Stats Chap10

8/3/2019 Stats Chap10

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Chapter 10

Correlation and Regression

McGraw-Hill, Bluman, 5th ed., Chapter 10 1


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Chapter 10 OverviewIntroduction

10-1 Correlation

10-2 Regression 10-3 Coefficient of Determination and

Standard Error of the Estimate

Bluman, Chapter 10 2


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Chapter 10 Objectives1. Draw a scatter plot for a set of ordered pairs.

2. Compute the correlation coefficient.

3. Test the hypothesis H0: = 0.

4. Compute the equation of the regression line.

5. Compute the coefficient of determination.

6. Compute the standard error of the estimate.

7. Find a prediction interval.



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Introduction

In addition to hypothesis testing andconfidence intervals, inferential statisticsinvolves determining whether a

relationship between two or morenumerical or quantitative variables exists.



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Introduction

CorrelationCorrelation is a statistical method usedto determine whether a linear relationshipbetween variables exists.

RegressionRegression is a statistical method usedto describe the nature of the relationship

between variablesthat is, positive ornegative, linear or nonlinear.



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Introduction

The purpose of this chapter is to answerthese questions statistically:

1. Are two or more variables related?

2. If so, what is the strength of therelationship?

3. What type of relationship exists?4. What kind of predictions can be

made from the relationship?



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Introduction

1. Are two or more variables related?

2. If so, what is the strength of the

relationship?

To answer these two questions, statisticians usethe correlation coefficientcorrelation coefficient, a numericalmeasure to determine whether two or more

variables are related and to determine thestrength of the relationship between or amongthe variables.



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Introduction

3. What type of relationship exists?

There are two types of relationships: simple andmultiple.

In a simple relationship, there are two variables: anindependent variableindependent variable (predictor variable) and adependent variabledependent variable (response variable).

In a multiple relationship, there are two or moreindependent variables that are used to predict onedependent variable.



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Introduction

4. What kind of predictions can be made

from the relationship?

Predictions are made in all areas and daily. Examplesinclude weather forecasting, stock market analyses,sales predictions, crop predictions, gasoline pricepredictions, and sports predictions. Some predictionsare more accurate than others, due to the strength of

the relationship. That is, the stronger the relationship isbetween variables, the more accurate the prediction is.



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10.1 Scatter Plots and Correlation

A scatter plotscatter plot is a graph of the orderedpairs (x, y) of numbers consisting of theindependent variablexand the

dependent variable y.



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Chapter 10Correlation and Regression

Section 10-1Example 10-1

Page #536



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Example 10-1: Car Rental Companies

Construct a scatter plot for the data shown for car rentalcompanies in the United States for a recent year.

Step 1: Draw and label thexand yaxes.

Step 2: Plot each point on the graph.



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Positive Relationship


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Correlation

The correlation coefficientcorrelation coefficient computed fromthe sample data measures the strength anddirection of a linear relationship between two

variables. There are several types of correlation

coefficients. The one explained in this sectionis called the Pearson product momentPearson product moment

correlation coefficient (PPMC)correlation coefficient (PPMC). The symbol for the sample correlation

coefficient is r. The symbol for the populationcorrelation coefficient is V.



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Correlation

The range of the correlation coefficient is from1 to 1.

If there is a strong positive linearstrong positive linear

relationshiprelationship between the variables, the valueofrwill be close to 1.

If there is a strong negative linearstrong negative linearrelationshiprelationship between the variables, the value

ofr will be close to 1.



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Correlation



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Correlation Coefficient

The formula for the correlation coefficient is

where n is the number of data pairs.

Rounding Rule: Round to three decimal places.


2 2

2 2

!

- -

n xy x yr

n x x n y y


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Section 10-1Example 10-2

Page #540



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Compute the correlation coefficient for the data inExample 101.


CompanyCarsx

(in 10,000s)Income y

(in billions) xy x 2 y2

A

BCDEF

63.0

29.020.819.113.48.5

7.0

3.92.12.81.41.5

441.00

113.1043.6853.4818.762.75

3969.00

841.00432.64364.81179.5672.25

49.00

15.214.417.841.962.25

x =

153.8

y =

18.7

xy =

682.77

x2 =

5859.26

y2 =

80.67


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Compute the correlation coefficient for the data inExample 101.


x = 153.8, y = 18.7, xy = 682.77, x2 = 5859.26,

y2 = 80.67, n = 6

2 22 2

! - -

n xy x yr

n x x n y y

2 2

6 682.77 153.8 18.7

6 5859.26 153.8 6 80.67 18.7

!

- -

r

0.982 (strong positive relationship)!r


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Hypothesis Testing

In hypothesis testing, one of the following istrue:

H0: V ! 0 This null hypothesis means that

there is no correlation between thexand yvariables in the population.

H1: V { 0 This alternative hypothesis meansthat there is a significant correlationbetween the variables in thepopulation.



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tTest for the Correlation Coefficient


2

2

1

with degrees of freedom equal to 2.

!

nt r

r

n


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Section 10-1

Example 10-3

Page #544



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Test the significance of the correlation coefficient found inExample 102. Use = 0.05 and r= 0.982.

Step 1: State the hypotheses.

H0: = 0 and H1: { 0

Step 2: Find the critical value.

Since = 0.05 and there are 6 2 = 4 degrees of

freedom, the critical values obtained from Table Fare 2.776.



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Step 3: Compute the test value.

Step 4:

Make the de

cision

.

Reject the null hypothesis.

Step 5: Summarize the results.

There is a significant relationship between thenumber of cars a rental agency owns and itsannual income.



2

2

1

!

nt r

r 2

6 20.982 10.4

1 0.982

! !


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Section 10-1

Example 10-4

Page #545



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Using Table I, test the significance of the correlationcoefficient r = 0.067, from Example 103, at = 0.01.

Step 1: State the hypotheses.

H0: = 0 and H1: { 0

There are 9 2 = 7 degrees of freedom. The value inTable I when = 0.01 is 0.798.

For a significant relationship, rmust be greater than 0.798or less than -0.798. Since r= 0.067, do not reject the null.

Hence, there is not enough evidence to say that there is asignificant linear relationship between the variables.



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Possible Relationships Between

VariablesWhen the null hypothesis has been rejected for a specifica value, any of the following five possibilities can exist.

1. There is a direct cause-and-effectrelationship

between the variables. That is,xcauses y.2. There is a reverse cause-and-effectrelationship

between the variables. That is, ycausesx.

3. The relationship between the variables may be

caused by a third variable.4. There may be a complexity of interrelationships

among many variables.

5. The relationship may be coincidental.



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Variables1. There is a reverse cause-and-effectrelationship


For example,

water causes plants to grow

poison causes death

heat causes ice to melt



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Variables2. There is a reverse cause-and-effectrelationship


For example,

Suppose a researcher believes excessive coffeeconsumption causes nervousness, but theresearcher fails to consider that the reverse

situation may occur. That is, it may be that anextremely nervous person craves coffee to calmhis or her nerves.



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Variables3. The relationship between the variables may be

caused by a third variable.

For example,

If a statistician correlated the number of deathsdue to drowning and the number of cans of softdrink consumed daily during the summer, he or

she would probably find a significant relationship.However, the soft drink is not necessarilyresponsible for the deaths, since both variablesmay be related to heat and humidity.



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Variables4. There may be a complexity of interrelationships

among many variables.

For example,

A researcher may find a significant relationshipbetween students high school grades and collegegrades. But there probably are many other

variables involved, such as IQ, hours of study,influence of parents, motivation, age, andinstructors.



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Variables5. The relationship may be coincidental.

For example,

A researcher may be able to find a significantrelationship between the increase in the number ofpeople who are exercising and the increase in thenumber of people who are committing crimes. But

common sense dictates that any relationshipbetween these two values must be due tocoincidence.



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10.2 Regression

If the value of the correlation coefficient issignificant, the next step is to determinethe equation of the regression lineregression line

which is the datas line of best fit.



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Regression


Best fitBest fit means that the sum of thesquares of the vertical distance fromeach point to the line is at a minimum.


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Regression Line


d! y a bx

2

22

22

where

= intercept

= the slope of the line.

!

!

d

y x x xya

n x x

n xy x yb

n x x

a y

b


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Section 10-2

Example 10-5

Page #553



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Find the equation of the regression line for the data inExample 102, and graph the line on the scatter plot.


x = 153.8, y = 18.7, xy = 682.77, x2 = 5859.26,

y2 = 80.67, n = 6

2

22

!

y x x xya

n x x

22

!

n xy x yb

n x x

2

18.7 5859.26 153.8 682.77

6 5859.26 153.8

!

0.396!

2

6 682.77 153.8 18.7

6 5859.26 153.8

!

0.106!

0.396 0.106d d! p ! y a bx y x


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Find two points to sketch the graph of the regression line.

Use anyxvalues between 10 and 60. For example, letxequal 15 and 40. Substitute in the equation and find thecorresponding yvalue.

Plot (15,1.986) and (40,4.636), and sketch the resultingline.



0.396 0.106

0.396 0.106 15

1.986

d!

!

!

y x

0.396 0.106

0.396 0.106 40

4.636

d!

!

!

y x


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Find the equation of the regression line for the data inExample 102, and graph the line on the scatter plot.


0.396 0.106d! y x

15, 1.986

40, 4.636


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Section 10-2

Example 10-6

Page #555



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Use the equation of the regression line to predict theincome of a car rental agency that has 200,000automobiles.

x = 20 corresponds to 200,000 automobiles.

Hence, when a rental agency has 200,000 automobiles, itsrevenue will be approximately $2.516 billion.



0.396 0.106

0.396 0.106 20

2.516

d!

!

!

y x


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Regression

The magnitude of the change in one variablewhen the other variable changes exactly 1 unitis called a marginal changemarginal change. The value ofslope b of the regression line equationrepresents the marginal change.

For valid predictions, the value of thecorrelation coefficient must be significant.

When ris not significantly different from 0, thebest predictor ofyis the mean of the datavalues ofy.



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Assumptions for Valid Predictions

1. For any specific value of the independentvariablex, the value of the dependent variableymust be normally distributed about theregression line. See Figure 1016(a).

2. The standard deviation of each of thedependent variables must be the same foreach value of the independent variable. See

Figure 1016(b).



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Extrapolations (Future Predictions)

ExtrapolationExtrapolation, or making predictions beyondthe bounds of the data, must be interpretedcautiously.

Remember that when predictions are made,they are based on present conditions or on thepremise that present trends will continue. Thisassumption may or may not prove true in the

future.



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Procedure Table

Step 1: Make a table with subject,x, y,xy,x2,and y2 columns.

Step 2: Find the values ofxy,x2, and y2. Place

them in the appropriate columns and sumeach column.

Step 3: Substitute in the formula to find the valueofr.

Step 4: When ris significant, substitute in theformulas to find the values ofa and b forthe regression line equation yd = a +bx.



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10.3 Coefficient of Determination

and Standard Error of the Estimate The total variationtotal variation is the

sum of the squares of the vertical

distances each point is from the mean. The total variation can be divided into two

parts: that which is attributed to the

relationship ofxand y, and that which isdue to chance.


2

y y


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Variation

The variation obtained from therelationship (i.e., from the predicted y'values) is and is called the

ex

plained variationex

plained variation. Variation due to chance, found by

, is called the unexplainedunexplainedvariationvariation. This variation cannot beattributed to the relationships.


2

d y y

2

d y y


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Variation



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Coefficient of Determiation

The coefficient of determinationcoefficient of determination is theratio of the explained variation to the totalvariation.

The symbol for the coefficient ofdetermination is r2.

Another way to arrive at the value forr2

is to square the correlation coefficient.


2 explained variation

total variation

!r


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Coefficient of Nondetermiation

The coefficient ofcoefficient of nondeterminationnondetermination isa measure of the unexplained variation.

The formula for the coefficient of

determination is 1.00 r2

.



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Standard Error of the Estimate

The standard error of estimatestandard error of estimate,denoted by sest is the standard deviationof the observed yvalues about the

predicted y'values. The formula for thestandard error of estimate is:


2

2

d

!

est

y ys

n


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Section 10-3

Example 10-7

Page #569



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A researcher collects the following data and determinesthat there is a significant relationship between the age of acopy machine and its monthly maintenance cost. Theregression equation is yd = 55.57 + 8.13x. Find thestandard error of the estimate.

Example 10-7: Copy Machine Costs



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MachineAgex(years)

Monthlycost, y yd y yd (y yd)2

ABCD

EF

1234

46

62787090

93103

63.7071.8379.9688.09

88.09104.35

-1.706.17

-9.961.91

4.91-1.35

2.8938.068999.20163.6481

24.10811.8225

169.7392

55.57 8.13

55.57 8.13 1 63.70

55.57 8.13 2 71.8355.57 8.13 3 79.96

55.57 8.13 4 88.09

55.57 8.13 6 104.35

d!

d! !

d

! !d! !

d! !

d! !

y x

y

yy

y

y

2

2

d!

est

y ys

n

169.73926.51

4! !

ests


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Section 10-3

Example 10-8

Page #570



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2

2

!

est

y a y b xys

n


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496 1778

MachineAgex(years)

Monthlycost, y xy y 2

ABCD

EF

1234

46

62787090

93103

62156210360

372618

3,8446,0844,9008,100

8,64910,609

2

2

!

est

y a y b xys

n

42,186

42,186 55.57 496 8.13 17786.48

4

! !

ests


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Formula for the Prediction Interval

about a Value yd


2

2 22

2

2 22

11

11

with d.f. = - 2

d

d

est

est

n x Xy t s y

n n x x

n x Xy t s

n n x x

n

E

E


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Section 10-3

Example 10-9Page #571



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For the data in Example 107, find the 95% predictioninterval for the monthly maintenance cost of a machinethat is 3 years old.

Step 1: Find

Step 2: Find yd forx = 3.

Step 3: Find sest.

(as shown in Example 10-13)



2, , and . x x X 2 2020 82 3.3

6! ! ! ! x x X

55.57 8.13 3 79.96d! !y

6.48!est

s


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Step 4: Substitute in the formula and solve.



2

2 22

2

2 22

11

11

d

d

est

est

n x Xy t s y

n n x x

n x Xy t s

n n x x

E

E

2

2

2

2

6 3 3.3179.96 2.776 6.48 1

66 82 20

6 3 3.3179.96 2.776 6.48 1

6 6 82 20

y


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Step 4: Substitute in the formula and solve.


2

2

2

2

6 3 3.3179.96 2.776 6.48 1

6 6 82 20

6 3 3.31

79.96 2.776 6.48 1 6 6 82 20

y

79.96 19.43 79.96 19.43

60.53 99.39

y

y

Hence, you can be 95% confident that the interval60.53 < y < 99.39 contains the actual value ofy.

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