corresponding y score. After obtaining all such products, find their sum.

r represents linear correlation coefficient for a sample

represents linear correlation coefficient for a population


Round to three decimal places so that it can be compared to critical values in Table A-6

Use calculator or computer if possible

Rounding the Linear Correlation Coefficient r


Interpreting the Linear Correlation Coefficient

If the absolute value of r exceeds the value in Table A - 6, conclude that there is a significant linear correlation.

Otherwise, there is not sufficient evidence to support the conclusion of significant linear correlation.


TABLE A-6 Critical Values of the Pearson Correlation Coefficient r

456789

101112131415161718192025303540455060708090

100

n

.999

.959

.917

.875

.834

.798

.765

.735

.708

.684

.661

.641

.623

.606

.590

.575

.561

.505

.463

.430

.402

.378

.361

.330

.305

.286

.269

.256

.950

.878

.811

.754

.707

.666

.632

.602

.576

.553

.532

.514

.497

.482

.468

.456

.444

.396

.361

.335

.312

.294

.279

.254

.236

.220

.207

.196

= .05 = .01


Example 1

• Construct a scatter plot for the given data

Age, x 43 48 56 61 67 70

Pressure, y 128 120 135 143 141 152


Example 1Solution: Draw and label the x and y axes. Plot each point on the graph below


Example 2

• A Statistics professor at a state university wants to see how strong the relationship is between a student’s score on a test and his or her grade point average. The data obtained from the sample follow:Test score, x

98 105 100 100 106 95 116 112

GPA, y 2.1 2.4 3.2 2.7 2.2 2.3 3.8 3.4


Subject Test Score GPA

x y xy x^2 y^2

1 98 2.1 205.8 9604 4.41

2 105 2.4 252 11025 5.76

3 100 3.2 320 10000 10.24

4 100 2.7 270 10000 7.29

5 106 2.2 233.2 11236 4.84

6 95 2.3 218.5 9025 5.29

7 116 3.8 440.8 13456 14.44

8 112 3.4 380.8 12544 11.56

SUM 832 22.1 2321.1 86890 63.83


Example 2

• Solve of SSxy, SSxx, and Ssyy;

• SSxy = ∑xy – [(∑x) (∑y )]/n

• = 2321.1 – [(832)(22.1)]/8 = 22.7

• SSxx = ∑x2 – (∑x)2/n = 86890 – [(832)2]/8

• = 362

• SSyy = ∑y2 – (∑y)2/n = 63.83 – [(22.1)2]/8

• = 2.78


Example 2

• Substitute in the formula and solve for r;

• r = SSxy/(SSxx * Ssyy)0.5

• = 22.7/[(362)(2.78)]0.5 = 0.716

• The correlation coefficient suggests a strong positive relationship between the test score and the grade point average.


Properties of the Linear Correlation Coefficient r

1. -1 r 1

2. Value of r does not change if all values of either variable are converted to a different scale.

3. The r is not affected by the choice of x and y. Interchange x and y and the value of r will not

change.

4. r measures strength of a linear relationship.


Common Errors Involving Correlation

1. Causation: It is wrong to conclude that correlation implies causality.

2. Averages: Averages suppress individual variation and may inflate the correlation coefficient.

3. Linearity: There may be some relationship between x and y even when there is no significant linear correlation.


0

50

100

150

200

250

0 1 2 3 4 5 6 7 8

Dis

tan

ce(f

eet)

Time (seconds)

FIGURE 9-2

Common Errors Involving Correlation

Scatterplot of Distance above Ground and Time for Object Thrown Upward


Formal Hypothesis Test

To determine whether there is a significant linear correlation between two variables

Two methods

Both methods let H0: = (no significant linear correlation)

H1: (significant linear correlation)


Method 1: Test Statistic is t(follows format of earlier chapters)

Test statistic:

1 - r 2

n - 2

r

t =



Test statistic:

1 - r 2

n - 2

r

Critical values:

use Table A-3 with degrees of freedom = n - 2

t =


Figure 9-4



Test statistic: r

Critical values: Refer to Table A-6 (no degrees of freedom)

Method 2: Test Statistic is r(uses fewer calculations)


Test statistic: r

Critical values: Refer to Table A-6 (no degrees of freedom)

Method 2: Test Statistic is r(uses fewer calculations)

Fail to reject = 0

0r = - 0.811 r = 0.811 1

Sample data:r = 0.828

-1

Figure 9-5

Reject= 0

Reject= 0


FIGURE 9-3

Testing for a

Linear Correlation

If H0 is rejected conclude that thereis a significant linear correlation.

If you fail to reject H0, then there isnot sufficient evidence to conclude

that there is linear correlation.

If the absolute value of thetest statistic exceeds the

critical values, reject H0: = 0Otherwise fail to reject H0

The test statistic is

t =1 - r 2

n -2

r

Critical values of t are from Table A-3

with n -2 degrees of freedom

The test statistic is

Critical values of t are from Table A-6

r

Calculate r using

Formula 9-1

Select asignificance

level

Let H0: = 0 H1: 0

Start

METHOD 1 METHOD 2


0.27

2

1.41

3

2.19

3

2.83

6

2.19

4

1.81

2

0.85

1

3.05

5

Data from the Garbage Projectx Plastic (lb)

y Household

Is there a significant linear correlation?


0.27

2

1.41

3

2.19

3

2.83

6

2.19

4

1.81

2

0.85

1

3.05

5

Data from the Garbage Projectx Plastic (lb)

y Household

n = 8 = 0.05 H0: = 0

H1 : 0

Test statistic is r = 0.842



n = 8 = 0.05 H0: = 0

H1 : 0

Test statistic is r = 0.842

Critical values are r = - 0.707 and 0.707(Table A-6 with n = 8 and = 0.05)

TABLE A-6 Critical Values of the Pearson Correlation Coefficient r

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101112131415161718192025303540455060708090

100

n.999.959.917.875.834.798.765.735.708.684.661.641.623.606.590.575.561.505.463.430.402.378.361.330.305.286.269.256

.950

.878

.811

.754

.707

.666

.632

.602

.576

.553

.532

.514

.497

.482

.468

.456

.444

.396

.361

.335

.312

.294

.279

.254

.236

.220

.207

.196

= .05 = .01



0r = - 0.707 r = 0.707 1


- 1


Fail to reject = 0

Reject= 0

Reject= 0


0r = - 0.707 r = 0.707 1


- 1

0.842 > 0.707, That is the test statistic does fall within the critical region.


Fail to reject = 0

Reject= 0

Reject= 0


0r = - 0.707 r = 0.707 1


- 1

0.842 > 0.707, That is the test statistic does fall within the critical region.

Therefore, we REJECT H0: = 0 (no correlation) and concludethere is a significant linear correlation between the weights ofdiscarded plastic and household size.


Fail to reject = 0

Reject= 0

Reject= 0


Justification for r Formula


r = (x -x) (y -y)(n -1) Sx Sy

Justification for r FormulaFormula 9-1 is developed from



r = (x -x) (y -y)(n -1) Sx Sy

(x, y) centroid of sample points

Formula 9-1 is developed from


x = 3

•••Quadrant 3

Quadrant 2 Quadrant 1

Quadrant 4

•

x

y

y = 11(x, y)

x - x = 7- 3 = 4

y - y = 23 - 11 = 12

0 1 2 3 4 5 6 70

4

8

12

16

20

24

r = (x -x) (y -y)(n -1) Sx Sy

(x, y) centroid of sample points

•(7, 23)

FIGURE 9-6

Formula 9-1 is developed from


Documents

1 Chapter 9. Section 9-1 and 9-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman C.M. Pascual E LEMENTARY S TATISTICS