© the mcgraw-hill companies, inc., 2000 11-1 chapter 11 correlation and regression

© The McGraw-Hill Companies, Inc., 2000

11-111-1

Chapter 11Chapter 11

Correlation and Correlation and RegressionRegression


11-211-2 OutlineOutline

11-1 Introduction

11-2 Scatter Plots

11-3 Correlation

11-4 Regression


11-311-3 OutlineOutline

11-5 Coefficient of

Determination and

Standard Error of

Estimate


11-411-4 ObjectivesObjectives

Draw a scatter plot for a set of ordered pairs.

Find the correlation coefficient. Test the hypothesis H0: = 0. Find the equation of the

regression line.


11-511-5 ObjectivesObjectives

Find the coefficient of determination.

Find the standard error of estimate.

Find a prediction interval.


11-611-6 11-2 Scatter Plots11-2 Scatter Plots

AA scatter plotscatter plot is a graph of the ordered pairs (x, y)(x, y) of numbers consisting of the independent variable, xx, and the dependent variable, yy.


11-711-7 11-2 Scatter Plots -11-2 Scatter Plots - Example

Construct a scatter plot for the data obtained in a study of age and systolic blood pressure of six randomly selected subjects.

The data is given on the next slide.



Subject Age, x Pressure, y

A 43 128

B 48 120

C 56 135

D 61 143

E 67 141

F 70 152



70605040

150

140

130

120

Age

Pre

ssur

e

70605040

150

140

130

120

Age

Pre

ssur

ePositive Relationship


11-1011-10 11-2 Scatter Plots -11-2 Scatter Plots - Other Examples

15105

90

80

70

60

50

40

Number of absences

Fin

al g

rade

15105

90

80

70

60

50

40

Number of absences

Fin

al g

rade

Negative Relationship


11-1111-1111-2 Scatter Plots -11-2 Scatter Plots - Other Examples

706050403020100

10

5

0

X

Y

706050403020100

10

5

0

x

yNo Relationship


11-1211-12 11-3 Correlation Coefficient11-3 Correlation Coefficient

The correlation coefficientcorrelation coefficient computed from the sample data measures the strength and direction of a relationship between two variables.

Sample correlation coefficient, r. Population correlation coefficient,


11-1311-1311-3 Range of Values for the 11-3 Range of Values for the

Correlation CoefficientCorrelation Coefficient

Strong negativerelationship

Strong positiverelationship

No linearrelationship


11-1411-1411-3 Formula for the Correlation 11-3 Formula for the Correlation

Coefficient Coefficient rr

r

n xy x y

n x x n y y

2 2 2 2

Where n is the number of data pairs


11-1511-1511-3 Correlation Coefficient - 11-3 Correlation Coefficient -

Example (Verify)

Compute the correlation coefficientcorrelation coefficient for the age and blood pressure data.

x y xy

x y

Substituting in the formula for r gives

r

345 819 47 634

20 399 112 443

0 897

2 2

,

, , , .

. .

= , = ,


11-1611-1611-3 The Significance of the 11-3 The Significance of the

Correlation Coefficient Correlation Coefficient

The population corelation population corelation coefficientcoefficient, , is the correlation between all possible pairs of data values (x, y) taken from a population.


11-1711-1711-3 The Significance of the 11-3 The Significance of the


H0: = 0 H1: 0 This tests for a significant

correlation between the variables in the population.


11-1811-1811-3 Formula for the 11-3 Formula for the t t tests for the tests for the


tn

rwith d f n

2

12

2

. .


11-1911-19 11-311-3 Example

Test the significance of the correlation coefficient for the age and blood pressure data. Use = 0.05 and r = 0.897.

Step 1:Step 1: State the hypotheses. H0: = 0 H1: 0


11-2011-20

Step 2:Step 2: Find the critical values. Since = 0.05 and there are 6 – 2 = 4 degrees of freedom, the critical values are t = +2.776 and t = –2.776.

Step 3: Step 3: Compute the test value. t = 4.059 (verify).

11-311-3 Example


11-2111-21

Step 4:Step 4: Make the decision. Reject the null hypothesis, since the test value falls in the critical region (4.059 > 2.776).

Step 5: Step 5: Summarize the results. There is a significant relationship between the variables of age and blood pressure.

11-311-3 Example


11-2211-22

The scatter plot for the age and blood pressure data displays a linear pattern.

We can model this relationship with a straight line.

This regression line is called the line of best fit or the regression line.

The equation of the line is y = a + bx.

11-4 Regression11-4 Regression


11-2311-2311-4 Formulas for the Regression 11-4 Formulas for the Regression

Line Line y = a + bx.

ay x x xy

n x x

bn xy x y

n x x

2

2 2

2 2

Where a is the y intercept and b is the slope of the line.


11-2411-24 11-411-4 Example

Find the equation of the regression line for the age and the blood pressure data.

Substituting into the formulas give a = 81.048 and b = 0.964 (verify).

Hence, y = 81.048 + 0.964x. Note, aa represents the interceptintercept and bb

the slopeslope of the line.


11-2511-25 11-411-4 Example

70605040

150

140

130

120

Age

Pre

ssur

e

70605040

150

140

130

120

Age

Pre

ssur

e

y = 81.048 + 0.964x


11-2611-2611-4 Using the Regression Line to11-4 Using the Regression Line to Predict Predict

The regression line can be used to predict a value for the dependent variable (y) for a given value of the independent variable (x).

Caution:Caution: Use x values within the experimental region when predicting y values.


11-2711-27 11-411-4 Example

Use the equation of the regression line to predict the blood pressure for a person who is 50 years old.

Since y = 81.048 + 0.964x, theny = 81.048 + 0.964(50) = 129.248 129.

Note that the value of 50 is within the range of x values.


11-2811-2811-5 Coefficient of Determination 11-5 Coefficient of Determination and Standard Error of Estimateand Standard Error of Estimate

The coefficient of determinationcoefficient of determination, denoted by r2, is a measure of the variation of the dependent variable that is explained by the regression line and the independent variable.



r2 is the square of the correlation coefficient.

The coefficient of coefficient of nondeterminationnondetermination is (1 – r2).

Example: If r = 0.90, then r2 = 0.81.



The standard error of estimatestandard error of estimate, denoted by sest, is the standard deviation of the observed y values about the predicted y values.

The formula is given on the next slide.


11-3111-3111-5 Formula for the Standard 11-5 Formula for the Standard

Error of Estimate Error of Estimate

s

y y

nor

sy a y b xy

n

est

est

2

2

2

2


11-3211-3211-5 Standard Error of Estimate -11-5 Standard Error of Estimate -

Example

From the regression equation, y = 55.57 + 8.13x and n = 6, find sest.

Here, a = 55.57, b = 8.13, and n = 6. Substituting into the formula gives sest

= 6.48 (verify).


11-3311-33 11-5 Prediction Interval11-5 Prediction Interval

A prediction intervalprediction interval is an interval constructed about a predicted y value, y , for a specified x value.


11-3411-34 11-5 Prediction Interval11-5 Prediction Interval

For given value, we can state with (1 – )100% confidence that the interval will contain the actual mean of the y values that correspond to the given value of x.


11-3511-3511-5 Formula for the Prediction 11-5 Formula for the Prediction Interval about a Value Interval about a Value yy

22

2)(11

2 xxn

Xxn

neststy

22

2)(11

2 xxn

Xxn

neststy

y

2.. nfdwith


11-3611-3611-5 Prediction interval -11-5 Prediction interval - Example

A researcher collects the data shown on the next slide and determines that there is a significant relationship between the age of a copy machine and its monthly maintenance cost. The regression equation is y = 55.57 + 8.13x. Find the 95% prediction interval for the monthly maintenance cost of a machine that is 3 years old.


11-3711-3711-5 Prediction Interval -11-5 Prediction Interval - Example

Machine Age, x (Years) Monthly cost, y

A 1 $62

B 2 $78

C 3 $70

D 4 $90

E 4 $93

F 6 $103


11-3811-38

Step 1: Step 1: Find x, x2 and . x = 20,

x2 = 82, Step 2: Step 2: Find y for x = 3.

y = 55.57 + 8.13(3) = 79.96 Step 3: Step 3: Find sest

sest = 6.48 as shown in previous example.

11-5 Prediction Interval -11-5 Prediction Interval - Example

X

X 3.36

20


11-3911-39

Step 4: Step 4: Substitute in the formula and solve. t/2 = 2.776, d.f. = 6 – 2 = 4 for 95%

60.53 < y < 99.39 (verify)

Hence, one can be 95% confident that the interval 60.53 < y < 99.39 contains the actual value of y.

11-5 Prediction Interval -11-5 Prediction Interval - Example

© the mcgraw-hill companies, inc., 2000 11-1 chapter 11 correlation and regression

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