Section 10-3

Chapter 10Correlation and Regression

Section 10-3Correlation

Section 10-3

Exercise #13


a. Draw the scatter plot for the variables.

b. Compute the value of the correlation coefficient.

c. State the hypotheses.

d. Test the significance of the correlation coefficient at = 0.05, using Table I.

e. Give a brief explanation of the type of relationship.

For the following exercise, complete these steps.

A researcher wishes to determine if a person’s age is related to the number of hours he or she exercises per week. The data for the sample are shown below.

Age x 18 26 32 38 52 59

Hours y 10 5 2 3 1.5 1


2

4

6

810

0 10 20 3040 50 60Age

Hou

rs

70

b. Compute the value of the correlation coefficient.Age x 18 26 32 38 52 59

Hours y 10 5 2 3 1.5 1

y = 22.5

y2 = 141.25

n = 6

x = 225

x 2 = 9653

xy = 625

y = 22.5 y2 = 141.2 n = 6

x = 225 x 2 = 9653 xy = 625

– =

2 22 2 –

n xy x yr

n x x n y y

y = 22.5 y2 = 141.2 n = 6

x = 225 x 2 = 9653 xy = 625

– 6 625 225 22.5=

2 2 – –6 69653 225 141.25 22.5

r

r = – 0.832

0 1: = 0 and : 0H Hc. State the hypotheses.

Age x 18 26 32 38 52 59

Hours y 10 5 2 3 1.5 1

0 1: = 0 and : 0H H


Age x 18 26 32 38 52 59

Hours y 10 5 2 3 1.5 1

n = 6 d.f . = 4 r = – 0.832

C.V . = ± 0.811

Decision: Reject H0 .

0 1: = 0 and : 0H H


Age x 18 26 32 38 52 59

Hours y 10 5 2 3 1.5 1

n = 6 d.f . = 4 r = – 0.832

There is a significant linear relationship between a person’s age and the number of hours he or she exercises per week.


Section 10-3

Exercise #15








The director of an alumni association for a small college wants to determine whether there is any type of relationship between the amount of an alumnus’s contribution (in dollars) and the years the alumnus has been out of school. The data are shown here.

Years x 1 5 3 10 7 6

Contribution y 500 100 300 50 75 80

a. Draw the scatter plot for the variables.Years x 1 5 3 10 7 6

Contribution y 500 100 300 50 75 80

100

200

300

400500

0 2 4 86 10 20Years

Con

trib

utio

n

30

b. Compute the value of the correlation coefficient.Years x 1 5 3 10 7 6

Contribution y 500 100 300 50 75 80

y = 1105

y2 = 364,525

n = 6

x = 32

x 2 = 220

xy = 3405


y = 1105 y2 = 364,52 n = 6 x = 32 x 2 = 220 xy = 3405

– =

2 22 2 –

n xy x yr

n x x n y y


y = 1105 y2 = 364,52 n = 6 x = 32 x 2 = 220 xy = 3405

3405 – 32 11056=

2 2 6 220 – 32 364,525 – 11056

r

r = – 0.883


0 1: = 0 and : 0H H

Years x 1 5 3 10 7 6

Contribution y 500 100 300 50 75 80


Years x 1 5 3 10 7 6

Contribution y 500 100 300 50 75 80

0 1: = 0 and : 0H H

n = 6 d.f . = 4 r = – 0.883


C.V . = ± 0.811


0 1: = 0 and : 0H H

Years x 1 5 3 10 7 6

Contribution y 500 100 300 50 75 80

n = 6 d.f . = 4 r = – 0.883

There is a significant linear relationship between a person’s age and his or her contribution.


Section 10-3

Exercise #17








A criminology student wishes to see if there is a relationship between the number of larceny crimes and the number of vandalism crimes on college campuses in Southwestern Pennsylvania. The data are shown. Is there a relationship between the two types of crimes?

Number of larceny crimes, x

24 6 16 64 10 25 35

Number of vandalism crimes y

21 3 6 15 21 61 20

a. Draw the scatter plot for the variables.Number of larceny

crimes, x24 6 16 64 10 25 35


21 3 6 15 21 61 20

20

4060

80

0 10 20 4030 50 60larceny crimes

vand

alis

m c

rimes

70 80

b. Compute the value of the correlation coefficient.Number of larceny

crimes, x24 6 16 64 10 25 35


21 3 6 15 21 61 20

y = 147

y2 = 5273

n = 7

x = 180

x 2 = 6914

xy = 4013

y = 147 y2 = 527 n = 7 x = 180 x 2 = 6914 xy = 4013

– =

2 22 2 –

n xy x yr

n x x n y y

y = 147 y2 = 527 n = 7 x = 180 x 2 = 6914 xy = 4013

4013 – 180 1476=

2 2 6 6914 – 180 5273 – 1476

r

r = 0.104


0 1: = 0 and : 0H H


24 6 16 64 10 25 35


21 3 6 15 21 61 20

n = 7 d.f .= 5 r = 0.104



24 6 16 64 10 25 35


21 3 6 15 21 61 20

C.V. = ± 0.754

Decision: Do not reject H0 .



24 6 16 64 10 25 35


21 3 6 15 21 61 20

There is not a significant linear relationship between the number of larceny crimes and the number of vandalism crimes.

Decision: Do not reject H0 .

n = 7 d.f .= 5 r = 0.104

Section 10-3

Exercise #23








The average daily temperature (in degrees Fahrenheit) and the corresponding average monthly precipitation (in inches) for the month of June are shown here for seven randomly selected cities in the United States. Determine if there is a relationship between the two variables.

Average daily temperature, x

86 81 83 89 80 74 64

Average monthly precipitation, y

3.4 1.8 3.5 3.6 3.7 1.5 0.2

a. Draw the scatter plot for the variables.Average daily temperature, x

86 81 83 89 80 74 64


3.4 1.8 3.5 3.6 3.7 1.5 0.2

1

2

3

4

0 60Temperature

Prec

ipita

tion

70 80 90 100

5

b. Compute the value of the correlation coefficient.Average daily temperature, x

86 81 83 89 80 74 64


3.4 1.8 3.5 3.6 3.7 1.5 0.2

y = 17.7

y2 = 55.99

n = 7

x = 557

x 2 = 44,739

xy = 1468.9

y = 17.7 y2 = 55.99 n = 7 x = 557 x 2 = 44,739 xy = 1468.9

– =

2 22 2 –

n xy x yr

n x x n y y

y = 17.7 y2 = 55.99 n = 7 x = 557 x 2 = 44,739 xy = 1468.9

r = 7(1468.9) – (557)(17.7)

7(44,739) – (557)2

7(55.99) – (17.7)2

r = 0.883



86 81 83 89 80 74 64


3.4 1.8 3.5 3.6 3.7 1.5 0.2

0 1: = 0 and : 0H H

d. Test the significance of the correlation coefficient at = 0.05, using Table I.Average daily temperature, x

86 81 83 89 80 74 64


3.4 1.8 3.5 3.6 3.7 1.5 0.2

0 1: = 0 and : 0H H

C.V. = ± 0.754


n =7 d.f . = 5 r = 0.883



86 81 83 89 80 74 64


3.4 1.8 3.5 3.6 3.7 1.5 0.2

0 1: = 0 and : 0H H

There is a significant linear relationship between temperature and precipitation.


n =7 d.f . = 5 r = 0.883


Section 10-4Regression

Section 10-4

Exercise #13


Find the equation of the regression line and find the y value for the specified x value. Remember that no regression should be done when r is not significant.

Ages and Exercise

11.532510Hours y

595238322618Age x

2

22

– =

–

y x x xya

n x x

Find y when x = 35 years.Ages and Exercise

11.532510Hours y

595238322618Age x

2

22.5 9653 – 225 625 =

6 9653 – 225a

a = 10.499

2

22

– =

–

y x x xya

n x x


11.532510Hours y

595238322618Age x


11.532510Hours y

595238322618Age x

22

– =

–

n xy x yb

n x x

2

6 625 – 225 22.5 =

6 9653 – 225b


11.532510Hours y

595238322618Age x

22

– =

–

n xy x yb

n x x

b = – 0.18


11.532510Hours y

595238322618Age x

a = 10.499 b = – 0.18

y = a + bx

y = 10.499 – 0.18x

y = 10.499 – 0.18(35)

y = 4.199 hours

Section 10-4

Exercise #15


Find the equation of the regression line and find the y value for the specified x value. Remember that no regression should be done when r is not significant.Years and Contribution

807550300100500Contribution y, $

6710351Years x

2

22

– =

–

y x x xya

n x x

2

1105 220 – 32 3405 =

6 220 – 32a

Find y when x = 4 years.Years and Contribution


6710351Years x

a =

243,100 – 108,9601320 – 1024

a = 134,140

296

2

1105 220 – 32 3405 =

6 220 – 32a

a = 453.176

b = 6(3405) – (32)(1105)

6(220) – (32)2

22

– =

–

n xy x yb

n x x



6710351Years x

b = 6(3405) – (32)(1105)

6(220) – (32)2

b = – 50.439

b =

20,430 – 35,360296

b =

–14,930296

b = – 50.439

y = a + bx

y = 453.176 – 50.439x

y = $251.42

y = 453.176 – 50.439(4)

a = 453.176



6710351Years x

Section 10-4

Exercise #23


Find the equation of the regression line and find the y value when x = 70 ºF. Remember that no regression should be done when r is not significant.

0.2

64

1.53.73.63.51.83.4Avg. mo. Precip. y

748089838186Avg. daily temp. x

Temperatures ( in. F ) and precipitation (in.)

x = 557

x2 = 44,739

y = 17.7xy = 1468.9

x = 557

x2 = 44,739

y = 17.7xy = 1468.9

2

22

– =

–

y x x xya

n x x

a = (17.7)(44,739) – (557)(1468.9)

7(44,739) – (557)2

a = – 8.994

22

– =

–

n xy x yb

n x x

x = 557

x2 = 44,739

y = 17.7xy = 1468.9

b = 7(1468.9) – (557)(17.7)

7(44,739) – (557)2

b = 0.1448

y = a+ bx

y = – 8.994 + 0.1448x

y = 1.1 inches

y = – 8.994+ 0.1448(70)

a = – 8.994

b = 0.1448

x = 557

x2 = 44,739

y = 17.7xy = 1468.9


Section 10-5Coefficient of Determination and Standard Error of the Estimate

Section 10-5

Exercise #9


Find the coefficients of determination and non-determination when and explain the meaning of:

r2 = 0.49

r = 0.70

49% of the variation of y is due to the variation of x.

Find the coefficients of determination and non-determination when and explain the meaning of:

1– r 2 = 0.51

r = 0.70

51% of the variation of y is due to chance.

Section 10-5

Exercise #15


Compute the standard error of the estimate.

x = 225

x2

= 9653

xy = 625

y = 22.5

y2 = 141.25

n = 6

a = 10.499 b = – 0.18

sest =

y 2 – a y – b xy

n – 2

s

est=

141.25 – 10.499(22.5) – (– 0.18)(625)

6 – 2

sest =

141.25 – 10.499(22.5) – (– 0.18)(625)

6 2

sest = 4.380625

sest = 2.09

Section 10-5

Exercise #19


Find the 90% prediction interval when x = 20 years.

x = 225

y = 22.5

x2

= 9653

y2 = 141.25

xy = 625

n = 6

a = 10.499

b = – 0.18

Age x 18 26 32 38 52 59

Hours y 10 5 2 3 1.5 1

y = 10.499 – 0.18x

= 10.499– 0.18(20) = 6.899

x = 225

y = 22.5

x2

= 9653

y2 = 141.25

xy = 625

n = 6

a = 10.499

b = – 0.18

y = 6.899

2

2 2( )11 + +

( )– < 2

n x Xyy t s nest n x x

2

2 2– + +

( )1< + 1 2( )

n x Xy t sest n n x x

1.60 < y < 12.20

6.899 – (2.132)(2.09) 1 + 16

+ 6(20 – 37.5)2

6(9653) 2252< y

< 6.899 + (2.132)(2.09) 1+ 16

+ 6(20 – 37.5)2

6(9653) – 2252

6.899– (2.132)(2.09)(1.19)< y

< 6.899 + (2.132)(2.09)(1.19)

Section 10-5

Exercise #21


Find the 90% prediction interval when x = 4 years.

Years x 1 5 3 10 7 6

Contributions y, $ 500 100 300 50 75 80

x = 35

y = 1105

x2

= 220

y2 = 364,525

xy = 3405

n = 6

a = 453.176

b = – 50.439

y = 453.176 – 50.439x

= 251.42 = 453.176 – 50.439(4)

2

2 2( )11 + + 2 ( )

–– <

–

n x Xyy t s nest n x x

2

2 2 + +

( – )1< + 1 2– ( )

n x Xy t sest n n x x

x = 35

y = 1105

x2

= 220

y2 = 364,525

xy = 3405

n = 6

a = 453.176

b = – 50.439

y = 251.42

$30.46 < y < $472.38

251.42 – (2.132)(94.22) 1 + 16

+ 6(4 – 5.33)2

6(220) – 322< y

< 251.42+ (2.132)(94.22) 1 + 16

+ 6(4 – 5.33)2

6(220) – 322

251.42 – (2.132)(94.22)(1.1) < y

< 251.42+ (2.132)(94.22)(1.1)


Section 10-6Multiple Regression

Section 10-6

Exercise #7


A manufacturer found that a significant relationship

exists among the number of hours an assembly line

employee works per shift x1, the total

number of items produced x2, and the

number of defective items produced y.

The multiple regression equation is

. Predict the

number of defective items

produced by an employee who has

worked 9 hours and produced

24 items.

y = 9.6 + 2.2x1 – 1.08x2

y = 3.48 or 3 items

y = 9.6 + 2.2x1 – 1.08x2

= 9.6 + 2.2 9 – 1.08 24y

Section 10-6

Exercise #9


An educator has found a significant relationship among a college graduate’s IQ x1, score on the verbal section

of the SAT x2, and income for the first year

following graduation from college y. Predict the income of a college graduate whose IQ is 120 and verbal SAT score is650. The regression equation is

y ' = 5000 + 97x1+ 35x2 .

y = 5000+ 97(120) – 35(650)

y = $39,390

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Section 10-3

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