Chapter 9wellsmat.startlogic.com/sitebuildercontent/sitebuilderfiles/...The results do not support any conclusion about the removal of trees affecting the risk of fires, because none

Chapter 9

Correlation and Regression

Lesson 9-1/9-2, Part 1

Correlation

Registered Florida Pleasure Crafts and Watercraft

Related Manatee Deaths

020

4060

80

100

1991 1993 1995 1997 1999

Year

Boats in Ten Thousands Manatee Deaths

9-1 Overview

This chapter introduces methods for making

inferences based on sample data that come

in pairs or bivariate data.

Is there a relationship between the data?

If so, what is the equation?

Use that equation for prediction.

Correlation

A correlation exists between two variables

when one of them is related to the other in

some way.

Scatter Diagram

A 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.

Table 9-1: Registered Florida Craft (in tens of thousands) and

Watercraft-Related Manatee Deaths

Year 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000

x: Boats 68 68 67 70 71 73 76 81 83 84

y: Manatee 53 38 35 49 42 60 54 67 82 78

Scatter Diagram

Scatter Plots – TI83

Scatter Plot

Registered Boats (tens of thousands)

Ma

nate

e D

ea

ths f

rom

Boats

Dangers of Scatter Diagrams

The horizontal or vertical scale of a scatter diagram can be

set so that the scatter diagram misleads a reader.

Dangers of Scatter Diagrams

Because of this danger you must use numerical summaries

of paired data in conjunction with graphs in order to determine

the type of relationship.

Positive Linear Correlations

Negative Linear Correlations

No Linear Correlations

Linear Correlation Coefficient

The linear correlation coefficient (r )

measures the direction and strength of the

linear relationship between paired x and y

values in a sample.

Assumptions

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

sample.

The pairs of (x, y) data have a bivariate

normal distribution.

Notation for the Linear

Correlation Coefficient n number of pairs of data present

x the sum of all x-values

2x the x-value should be squared and then those

squares added.

xy

linear correlation coefficient for a sample

2

x the sum of all x-values and the total then squared

r

the x-value should be multiplied by its corresponding

y-value

ρ linear correlation coefficient for a population

Formula for the Linear

Correlation Coefficient

2 22 2

n xy x yr

n x x n y y

Figure 9-7

Justification for r formula

1 x y

x x y yr

n s s

,x y centriod

Interpreting the Linear

Correlation Coefficient1) Linear correlation coefficient r is always between 1 and

– 1.

2) The closer r is to 1, the stronger the evidence of positive

association between the two variables.

3) The closer r is to a – 1, the stronger the evidence of

negative relationship between two variables.

4) If r is closer to 0, does not rule out any strong relationship

between x and y, there could still be a strong relationship

but one that is not linear.

5) The units of x and y plays no role in the interpretation of r.

6) Correlation is strongly effected by outlying observations.

Example – Page 511, #6

Use a scatter plot and the linear correlation r to determine

whether there is a correlation between the two variables.

1 2 2 5 6

2 5 4 15 15

x

y

2 22 2

n xy x yr

n x x n y y

x y x y x2 y2

1 2

2 5

…… ……


2 22 2

n xy x yr

n x x n y y

x y xy x2 y2

1 2 2 1 4

2 5 10 4 25

2 4 8 4 16

5 15 75 25 225

6 15 90 36 225

x = 16 y = 41 xy = 185 x = 70 x = 495


2 22 2

n xy x yr

n x x n y y

x = 16 y = 41 xy = 185 x2 = 70 y2 = 495

2 2

5(185) 16 410.985

5 70 16 5 495 41r


x = 16 y = 41 xy = 185 x2 = 70 y2 = 495



Using Table A-6, n = 5 and assuming = 0.05

Critical Value = 0.878; therefore r = 0.984 indicates

significant (positive) linear correlation.

Lesson 9-1/9-2, Part 2

Correlation

Interpreting the Linear

Correlation Coefficient The absolute value of the computed value

of r exceeds the value in Table A-6,

conclude that there is significant linear

correlation.

Otherwise there is not significant evidence

to support the conclusion of significant

linear correlation.

– 1 r 1

Value of r does not change if all values of

either variable are converted to a different

scale.

The r is not affected by the choice of x and y.

Interchange x and y and the value r will not

change.

r measures the strength of a linear

relationship.

Properties of the Linear

Correlation Coefficient

Interpreting r: Explain Variation

The value of r2 is the proportion of the

variation in y that is explained by the linear

relationship between x and y.

r2 is the square of the linear correlation

coefficient r.



Year 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000

x: Boats 68 68 67 70 71 73 76 81 83 84

y: Manatee 53 38 35 49 42 60 54 67 82 78

Example: Boats and Manatees

A. Find the value of the linear correlation coefficient r.

B. What proportion of the variation of the manatee deaths can

be explained the variation in the number of boat

registrations?


with r = 0.922, we get r2 = 0.850.

We can conclude that 0.850 (or about 85%) of the

variation in manatee deaths can be explained by the

linear relationship between the number of boat

registrations and the number of manatee deaths from

boats. This implies that 15% of the variation of manatee

deaths cannot be explained by the number of boat

registrations.

Hypothesis Test – Method 1

We wish to determine whether there is a

significant linear correlation between two

variables.

Let Ho: = 0 No significant linear

correlation

Let H1: ≠ 0 Significant linear correlation

Let H1: > 0 Significant positive linear

correlation

Let H1: < 0 Significant negative linear

correlation

Test Statistic

21

2

rt

r

n

Critical Values: Use table A-3 with df = n - 2



Year 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000

x: Boats 68 68 67 70 71 73 76 81 83 84

y: Manatee 53 38 35 49 42 60 54 67 82 78

Example: Boats and ManateesTest the claim that there is a linear correlation between

the number of registered boats and the number of manatee

deaths from boats.

Ho: = 0

H1: ≠ 0


21

2

rt

r

n

2

0.9226.71

1 0.922

10 2

p-value = 1.51E – 4

Example: Boats and Manatees6.71

0.0001

0.05

2 8

t

p value

df n

0.025 0.025

– 2.306 2.306

Critical Values


There is sufficient evidence to reject Ho, since

p-value = 0.0001 < = 0.05 and conclude that

there is significant correlation between the

number of registered boats and the number of

manatee deaths from boats.

Hypothesis Test – Method 2

We wish to determine whether there is a

significant linear correlation between two

variables.

Test statistic is r

Critical values refer to Table A-6 with no degrees

of freedom.


Using data collected from the FBI and the Bureau of Alcohol,

Tobacco, and Firearms, the number of registered automatic

Weapons and the murder rate (in murders per 100,000

people) was obtained for each of eight randomly selected

states. Statdisk was used to find the value of the linear

correlation Coefficient is r = 0.885.

a) Is there a significant linear correlation between the

number of registered automatic weapons and the

murder rate?

8

0.885

n

r

Use table A-6 assume = 0.05


Since the absolute value of 0.885 is greater

than 0.707 there is significant positive

correlation

critical values = 0.707


0.885r 2 2(0.885) 0.783r

b) What proportion of the variation in the murder rate can be

explained by the linear relationship between the murder

rate and the number of registered automatic weapons?

The proportion of the variation in murder rate

that can be explained in terms of variation in

registered automatic weapons is 78.3%.


Listed below are the numbers of fires (in thousands) and the

acres that were burned (in millions) in 11 western states in

year of the last decade (based on the data from USA Today.

Is there a correlation?

73 69 58 48 84 62 57 45 70 63 48

6.2 4.2 1.9 2.7 5.0 1.6 3.0 1.6 1.5 2.0 3.7

Fires

Acres

Burned


LinRegTTest

1

: 0

: 0

0.05

oH ρ

H ρ

α

2

0.1031

0.5173

26.8%

p value

r

r

There is sufficient evidence to fail to reject Ho since

the p-value > α and are unable to conclude that there

is significant linear correlation between the number

of fires and the numbers of acres burned.


The results do not support any conclusion about

the removal of trees affecting the risk of fires,

because none of the variables addresses the

removal/density of trees.

The data were listed under a headline of “Loggers seize

on fires to argue for more cutting.” Do the data support the

argument that as loggers remove more trees, the risk of fire

decreases because the forests are less dense?

Common Errors Involving

Correlation Causation: It is wrong to conclude that

correlation implies causality.

Averages: Averages suppress individual

variation and may inflate the correlation

coefficient.

Linearity: There may be some relationship

between x and y even when there is no

significant linear correlation.


Describe the error in the stated conclusion.

Given: There is significant linear correlation between

personal income and years of education.

Conclusion: More education causes a person‟s income to

rise.

The presence of a significant linear correlation does

not necessarily mean that one of the variables

causes the other. Correlation does not imply

causality.


Describe the error in the stated conclusion.

Given: There is significant linear correlation between

state average tax burdens and state average incomes.

Conclusion: There is significant linear correlation between

Individual tax burdens and individual incomes rise.

Averages tend to suppress variation among

individuals, so a correlation among averages does

not necessary mean that there is correlation among

individuals.

Lesson 9-3, Part 1

Regression

Objective

In lesson 9-2 we use a scatter diagram and

linear correlation coefficient to indicate that a

linear relation exists between two variables.

In lesson 9-3 we are going to find a linear

equation that describes the relation between

the two variables.

Regression Equation

The regression equation expresses a

relationship between x (called the

independent variable, predictor variable or

explanatory variable), and y (called the

dependent variable or response variable)

The typical equation of a straight line

is expressed in form , where bo is

the y-intercept and b1 is the slope.

y mx b

1ˆ

oy b b x

Assumptions

We are investigating only linear relationships

For each x-value, y is a random variable having a

normal (bell-shaped) distribution.

All of these y distributions have the same

variance.

For a given value of x, the distribution of y-values

has a mean that lies on the regression line.

Results are not seriously affected if departures

from normal distributions and equal variances are

not too extreme.

Regression Line

Regression Equation

Given a collection of paired data, the regression

equation

Algebraically describes the relationship between two

variables.

Regression Line

The graph of the regression equation is called the

regression (or line of best fit, or least square line).

1ˆ

oy b b x

Notation for Regression Equation

Population

Parameter

Sample

Statistic

TI

y-intercept of

Regression

equationo bo a

Slope of

regression

equation1 b1 b

Equation of the

regression line y = o + 1x y = a + bx1

ˆoy b b x

Formulas for bo and b1

Formula 9-2 (Slope)

Formula 9-3 (y-intercept)

1 22

n xy x yb

n x x

1ob y b x



Year 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000

x: Boats 68 68 67 70 71 73 76 81 83 84

y: Manatee 53 38 35 49 42 60 54 67 82 78

Example: Boats and Manatee

Find the regression equation

Enter the values in L1 and L2 and use LinRegTTest


Find the equation of the regression line.

y a bx

ˆ 113 2.27y x


Regression Line


Find the equation of the regression line. Use the given data.

1 2 2 5 6

2 5 4 15 15

x

y

Enter the data into L1 and L2 and use LinRegTTest


ˆ 0.957 2.86y x

Using the Regression Equation

for Predictions

In predicting a value of y based on some given

value of x…

If there is not a significant linear correlation, the

best predicted y-value is

If there is significant linear correlation, the best

predicated y-value is found by substituting the

x-value into the regression equation.

( ).y

Guidelines for Using TheRegression Equation

1. If there is no linear correlation, don‟t use the regression

to make predictions.

2. When using the regression equation for predictions,

stay within the scope of the available sample data.

3. A regression equation on old data is not necessary

valid now.

4. Don‟t make predictions about a population that is

different from the population from which the sample data

were drawn.

Example – page 527, #2

In each of the following cases, find the best predicted

value of y given that x = 2.00. The given statistics are

summarized from paired sample data.

0.123 8.00 30,r y n a. and the equation of the

regression line is ˆ 7.00 2.00y x

Step 1 – Check for linear correlation

Since the absolute value of 0.123 is less than 0.361.

There is no linear correlation.

2.00ˆ 8.00y y

Example – page 527, #2

0.567 8.00 30,r y n b. and the equation of the

regression line is ˆ 7.00 2.00y x

Step 1 – Check for linear correlation

Since the absolute value of 0.567 is greater than 0.361.

There is linear correlation.

2.00

ˆ 7.00 2.00

ˆ 7.00 2.00(2.00)

y x

y

2.00ˆ 3.00y

Example – Page 528, #8Find the best predicted value for the number of acres burned

given that there was 80 fires.

73 69 58 48 84 62 57 45 70 63 48

6.2 4.2 1.9 2.7 5.0 1.6 3.0 1.6 1.5 2.0 3.7

Fires

Acres

Burned

Enter data into L1 and L2 and use LinRegTTest to check

for linear correlation.

Lesson 9-3, Part 2

Regression


P-value > α there is no significant linear correlation.

Use for predication when

ˆ 1.13 0.0677y x Regression Equation:

y 80.x


80ˆ 3.04y y

Go to 2nd STAT

Interpreting the Regression Equation

Marginal change is the amount that a variable

changes when the other variable changes by

exactly one unit.

Outlier is a point lying far away from the other

data points.

Influential point strongly effects the graph of

the regression line.


Refer to the sample data listed in Table 9-1. If we include

another pair of values consisting of x = 120 (for 1,200,000

boats) and y = 10 (manatee deaths from boats), is this new

point an outlier? Is it an influential point?



Year 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000

x: Boats 68 68 67 70 71 73 76 81 83 84

y: Manatee 53 38 35 49 42 60 54 67 82 78


Yes; the point is an outlier,

since it is far from the other

data points – 120 is far from

other boat values, and 10 is

far from the other manatee

death values.

Yes; the point is an influential one, since it will

dramatically alter the regression line – the original regression

line predicts and so the new

regression line will have to change considerably to come close

to (120, 10).

ˆ 113 2.27(120) 159.4,y

Residuals and Least-Square Property

Residual

For a sample paired (x, y) data, the difference

between the observed sample y-value and y-value

that is predicted by using the regression equation.

Least-Square Property

A straight line satisfies this property if the sum of the

squares of the line residuals is the smallest sum

possible.

ˆy y

Example: Residuals

1 2 4 5

4 24 8 32

x

y

ˆ 5 4y x

1 2 4 5

ˆ 9 13 21 25

x

y

The Method Least-Square

The line that “best” describes the relation

between two variables is the one that makes

the residuals “as small as possible.”

The least-square regression line minimizes

the square of the vertical distance between

observed values and predicated values.

Minimize (residuals)2

Lesson 9-4, Part 1

Variation and

Prediction Intervals

Different types of Variation

We consider different types of variation that

can be used for two major applications:

To determine the proportion of the variation in y

that can be explained by the linear relationship

between x and y.

To construct interval estimates of predicted y-

values such intervals are called prediction

intervals.

Total and Explained Deviations

Total Deviation from the mean of the particular

point (x, y) is the vertical distance which

is the distance between the point (x, y) and the

horizontal line passing through the sample

mean .

Explained Deviation is the vertical distance

which is the distance between the

predicted y-value and the horizontal line

passing through the sample mean .

y y

y

y y

y

Total and Explained Deviations

Unexplained Deviation is the vertical

distance which is the vertical distance

between the point (x, y) and the regression

line.

The distance is also called residual, as

defined by Lesson 9-3.

ˆ,y y

ˆ,y y

Deviations

Total Deviation

(total deviation) = (explained deviation) + (unexplained deviation)

y y y y ˆy y

2

y y 2

y y 2

ˆy y

Formula 9-4

Coefficient of Determination

The coefficient of determination measures

the percentage of total variation in the y-value

that is explained by the regression line.

2

2

2

y yr

y y

explained variation

total variation


Use the value of the linear correlation coefficient r to find the

coefficient of determination and the percentage of the total

variation that can be explained by the linear relationship

between the two variables.

0.6r

2 2( 0.6) 0.36r

The portion of the total variation explained by the regression

line is 36%

Standard Error of Estimate

The standard error of estimate is a measure

of the differences (or distances) between the

observed sample y-values and the predicted

y-values that are obtained using the

regression equation.

2 2

1ˆ

2 2

o

e

y y y b y b xys

n n

y


Fourteen different second-year medical students took

blood pressure measurements of the same patient and

the results are listed below.

138 130 135 140 120 125 120 130 130 144 143 140 130 150

82 91 100 100 80 90 80 80 80 98 105 85 70 100

Systolic

Diastolic

Enter the data into L1 and L2 and use LinRegTtest


A. Find the explained variation .

Step 1 – Find the regression equation

ˆ 14.38 0.769y x

2

y y


Step 2 – Use L3 to find y


Step 3 – Use L4 to find y

Use 2nd Stat


Step 4 – Use L4 to find 2

y y


Step 5 – Find

Explained Variation 2

y y

2

y y

Use 2nd Stat

628.58


B. Find the unexplained variation . 2

ˆy y

Step 1 – Use L5 to find 2ˆ( )y y


Step 3 – Find

Unexplained Variation 2

ˆy y

2

ˆy y

Use 2nd Stat

824.27


C. Find the total variation

TV EV UV

628.58 824.27

1452.85

2

y y


D. Find the coefficient of determination.

2 0.4328r


E. Find the standard error of estimate

8.2878e

s

Lesson 9-4, Part 2

Variation and



Prediction Intervals are intervals constructed

about the predicted value of y that are used

to measure the accuracy of a single predicted

value.

Predication Interval

for an Individual y

ˆ ˆy E y y E

2

222

( )11 o

α e

n x xE t s

n n x x

xo = the given value of x

t/2 = has n – 2 degree of freedom in table A-3


Refer to problem #10 and assume necessary conditions

of normality and variance are met.

138 130 135 140 120 125 120 130 130 144 143 140 130 150

82 91 100 100 80 90 80 80 80 98 105 85 70 100

Systolic

Diastolic

A. Find the predicted diastolic reading given that the

systolic reading is 120.

Enter the data into L1 and L2 and use LinRegTtest


Step 1 – Find the regression equation

ˆ 14.38 0.769y x

14.38 0.769(120)

77.9 78


B. Find a 95% predication interval estimate of the diastolic

reading given that the systolic reading is 120.

Step 1 – Identify the calculations needed for the

margin of error formula

n 14

x 18752x 252179

x 133.929

2

2n x x 14881


n 14

x 18752x 252179

x 133.929

2

2 14881n x x

12,0.0252.179t 8.2878

es

2

22

2

( )11 o

α e

n x xE t s

n n x x

21 14(120 133.929)

2.179 8.2878 1 20.2214 14881


Step 2 – Use and margin of errory

ˆ ˆy E y y E

78 20.22 78 20.22y

12057.78 98.22y

When x = 120, (78) we are 95% certain that the

corresponding y-value is between 57.78 and 98.22

Interpreting a Computer DisplayMinitabThe regression equation isNicotine = 0.154 + 0.0651 Tar

Predictor Coef SE Coef T PConstant 0.15403 0.04635 3.32 0.003Tar 0.065052 0.003585 18.15 0.000

S = 0.08785 R-Sq = 92.4% R-Sq(adj) = 92.1%

Predicted values for New ObservationsNew Obs Fit SE Fit 95.0% CI 95.0% PI1 1.2599 0.0240 (1.2107, 1.3091) (1.0731, 1.4468)

Example – Page 539, #6 and 86. Identify Total Variation – What percent of the total

variation in nicotine can be explained by the linear

relationship between tar and nicotine?

92.4%

8. Finding the Prediction Interval – For given tar

amount 17mg. Identify the 95% confidence interval

estimate of the amount of nicotine, and write a statement

interpreting that interval.

1.1 < y < 1.4; we have 95% confidence that the limits

1.1 and 1.4 contain the true amount of nicotine.

Lesson 9-6

Modeling

Mathematical Modeling

A mathematical model is a mathematical

function that „fits‟ or describes real-world data

y a bx

TI Generic Models2y ax bx c lny a b x

xy abby ax 1 bx

cy

ae

TI Key Strokes

Use the Stat key

Development of a

Good Mathematics Model Look for a Pattern in the Graph

Examine the graph of the plotted points and compare

the basic pattern to the known generic graphs.

Find and Compare Values of R2

Select functions that result in larger values of R2,

because such larger values correspond to functions

that better fit the observed points.

Think

Use common sense. Don‟t use a model that lead to

predicted values known to be totally unrealistic.


Construct a scatterplot and identify the mathematical

model that best fits the given data. Assume that the model

is to be used only for the scope of the given data, and

considered only linear, quadratic, logarithmic, exponential

and power models.

1 2 3 4 5 6

3 8 13 18 23 28

x

y


GRAPH Zoom #9:Zoomstat


Step 3 – Decide which mathematical model

Linear: ˆ 5 2y x


Construct a scatterplot and identify the mathematical

model that best fits the given data. Assume that the model

is to be used only for the scope of the given data, and

considered only linear, quadratic, logarithmic, exponential

and power models.

1 2 3 4 5 6

2 2.828 3.464 4 4.472 4.899

x

y


GRAPH Zoom #9:Zoomstat


Step 3 – Decide which mathematical model

Power: 12ˆ 2y x


Refer to the annual high values of the Dow-Jones Industrial

Average that are listed in Data Set 25 in Appendix B. What is

the best predicted value for the year 2001? Given that the

Actual high value in 2001 was 11,350, how good was the

predicted value? What does the pattern suggest about the

stock market for investment purposes?



Decide which mathematical model

Exponential: ˆ 762.701 1.13456x

y


What is the best predicted value for the year 2001?

Given that the actual high value in 2001 was 11,350, how

good was the predicted value?

ˆ 762.701 1.13456x

y

2222

ˆ 762.701(1.13456) 12,262y

It misses by a fairly large amount.

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Chapter 9wellsmat.startlogic.com/sitebuildercontent/sitebuilderfiles/...The results do not support any conclusion about the removal of trees affecting the risk of fires, because none