Chapter 6

Chapter 6 : Simple Linear Regression

CHAPTER 6 : SIMPLE LINEAR REGRESSION Sub-Topic  Introduction.  Scatter plots.  Simple linear regression model.  The least square method.  Inference of regression of coefficient.  Confidence intervals of the regression line.  Coefficient of determination.  Coefficient of pearson correlation. Chapter Learning Outcome  Solve the problems involve the simple linear regression. Learning Objective

By the end of this chapter, students should be able to  Draw the scatter plots.  Plot the regression line through the least square method.  Make inference concern to regression coefficients.  Find and interpret the determination coefficient and co rrelation coefficient. Key Term (English to Bahasa Melayu)

English

Bahasa Melayu

1.

Independent variable

→

Pemboleh ubah tidak bersandar

2.

Dependent variable

→

Pemboleh ubah bersandar

3.

Scatter plot

→

Plot serakan

4.

Intercept

→

Pintasan

5.

Slope

→

Kecerunan

6.

Simple linear regression

→

Regresi linear ringkas

7.

Least square method

→

Kaedah kuasa dua terkecil

8.

Correlation

→

Hubungan

9.

Confidence interval

→

Selang keyakinan

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6.1 Introduction

A major objective of many statistical investigations is to establish relationships that make it possible to predict one or more variables in terms of others. Thus, studies are made to predict the potential sales of a new product in terms of its price, a patient’s weight in terms of the number of weeks he or she has been on diet, family expenditures on entertainment in terms of family income, the per capita consumption of certain foods in terms of their nutritional values and the amount of money spent advertising them on television, and so forth. Although it is desirable to be able to predict one quantity exactly in terms of others, this is seldom possible, and in most instances we have to be satisfied with predicting averages or expected values. We may not be able to predict exactly how much money Aida will make 10 years after graduating from college, but if we are given suitable data, we can predict the average income of a college graduate in terms of the number of years she has been out of college.

6.2 Scatter Plots

Definition 1

A scatter plot is a graph of the ordered o rdered pairs ( x ( x,, y ) of numbers consisting of the independent variable x variable x and and the dependent variable y variable y..

Theory 1

In simple correlation and regression studies, the researcher collects data on two numerical or quantitative variables to see whether a relationship exists between the variables. The two variables for this study are called the independent variable and the dependent variable. The independent variable , x , is the variable in regression that can be controlled. It is also a variable used to predict or model. The dependent variable , y , is the variable in regression that cannot be controlled. It is a variable to

be predicted or modeled.

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For example, if the researcher wishes to see whether there is a relationship between number of hours study and test scores in exam. In this case, the independent variable is “number of “number of hours study” while the dependent depen dent variable is the “test scores in the students’ exam”. The reason for this selection is th e test scores depends on the number of hours he studied. He can control the number of hours he studies for exam. ex am.

Example 1

Construct a scatter plot for the data obtained in a study on the number of absences and the final grades of seven randomly selected students from a statistics class. The data are shown below : Student A B C D E F G

Number of absences, x absences, x 6 2 15 9 12 5 8

Final grade, y grade, y (%) (%) 82 86 43 74 58 90 78

Answer Example 1

1. Draw and label the x the x and and y y axes. axes. 2. Plot each point on the graph. Scatter Plot for Example 1 100 80 y

d

e

,

60 r

a a

l

g

40 ni F

20 0 0

5

10

Number of absences,

251

15 x

20


Example 2

Suppose an experiment involving five subjects is conducted to determine the relationship between the percentage of a certain drug in the bloodstream and the length of time it takes to react to a stimulus.

Subject 1 2 3 4 5

Amount of drug, x drug, x (%) 1 2 3 4 5

Reaction time, y time, y(seconds) (seconds) 1 1 2 2 4

Answer Example 2

1. Draw and label the x the x and and y y axes. axes. 2. Plot each point on the graph.

Scatter Plot for Example Example 2 4.5 4 y

,

3.5

e mi

)s n o

3

d t oi

n

2.5

c

2

(

1.5

es R

e

a

c

t

1 0.5 0 0

1

2

3

Amount of drug,

252

4 x

(%)

5

6


Exercise 6.2

1.

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

2.

Age, x Age, x

18

26

32

38

52

59

Hours, y Hours, y

10

5

2

3

1.5

1

The number of calories and the number of milligrams of cholesterol for a random sample of fast-food chicken sandwich from six café are shown below. Draw a scatter plot for the variables.

3.

Calories, x Calories, x

390

535

720

300

430

500

Cholesterol, y Cholesterol, y

43

45

80

50

55

52

Various doses of a poisonous substance were given to five mice and following results were observed. Draw a scatter plot for the variables.

4.

Dose, x Dose, x (mg)

4

6

8

10

12

14

16

No. of deaths, y

1

3

6

8

14

16

20

A researcher desires to know whether the typing speed of a secretary (in words per minute) is related to the time (in hours) that it takes the secretary to learn to use a new word processing program. Draw a scatter plot for the variables. The data are shown below.

5.

Speed, x Speed, x

48

74

52

79

83

56

85

63

88

74

90

92

Time, y Time, y

7

4

8

3.5

2

6

2.3

5

2.1

4.5

1.9

1.5

The following data pertain to the chlorine residual in a swimming pool at various times after it has been treated with chemicals. Draw a scatter plot for the variables.

No. of hours, x hours, x Chlorine residual (parts per million), y million), y

253

2

4

6

8

10

12

1.8

1.5

1.4

1.1

1.1

0.9


6.

Mehta and Deopura (1995) studied the mechanical properties of spun PETLCP blend fibers. They believe that the modulus (the response) depends on the percent of PET in the blend. The data is given by the table below. Make a scatter plot of the data.

7.

PET %, x

100

97.5

95

90

80

50

0

Modulus, y

2.12

2.26

2.57

3.26

3.46

4.54

8.5

The job placement center at State University wants to determine whether student’s grade point averages (GPAs) can explain the number of job offers they receive upon graduation. The data seen here are for 10 recent graduates. Draw a scatter plot for the variables.

8.

GPA, x

3.25

2.35

1.02

0.36

3.69

2.65

2.15

1.25

3.88

3.37

Offers, y

3

3

1

0

5

4

2

2

6

2

Dr. Ahmad has noticed many of his students have been absent from class this semester. He feels that he can explain this sluggish attendance by the distance his students live from campus. Eleven students are selected as to how many miles they must travel to attend class and the number of classes they have missed. Draw a scatter plot for the variables given in the table below.

9.

Miles, x

5

6

2

0

9

12

16

5

7

0

8

Misses, y

2

2

4

5

4

2

5

2

3

1

4

Ten sales people were surveyed and the average number of clients contacts per month , x, and the sales volume, y (in thousands), were recorded for each. Draw the scatter plot for the variables. X

12

14

16

20

23

46

50

48

50

55

Y

15

25

30

30

30

80

90

95

110

130

254


10.

The following are loads (grams) put on the ends of like plastic rods with the resulting deflections (cm). Draw the scatter plot for the variables.

11.

Load ( x)

25

30

35

40

55

45

50

60

Deflection ( y)

1.58

1.39

1.41

1.60

1.81

1.78

1.65

1.94

The following are the sample data provided by a moving company on the weights of six shipments and the damage that was incurred. Draw the scatter plot for the variables. Weight (1000 pounds) ( x) Damage (dollars) ( y)

12.

4

3

1.6

1.2

3.4

4.8

160

112

69

90

123

186

The following data pertain to the demand for a product (in thousands of unit) and its price (in cents) charged in five different market areas. Draw the scatter plot for the variables.

13.

Price, x

20

16

10

11

14

Demand, y

22

41

120

89

56

To reduce crimes, the president has budgeted more money to put more police on our city streets. Use the data below to draw a scatter plot for the variables.

14.

Police, x

13

15

23

25

15

10

9

20

No. of reported crimes, y

8

9

12

18

8

6

5

10

Aunt Reeta wants to get more yields from her tomato plants this summer by increasing the number of times she uses fertilizer. Based on the data below, draw a scatter plot for the variables. Use of fertilizer, x

4

9

5

8

2

Yield (pounds), y

12

20

15

17

7

255


15.

The resident of Taman Seri are worried about a rise in housing costs in the area. The head of the people think that home prices fluctuate with the land values. Data on 10 recently sold homes and the cost of the land on which they were built are seen here in thousands of ringgit. Draw a scatter plot for the variables.

Land values, x

7.0

6.9

5.5

3.7

5.9

3.8

8.9

9.6

9.9

10.0

Cost of the house, y

67

63

60

54

58

36

76

87

89

92

40

50

Answer Exercise 6.2

1. 12 10 8

y , s r u

6

o H

4 2 0 0

10

20

30

60

70

Age, x

2. 90 80 70

y

,l 60 o

r 50 e ts

el 40 o

h 30 C 20 10 0 0

200

400

Calories, x

256

600

800


3.

25 20 , s

th 15 a e d f

o 10 . o N

5 0 0

5

10

15

20

Dose, x

4. 9 8 7 6

y ,

e 5 mi

T 4 3 2 1 0 0

20

40

60

80

100

Speed, x

5. 2 1.8 y 1.6 , l

a 1.4 u di e

s 1.2

1 ni

0.8 ol

0.6 h

r

e

r

C 0.4

0.2 0 0

2

4

6

8

No. of hours, x

257

10

12

14


6. 9 8 7

y, 6 s

ul 5 u

d 4 o

M 3 2 1 0 0

20

40

60

80

100

120

PET %,x

7. 7 6 5

y

, 4 s r ef

f 3 O

2 1 0 0

1

2

3

4

5

GPA, x

8. 6 5

y 4 , s e

s 3 M

si

2 1 0 0

5

10

Miles, x

258

15

20


9. 140 120

y

, 100 e m

80

v

60

el

40

o

ul s a S

20 0 0

10

20

30

40

50

60

No. of clients,x

10. 2.5

2 ) y(

n 1.5 oi t c lef

1 e D

0.5

0 0

10

20

30

40

50

60

70

Load, x

11. 200 180

sr

y, 160 ) 140

lal

o 120 e

d( 100 80

m

60

D

a

a

g

40 20 0 0

1

2

3

Wei ght (pounds),x

259

4

5

6


12. 140 120 100

y , d

80

n a

60

D

40

e

m

20 0 0

5

10

15

20

25

Price, x

13. 20

y

, 18 s

e 16 mi

r 14 c

d 12 et 10 r o

8

p er

6

f

4

o

2

.

o N

0 0

5

10

15

20

25

30

Police, x

14. 25

y 20 ), s d

n 15 u p(

o 10

dl Y

ei 5

0 0

2

4

6

Use of fertilizer, x

260

8

10


15. 100 90

y ,

80

e s

70

h

60

e

50

f

40

t

30

o

20

o

u

th o s C

10 0 0

2

4

6

8

10

12

Land values, x

6.3 Simple Linear Regression Model

Definition 2

A simple linear regression is a statistical technique used to find relationships between variables for the purpose of predicting future values. It enables us to see the trend and make predictions on the basis of data. Theory 2

Given a scatter plot, one must be able to draw the line best fit. Best fit means that the sum of the squares of the vertical distances from each point to the line is at a minimum. The closer the points are to the line, the better the fit and the prediction will be.

261


y Observed value, y ( x, y) Line of best fit y   0   1 x

e

ˆ

( x, y ) Predicted value of y, y ˆ

ˆ

x

From the graph above, the error  is approximated by e  y  y , the difference ˆ

between the observed value of y and the predicted value of y, y , at a given value of ˆ

x. The model for simple linear regression is y   0 ˆ

  1 x   , where ˆ

y = dependent or response variable (variable to be modeled)

x = independent or predictor variable (variable used as a predictor of y)

 0 ˆ

 y intercept of the line ( the point at which the line intersects or cuts through the y-axis)

 1  slope of the line (the amount of increase (or decrease) in the ˆ

deterministic component of y for every 1-unit increase in x)

  statistical error (random variable that accounts for the failure of the model to fit the data exactly ) This regression model is said to be simple, linear in the parameters, and linear in the predictor variable. It is “simple” in that there is only one predictor variable, “linear in the parameters,” because no parameter appears as an exponent or is multiplied or divided by another parameter, and “linear in the predictor variable,” because this

262


variable appears only in the first power. 6.4 The Least Square Method

Definition 3

One way to know how well a straight line fits a set of data is to note the extent to which the data points deviate from the line. The deviations (the difference between the observed and the predicted values of y) or the errors of prediction are the vertical distances between observed and predicted values. The sum of errors and the sum of squares of the errors (SSE ) gives greater emphasis to large deviations of the points from the line. It is possible to find many lines for which the sum of errors is equal to 0, but it can be shown that there is one (and only one) line for which the value of sum of squares of the errors is a minimum. This line is called the least squares line or the regression line . The methodology used to obtain this regression line is called the least squares method .

Theory 3

Given the sample data ( xi , yi ); i  1,2,...., n, the coefficients of the least squares line, for y   0 ˆ

ˆ

 1  ˆ

Sxy

  1 x , the coefficients are; ˆ

(slope) and  0 ˆ

Sxx

 y   1 x , ( y-intercept) where ˆ

 n  Sxy   ( xi  x )( yi  y )   xi yi    xi   yi  , n  i 1  i 1  i 1 i 1 n

Sxx 

1  n

n

n

n

 ( x  x )   x

2

2

i

i 1

i 1

i

1  n

2

    xi  , n  i1 

and n = sample size.

Example 3

Raw material used in the production of a synthetic fiber is stored in a place that has no humidity control. Measurements of the relative humidity and the moisture content of samples of the raw material (both in percentage) on 12 days yielded the following

263


results:

(a)

Humidity ( x)

Moisture content ( y)

46 53 37 42 34 29 60 44 41 48 33 40

12 14 11 13 10 8 17 12 10 15 9 13

Fit a least squares line that will enable us to predict the moisture content in terms of the relative humidity. Interpret the result.

(b)

Estimate the moisture content when the relative humidity is 38 percent.

Answer Example 3

(a)

 x  507,  x  22265,  y  144 ,  y  1802,  xy  6314, n  12 2

We get 2

from the data. Thus, Sxy  6314 

Thus,  1 ˆ



1 12

Sxy Sxx

 0  y   1 x  ˆ

(507)(144)  230, and Sxx  22,265 



230 844.25

144

ˆ

12

1 12

(507) 2

 844.25

 0.2724 and

 (0.2724 )

507 12

 0.4911 ,

and the equation of the least squares line is y   0 ˆ

ˆ

  1 x , ˆ

y  0.4911  0.2724 x . ˆ

When the humidity is increase by one percent, the moisture content will also increase by 0.2724 percent.

264


(b)

Substituting x = 38 into the equation obtained in answer (a), we get y  0.4911  0.2724 (38) ˆ

y  10.8423 or y  11, rounded to the nearest unit. ˆ

ˆ

Example 4

The following are the scores that 12 students obtained on the mid-term and final examinations in a course in statistics. Mid-term examination, x

Final examination, y

71 49 80 73 93 85 58 82 64 32 87 80 (a)

83 62 76 77 89 74 48 78 76 51 73 89

Find the equation of the least squares line that will enable us to predict a student’s final examination score in this course on the basis of his or her score in the mid-term examination. Interpret the result.

(b)

Predict the final examination score of a student who score 84 in the mid-term examination.

Answer Example 4

(a)

 x  854,  x  64222,  y  876  y  65850,  xy  64346, n  12 2

We get 2

from the data. Thus, Sxy  64346 

1 12

(854)(876)  2004 , and

265


Sxx  64222 

Thus,  1 ˆ

 0 ˆ



1

(854) 2

12

Sxy Sxx

 y   1 x 



 3445 .67

2004 3445 .67

876

ˆ

12

 0.5816 and

 (0.5816 )

854 12

 31.609 ,

and the equation of the least squares line is y   0 ˆ

ˆ

  1 x , ˆ

 0.5816 x . y  31.609 ˆ

When the score in mid-term examination is increase by one mark, the score in final examination will also increase by 0.5816 marks.

(b)

Substituting x = 84 into the equation obtained in answer (a), we get y  31.609  0.5816 (84) ˆ

y  80.4634 or y  80, rounded to the nearest unit. ˆ

ˆ

Exercise 6.4

1.

From the Exercise 6.2(1) , find the regression line using the least squares method. Interpret the result. Then, estimate the number of hours he or she exercises per week when his or her age is 50 years old.

2.

From the Exercise 6.2(2) , find the regression line using the least squares method. Interpret the result. Then, estimate the number of milligrams of cholesterol when the number of calories is 650.

3.

From the Exercise 6.2(3) , find the regression line using the least squares method. Interpret the result. Then, estimate the number of deaths when the 5 mg dose of a poison is given to the mice.

4.

From the Exercise 6.2(4) , find the regression line using the least squares method. Interpret the result. Then, estimate the time that it takes the secretary

266


to learn when the typing speed is 100 words per minute.

5.

From the Exercise 6.2(5) , find the regression line using the least squares method. Interpret the result. Then, estimate the chlorine residual in a swimming pool when the various times after it has been treated with chemicals is 13 hours.

6.

From the Exercise 6.2(6) , find the regression line using the least squares method. Interpret the result. Then, estimate the modulus when the PET in the blend is 88%.

7.

From the Exercise 6.2(7) , find the regression line using the least squares method. Interpret the result. Then, estimate the number of job offers when the GPA of a student is 2.98.

8.

From the Exercise 6.2(8) , find the regression line using the least squares method. Interpret the result. Then, estimate the number of class a student will misses when he or she lives 15 miles from campus.

9.

From the Exercise 6.2(9) , find the regression line using the least squares method. Interpret the result. Then, estimate the sales volumes when the number of clients is 60.

10.

From the Exercise 6.2(10) , find the regression line using the least squares method. Interpret the result. Then, estimate the deflections when the load is 65 grams.

11.

From the Exercise 6.2(11) , find the regression line using the least squares method. Interpret the result. Then, estimate the damage incurred when the weight is 5500 pounds.

267


12.

From the Exercise 6.2(12) , find the regression line using the least squares method. Interpret the result. Then, estimate the demand for a product when the price is 50 cents.

13.

From the Exercise 6.2(13) , find the regression line using the least squares method. Interpret the result. Then, estimate the number of reported crimes when there are 19 policemen.

14.

From the Exercise 6.2(14) , find the regression line using the least squares method. Interpret the result. Then, estimate the yields of her tomato plants when she uses the fertilizer 10 times.

15.

From the Exercise 6.2(15) , find the regression line using the least squares method. Interpret the result. Then, estimate the cost of the house when the land value is RM 73000.

Answer Exercise 6.4

1.

(a) y

 10.4989  0.17997 x ,

(b) y  1.5004

2.

(a) y

 20.2369  0.07081 x ,

(b) y  66.2634

3.

(a) y

 6.5357  1.625 x ,

(b) y  1.5893

4.

(a) y

 0.1371 x ,  14.083

(b) y  0.373

5.

(a) y  1.8999  0.0857 x ,

(b) y  0.7858

6.

(a) y  8.221  0.0602 x ,

(b) y  2.9234

7.

(a) y  0.248  1.272 x ,

(b) y  3.5426

8.

 0.06974 x , (a) y  2.647

(b) y  3.6931

9.

(a) y  13.4202  2.303 x ,

(b) y  124.7598

10.

(a) y  1.086  0.01314 x ,

(b) y  1.9401

11.

(a) y  34.146  29.729 x ,

(b) y  197.6555

12.

(a) y  196.775  9.238 x , (b) y  265.125

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

268


13.

(a) y  0.930  0.642 x ,

(b) y  11.268

14.

(a) y  4.8592  1.668 x ,

(b) y  21.5392

15.

(a) y  17.855  7.071 x ,

(b) y  534.038

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

6.5 Inference of Regression Coefficient

Definition 4

The in ference of regression coeff ici ents describes how to conduct a hypothesis test to determine whether there is a significant linear relationship between an independent variable x and a dependent variable y. The test focuses on the slope of the regression line y ˆ

  0   1 x , where  0 is a constant,  1 is the slope (also called the regression ˆ

ˆ

coefficient), x is the value of the independent variable, and y is the value of the dependent variable.

6.5.1

Hypothesis testing on slope ,  1

Definition 5

Hypothesis testing concerning  1 and  0 requires the additional assumption that the model errors  i are normally distributed. Thus the complete assumptions are that the errors are normally and independently distributed ( NID) with mean 0 and variance

 2 ,  i ~ (0,  2 ) . Theory 4

To test the hypothesis that the slope equals a constant, say  C . The appropriate hypothesis is

  C H 1 :  1   C , or  1   C , or  1   C H 0 :  1

Therefore, the statistics Z test



  2   . with  1 ~   1 , Sxx   ˆ

 1   C ˆ

 2 / Sxx

is distributed with N (0, 1) if the null

269


hypothesis, H 0 :  1

  C is true.

However, the residual mean square, MSE is an unbiased estimator of  2 , and the distribution of (n-2) MSE /  2 is  n22 . Both MSE and  1 are independent variables, ˆ

so these conditions imply that if we replace  2 in Z test by  2  MSE , the statistics ˆ

T test

 1   C ˆ



is distributed as T

MSE / Sxx

hypothesis, H 0 :  1

with n-2 degrees of freedom if the null

  C is true. The statistics of T is used to test the null hypothesis

by comparing the observed value of T with the upper  / 2 percentage point of the t n  2 distribution (t  / 2,n2 ) and rejecting the null hypothesis if T test

 t  / 2,n2 .

Example 5

Based on the Example 3, test the hypothesis concerning H 0 :  1 H 1 :  1

 1 against the

 1 at the 0.05 level of significance.

Answer Example 5

Step 1 : State the hypothesis H 0 :  1 H 1 :  1

1 1

Step 2:   0.05 , v  n  2  12  2  10 , this is a one-tailed test (right) T table= t  ,v

 t 0.05,10  1.812 , reject H 0 when T test is more than T table.

Step 3: Compute MSE , and T test . Sxy  230, Sxx  844.25 ,  1  0.2724 and  0  0.4911 . ˆ

ˆ

Syy  1802 

144 2 12

SSE = Syy   1 S xy ˆ

MSE 

SSE n2



 74

 74  0.2724(230)  11.348

11.348 10

 1.1348

270


 1   C ˆ

T test =

MSE S xx



0.2724  1 1.1348 844.25

 19.8458

Step 4 : Make decision Do not reject H 0 since T test is less than T table Step 5 : Make conclusion We can conclude that the slope is equal to one. Example 6


 5 against the

 5 at the 0.05 level of significance.

Answer Example 6

Step 1 : State the hypothesis H 0 :  1

5

H 1 :  1

5

Step 2:   0.05 ,  / 2  0.025 v  n  2  12  2  10 , this is a two-tailed test T table = t  / 2,v

 t 0.025,10  2.228 , reject H 0 when T test is more than 2.228 or less

than -2.228. Step 3: Compute MSE , and T test. Sxy  2004, Sxx  3445.67 ,  1  0.5816 and  0 ˆ

876 2

Syy  65850 

12


MSE 

SSE n2



 1   C MSE S xx

 31.609 .

 1902

 1902  0.5816 (2004 )  736.4736

736.4736

ˆ

T test =

ˆ

10



 73.64736 0.5816  5

73.64736 3445.67

 30.222

Step 4 : Make decision Reject H 0 since T test is less than -2.2282.

271


Step 5 : Make conclusion We can conclude that the slope is not equal to five.

Exercise 6.5.1

1.

Based on the Exercise 6.2(1), test the hypothesis concerning H 0 :  1 against the H 1 :  1

2.

3.

7.

 0.5 at the 0.1 level of significance.

1


1

 1 at the 0.005 level of significance.


1



8.

 0.5


2



6.

 1 at the 0.01 level of significance.


5.

 1


4.



1


272

1


9.


10.

11.

13.

14.

15.

1


Based on the Exercise 6.2(15) , test the hypothesis concerning H 0 :  1 against the H 1 :  1

1



1



1



1



12.



1


Answer Exercise 6.5.1

1.

T test = -19.664, do not reject H 0.

2.

T test = 35.5618, do not reject H 0.

3.

T test = -3.7238, reject H 0.

4.


5.

T test = -113.1043, reject H 0.

6.


273

1


7.


8.


9.


10.

T test =-281.0637, reject H 0.

11.

T test =5.6907, reject H 0.

12.

T test =-5.9512, do not reject H 0.

13.

T test =-3.3558, reject H 0.

14.

T test =2.6367, do not reject H 0.

15.

T test =8.0134, reject H 0.

6.5.2

Hypothesis testing on intercepts ,  0

Theory 5

A similar procedure can be used to test hypothesis about the intercept. To test the hypothesis that the intercept equals a constant, say  C . The appropriate hypothesis are

  C H 1 :  0   C , or  0   C , or  0   C H 0 :  0

The statistics T test



 1 x 2   . with  0 ~   0 ,  n Sxx   ˆ

 0   C ˆ

MSE (1 / n  x / Sxx) 2

is distributed as T with n-2 degrees of

freedom if the null hypothesis is true. The statistics of T is used to test the null hypothesis by comparing the observed value of T with the upper  / 2 percentage point of the t n  2 distribution (t  / 2,n2 ) and rejecting the null hypothesis if T test

 t  / 2,n2 .

Example 7



274

 1 against the


Answer Example 7


1

H 1 :  0

1

Step 2:   0.05 , v  n  2  12  2  10 , this is a one-tailed test (right) T table = t  ,v

 t 0.05,10  1.812 , reject H 0 when T test is more than T table.

Step 3: Compute MSE , and T test.. Sxy  230, Sxx  844.25 , x Syy  1802 

144 2 12


MSE 

SSE n2



2

ˆ

 74

 74  0.2724(230)  11.348

11.348 10

 1.1348

 0   C ˆ

T test =

 1785.06 and  0  0.4911 .

MSE (1 / n  x / Sxx) 2



0.4911  1 1.1348 (1 / 12  1785 .06 / 844.25)

 0.3222 Step 4 : Make decision Do not reject H 0 since T test is less than T table.. Step 5 : Make conclusion We can conclude that the intercept is equal to one.

Example 8



Answer Example 8


5

275

 5 against the


5

H 1 :  0

Step 2:   0.05 ,  / 2  0.025 v  n  2  12  2  10 , this is a two-tailed test T table = t  / 2,v

 t 0.025,10  2.228 , reject H 0 when T test

is more than 2.228 or less

than -2.228. Step 3: Compute MSE , and T test . Sxy  2004, Sxx  3445.67 , x 876 2

Syy  65850 

12


SSE

MSE 

n2

ˆ

 1902  0.5816 (2004 )  736.4736 10

 73.64736

 0   C ˆ

T test =

 5064.69 and  0  31.609 .

 1902

736.4736



2

MSE (1 / n  x / Sxx) 2



31.609  5 73.64736 (1 / 12  5064 .69 / 3445 .67)

 2.488 Step 4 : Make decision Reject H 0 since T test is more than 2.2282. Step 5 : Make conclusion We can conclude that the intercept is not equal to five.

.6.2 Confidence Intervals for intercept,  0 . Theory 7

The slope  0 of the regression line of the population can be estimated by means of a confidence interval.

 0 ˆ

 t  / 2,v

 1 x 2     0   0  t  / 2,v MSE    n Sxx  ˆ

where v = n-2

276

 1 x 2   MSE    n Sxx 


Example 11

Based on the Example 3, find the 95% confidence interval for the population’s intercept,  0 . Answer Example 11

Step 1 : n = 12, v  n  2  12  2  10,

  0.05, t  / 2,v

 t 0.025,10  2.228

Step 2: The value of  0 =0.4911, MSE = 1.1348, x 2 ˆ

Step 3:  0 ˆ

 t  / 2,v

 1785.06 and Sxx = 844.25

 1 x 2     0   0  t  / 2,v MSE   n Sxx   ˆ

  

 1 x 2   MSE   n Sxx  

1 1785 .06    1  1785 .06     0.4911  2.228  1.1348     0    12 844 . 25  12 844.25      

0.4911  2.228 1.1348

0.4911  3.5185   0

 0.4911  3.5185

 3.0274   0  4.0096 Step 4 : We are 95% confident that on average, the mean moisture is between -30274 and

4.0096 percent.

Example 12

Based on the Example 4, find the 90% confidence interval for the population’s intercept,  0 .

Answer Example 12

Step 1 : n = 12, v  n  2  12  2  10,

  0.10, t  / 2,v

 t 0.05,10  1.812

Step 2: The value of  0 =31.609, MSE = 73.64736 x 2 ˆ

277

 5064.69 and Sxx = 3445.67


Step 3:  0 ˆ

 t  / 2,v

  

 1 x 2     0   0  t  / 2,v MSE   n Sxx   ˆ

 1 x 2   MSE   n Sxx  

 1  5064 .69      31.609  1.812  1  5064 .69    73 . 64736    0  12 3445 . 67  12 3445 .67      

31.609  1.812 73.64736 

31.609  19.3799   0 12.2291   0

 31.609  19.3799  50.9889

Step 4 : We are 90% confident that on average, the mean scores in final examination is between 12.2291 and 50.9889.

Exercise 6.5.2

1.


2.

3.

 0.5 at the 0.1 level of significance. 1



2



6.

 1 at the 0.01 level of significance.


5.

 1


4.



1


278

1

 0.5


7.


8.

9.

11.

12.

13.

14.

15.

1



1



1



1



1



1



1



10.



1


279

1


Answer Exercise 6.5.2

1.

T test = 3.9466, reject H 0.

2.


3.


4.


5.


6.


7.


8.


9.

T test =-0.0669 , do not reject H 0.

10.


11.

T test =2.0166 , do not reject H 0.

12.


13.


14.


15.


6.6 Coefficient of Determination

Definition 7

The coeff ici ent of deter min ation measures the variation of the dependent variable that is explained by the regression line and the independent variable, x. the symbol for the coefficient of determination is r 2.

Theory 8

If ( xi , yi ); 1,2,...n are the values of a random sample from a bivariate population,

280


then 2

r



Syy  SSE Syy

 1

SSE Syy

. Notes that r 2 is always between 0 and 1, because r

(correlation coefficient) is between -1 and +1. In simple linear regression, it may also be computed as the square of the coefficient of correlation, r .

Example 13

Refer to data in Example 5, find and interpret the coefficient of determination.

Answer Example 13

From the example 5, Sxy  230, Sxx  844.25 , Syy  74 and SSE  11.348 , therefore the coefficient of determination is r 2

 1

SSE Syy

 1

11.348 74

 0.8466 .

85% of the total variation is explained by the regression line using the independent variable. Example 14

Refer to data in Example 6, find and interpret the coefficient of determination.

Answer Example 14

From the Example 6, Sxy  2004, Sxx  3445 .67 , Syy  1902 and SSE  736.4736 , therefore the coefficient of determination is r 2

 1

SSE Syy

 1

736.4736 1902

 0.6128 .

61% of the total variation is explained by the regression line using the independent variable. Exercise 6.7

1.

From the Exercise 6.2(1), find and interpret the coefficient of determination.

2.


3.


281


4.


5.


6.


7.


8.


9.


10.

From the Exercise 6.2(10) , find and interpret the coefficient of determination.

11.


12.


13.


14.


15.


Answer Exercise 6.7

1.

r 2

 0.6922

2.

r 2

3.

r 2

4.

r 2

 0.5803  0.9812  0.9488

5.

r 2

 0.9522

6.

r 2

 0.974

7.

r

2

8.

r 2

 0.712  0.061

9.

r

10.

r 2

 0.959  0.700

11.

r 2

 0.897

12.

r 2

 0.906

13.

r 2

 0.858

14.

r 2

 0.915

15.

r 2

 0.916

2

282


6.7 Coefficient of Pearson Correlation

Definition 8

A correlation exists between two variables when one of them is related to the other in some way. The coefficient of Pearson correlation measures the strength and direction of a linear relationship between the two variables. The symbol for the sample Pearson correlation coefficient is r . The symbol for the population correlation coefficient is  . Theory 9

If ( xi , yi ); 1,2,...n are the values of a random sample from a bivariate population, then r 

Sxy Sxx  Syy

. When r is between 0 to

 0.5, the correlation between the variables is

positively weak or negatively weak. But when r is between

 0.5 to  1, the

correlation between the variables is positively strong or negatively strong. There is no correlation between the variables if r = 0. Example 15

Refer to data in Example 5, find and interpret the Pearson correlation coefficient.

Answer Example 15

From the Example 5, Sxy  230, Sxx  844.25 and Syy  74 , therefore the Pearson correlation coefficient is r 

Sxy Sxx  Syy



230 (844.25)(74)

 0.9202 .

A Pearson correlation coefficient of 0.9202 indicates a strong positive linear relationship between the variables.

283


Example 16

Refer to data in example 6, find and interpret the Pearson correlation coefficient.

Answer Example 16

From the Example 6, Sxy  2004, Sxx  3445.67 and

Syy  1902 , therefore the

Pearson correlation coefficient is r 

Sxy Sxx  Syy



2004 (3445.67)(1902 )

 0.7828 .

A Pearson correlation coefficient of 0.7828 indicates a strong positive linear relationship between the variables.

Exercise 6.8

1.

From the Exercise 6.2(1), find and interpret the Pearson correlation coefficient.

2.


3.


4.


5.


6.


284


7.


8.


9.


10.

From the Exercise 6.2(10) , find and interpret the Pearson correlation coefficient.

11.


12.


13.


14.


15.


285


Answer Exercise 6.8

1.

r  0.832

2.

r  0.7618

3.

r  0.9905

4.

r  0.9742

5.

r  0.9759

6.

r  0.987

7.

r  0.844

8.

r  0.248

9.

r  0.979

10.

r  0.837

11.

r  0.947

12.

r  0.952

13.

r  0.926

14.

r  0.957

15.

r  0.957

EXERCISE CHAPTER 6 1.

The table shows the elongation (in thousands of an inch) of steel rods of nominally the same composition and diameter when subjected to various tensile forces (in thousands of pounds).

(a)

Force ( x)

Elongation ( y)

1.2

15.6

5.3

80.3

3.1

39.0

2.2

34.3

4.1

58.2

2.6

36.7

6.5

88.9

8.3

111.5

7.6

99.8

4.9

65.7

Assuming a linear relationship, use the least-squares method to find the regression coefficients of  0 and  1.

(b)

Interpret the meaning of the slope  1 in this problem.

(c)

Predict the elongation of steel rods when the various tensile forces are

286


5000 pounds. (d)

Find the coefficient of determination and coefficient of Pearson correlation. Interpret the results.

2.

The owner of MSR Enterprise would like to study the effect of number of sold cars (in 1000 units) on economy growth per year (in %) as stated on table below. Economy

1.3

1.8

2.5

3.5

4.8

6.5

7.7

1.2

1.5

1.8

2.3

2.2

2.5

2.7

Growth No. of Sold Cars

(a)

Find the Pearson’s correlation coefficient between economy growth X , and the number of sold cars Y . Interpret your result.

(b)

Obtain the linear regression model on number of sold cars against economy growth.

(c)

3.

Predict the number of sold cars in which the economy growth is 6 %.

During the harvest season in Malaysia, paddies are sold in large quantities at farm. One researcher wanted to study a relationship between calcium and the yield of paddy. To determine whether this was really true, a sample of 7 plots of paddy was measured for the weight of calcium and the weight of paddy. The following results shown table below. Calcium (mg)

Weight (kg/1m2)

50

2.2

55

3.0

54

2.5

52

2.7

37

1.5

287


(a)

52

2.0

53

2.5

Assuming a linear relationship, use the least-squares method to find the regression coefficients  0 and  1.

(b)

Interpret the meaning of the slope  1 in this problem.

(c)

Predict the weight for paddy where the paddy plot consists of 60 mg calcium.

4.

Crickets make a chirping sound with their wing covers. Scientists have recognized that there is relationship between the frequency of chirps and the temperature. 15 data had been observe from the study, are as below : Chirps, x

20

16

19.8

18.4

17.1

15.5

14.7

17.1

Temperature, y

88.6

71.6

93.3

84.3

80.6

75.2

69.7

82

Chirps, x

15.4

16.3

15

17.2

16

17

14.4

Temperature, y

69.4

83.3

79.6

82.6

80.6

83.5

76.3

(a)

sketch a scatter plot for the data above.

(b)

use the method of least squares to estimate the regression line. Interpret the result.

(c)

predict the temperature when x = 15 chirps per second.

(d)

test

the

null

hypothesis

 1  3

against

the

alternative

hypothesis  1  3 at the 0.01 level of significance.

5.

An engineer conducted a study to determine whether there is a linear relationship between the breaking strength, y, of wooden beams and the specific gravity, x, of the wood. Ten randomly selected beams of the same crosssectional dimensions were stressed until they broke. The breaking strength and the specific gravity of the wood

288


are shown in table below for each of the ten beams. Beam

Breaking strength, y

Specific gravity, x

1

11.14

0.499

2

12.74

0.558

3

13.13

0.604

4

11.51

0.441

5

12.38

0.550

6

12.60

0.528

7

11.13

0.418

8

11.70

0.480

9

11.02

0.406

10

11.41

0.467

(a)

Construct a scatter plot of the data.

(b)

Assuming the relationship between the variables is best described by a straight line, y   0  1 x, use the method of least squares or maximum likelihood to estimate the value of y-intercept,  0 and slope ˆ

of the line, 1 . Interpret the results. ˆ

(c)

Estimate the average of breaking strength when specific gravity is 0.455.

(d)

Test the hypothesis H 1 :  1

 0 by taking level of significance, α =

0.05. (e)

Find the correlation coefficient r , and coefficient of determination, r 2 and then interpret the results.

6.

An officer wants to study the relationship between biomass productions of orange and cumulative intercepted solar radiation (Wh/m2) over a six-week period following emergence. Biomass production is the mean dry weight in grams of independent samples of four plants which is collected at XY Plantation. The data of this study are

289


shown in table below. Solar Radiation ( X )

Plant Biomass (Y )

28.8

15.8

48.5

48.2

68.3

71.1

90.5

95.7

120.2

150.4

170.5

210.5

(a)

Sketch the scatter diagram for the above data.

(b)

Compute  0 and  1 for the linear regression of plant biomass on ˆ

ˆ

intercepted solar radiation. Write the regression equation and interpret the result. (c)

Calculate the sample correlation coefficient, r and interpret your result.

(d)

Predict the plant biomass for 300 Wh/m2 solar radiations.

(e)

Test the null hypothesis,  1 = 0 against the alternative hypothesis,

 1 ≠ 0 at 5% level of significance.

7.

Thermal conductivity of a material is the quantity of heat, transmitted through a thickness in a direction normal to a surface of area. The thermal conductivity is due to a temperature gradient under a steady state conditions. The materials with high thermal conductivities are good conductors of heat, whereas materials with low thermal conductivities are good thermal insulator. A test has been conducted to investigate the relationship between thickness of a material (millimeter) and the thermal conductivity of the material (Watt per meter Kelvin). Assume that there is a linear relationship between the thermal conductivity of a material and the thickness of the material. Seven materials are chosen at random where the pressure and temperature are at normal rate. The thicknesses of 7 materials are measured and the thermal conductivity of each material is recorded as shown in the table below.

290


Thickness ( x)

21

26 28

31

25

19

35

Thermal Conductivity ( y)

12

16 19

21

14

11

24

7

 x i 1

7

i

7

7

7

 185,  yi  117,  x  5073,  y  2095,  xi yi  3250 2 i

i 1

2 i

i 1

i 1

i 1

(a)

Plot the data on a scatter diagram.

(b)

Estimate the regression line by using the method of least square. Interpret your result.

(c)

Estimate the average of the thermal conductivity if the thickness of a material is 29mm.

(d)

8.

Calculate the coefficient of correlation r and r 2. Interpret their values.

From the past experience, a certain type of plastic indicates that a relation exists between the mean hardness (measured in Brinell units) of items molded from the plastic ( Y ) and the elapsed time (hours) since termination of the molding process ( X ). Twelve batches of the plastic were made, and from each batch one test item was molded and the hardness measured at some specific point in time. The results are shown in following table.

Batch

1

2

3

4

5

6

7

8

9

10

11

12

X

32

48

72

64

48

16

40

48

48

24

80

56

Y

230

262

323

298

255

199

248

279

267

214

359

305

(a)

Draw a scatter plot.

(b)

Find the estimated regression line by using the least square method. Interpret the result.

(c)

Estimate the mean hardness when the elapsed time is 48 hours. Interpret the result.

(d)

Calculate the coefficient of correlation, r and coefficient of determination. Interpret these results.

291


9.

Zaiton wishes to buy a car. She read a news paper to find the price of the used car for a local compact car. The data of the age (in years) and the prices (RM in thousand) are shown in table below.

Age (x) Price (y)

1

2

33.4 29.3

3

4

29.0 28.1

5

6

27.5 26.0

7

8

9

24.2 19.5

10

14.7 14.0

11

13.4 13.0

(a)

Sketch a scatter plot for the data.

(b)

Use the method of least squares to estimate the regression line. Interpret the results.

10.

12

(c)

Test the slope, β 1= -1 at 5% level of significance.

(d)

Estimate the car price when the cars are 14 years old.

Consider the following data for 10 such samples.

Soil Sample

Strontium Distribution Coefficient

Total Aluminium

1

100

200

2

120

225

3

300

325

4

250

310

5

400

350

6

500

400

7

450

375

8

445

385

9

310

350

10

200

290

Let Y represent the strontium distribution coefficient and X represent the total aluminium.

292


11.

(a)

Find the equation of the line of best fit.

(b)

Find a 95% confidence interval of  1 .

Suppose a fire insurance company wants to relate the amount of fire damage in major residential fires to the distance between the burning house and the nearest fire station. The study is to be conducted in a large suburb of a major city. A sample of 10 recent fires in this suburb is selected. The distance between the fire and the nearest fire station, x, and the amount of damage, y, are recorded for each fire. The results are given in the table below. Distance from Fire Station x (miles) 3.4 1.8 4.6 2.3 3.1 5.5 0.7 3.0 2.6 4.3

Fire Damage y (thousands of dollars) 26.2 17.8 31.3 23.1 27.5 36.0 14.1 22.3 19.6 31.3

(a)


(b)

Find and interpret the coefficient of determination and the Pearson correlation coefficient.

(c)

Find the regression line using the least squares method. Interpret the result.

(d)

Test the hypothesis concerning H 0 :  1

 5 against the H 1 :  1  5 at

the 0.05 level of significance.

12.

A manager of a car dealership believes that there is a relationship between the number of salespeople on duty and the number of cars sold in a week. The following data in table is used to develop a simple regression model.

293


Week

Number of Sales People, x

Number of Cars Sold, y

1

6

79

2

6

64

3

4

49

4

2

23

5

3

52

n



xi

 21 ,

n



y i

 267 ,

i 1

i 1

n



xi yi

 1256 ,

n



xi

i 1

i 1

2

 101 ,

n

 y

2 i

 15971

i 1

(a)


(b)

Calculate the sample correlation coefficient and interpret the result.

(c)

By using the least square method, estimate the regression line. Interpret the result.

(d)

Estimate the number of sales people when the number of cars sold is 41. Interpret the result.

(e)

Test the slope whether it is greater than ten at 5% level of significance.

ANSWER EXERCISE CHAPTER 6

1.

(a) 2.1978 + 13.2756 x, (c) 66380.1978, (d) r = 0.9939, r 2 = 0.9878

2.

(a) 0.9319, (b) y = 1.1841 + 0.2104 x, (c) 2446

3.

(a) y

4.

(b) y  24 967  3 3057 x , (c) 74.5524 , (d) T = 0.512, do not reject H 0

5.

(b) y

ˆ

ˆ

 1.0913  0.0681 x (c) 2.9947

ˆ

ˆ

.

.

 6.47 + 10.901 x, (c) 11.43, (d) T = 1.824, do not reject H 0,

(e) r = 0.913, r 2= 0.834 6.

(b) y  22.372  1.378 x , (c) r = 0.9977, (d) 391.028, (e) T = 30.144, ˆ

reject H 0 7.

(b) y  5.9958  0.8593 x , (c) 18.9239, (d) r = 0.9863, r 2 = 0.9728

8.

(b) y  153.915  2.416 x , (c) 269.883 , (d) r = 0.97942 , r 2

9.

(b) y

  

 0.95926

 35.5225  1.9766 x, (c) T = -6.2516, reject H 0 , (d) 7.8505

294


10.

(a) y  348.3351  2.0431x, (b) 1.6063   1  2.4799

11.

(b) r 2  0.9380 , r  0.9685 , (c) y



ˆ

 10.250  4.6868 x , (d) T =  0.7354 , do

not reject H0. ^

12.

 10.515625 x , (d) 3.2258, (e) T = 0.1852, do (b) r = 0.9089, (c) y  9.234375

not reject H0. SUMMARY CHAPTER 6

1.

Simple Linear Regression Model (i)

Least Squares Method y   0

The model :

 1  ˆ

Sxy Sxx

ˆ

ˆ

  1 x ˆ

(slope) and  0 ˆ

 y   1 x , ( y-intercept) where ˆ

 n  Sxy   ( xi  x )( yi  y )   xi yi    xi   yi  , n  i 1  i 1  i 1 i 1 n

Sxx 

n

n

 ( x  x )   x

2

2

i

i 1

Syy 

i

i 1

n

 ( y i 1

1  n

n

n

 y )   y i 2

i

i 1

1  n

2

    xi  , n  i1  2

1  n

    yi  n  i 1 

2

and n = sample size

2.

Inference of Regression Coefficients (i)

Slope SSE = Syy   1 S xy ˆ

(ii)

, MSE 

Intercept

 0   C ˆ

T test =

MSE (1 / n  x / Sxx) 2

295

SSE n2

 1   C ˆ

,

T test =

MSE S xx


3.

Confidence Intervals of the Regression Line Slope,  1

(i)

 1  t  / 2,v MSE / Sxx   1   1  t  / 2,v MSE / Sxx , ˆ

ˆ

where v = n-2 Intercept,  0

(ii)

 0 ˆ

 t  / 2,v

 1 x 2     0   0  t  / 2,v MSE   n Sxx   ˆ

where v = n-2

4.

Coefficient of Determination, r 2. 2

r

5.



Syy  SSE Syy

1

SSE Syy

Coefficient of Pearson Correlation, r . Sxy r  Sxx  Syy

296

 1 x 2   , MSE   n Sxx  


CORRECTION PAGE CHAPTER 6

297

Chapter 6

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