CHAPTER 6_correlation and Regression

8/3/2019 CHAPTER 6_correlation and Regression

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CHAPTER 6 : CORRELATION - REGRESSION

6.1 Introduction

So far we have considered only univariate distributions. By the averages, dispersion and

skewness of distribution, we get a complete idea about the structure of the distribution. Many atime, we come across problems which involve two or more variables. If we carefully study thefigures of rain fall and production of paddy, figures of accidents and motor cars in a city, of

demand and supply of a commodity, of sales and profit, we may find that there is some

relationship between the two variables. On the other hand, if we compare the figures of rainfall

in America and the production of cars in Japan, we may find that there is no relationship betweenthe two variables. If there is any relation between two variables i.e. when one variable changes

the other also changes in the same or in the opposite direction, we say that the two variables are

correlated.

W. J. King : If it is proved that in a large number of instances two variables, tend always to

fluctuate in the same or in the opposite direction then it is established that a relationship existsbetween the variables. This is called a "Correlation."

6.2 Correlation

It means the study of existence, magnitude and direction of the relation between two

or more variables. in technology and in statistics. Correlation is very important. The

famous astronomist Bravais, Prof. Sir Fancis Galton, Karl Pearson (who used this

concept in Biology and in Genetics). Prof. Neiswanger and so many others have

contributed to this great subject

6.3 Types of Correlation

1. Positive and negative correlation2. Linear and non-linear correlation

A) If two variables change in the same direction (i.e. if one increases the other also

increases, or if one decreases, the other also decreases), then this is called a positive

correlation. For example : Advertising and sales.

B) If two variables change in the opposite direction ( i.e. if one increases, the other

decreases and vice versa), then the correlation is called a negative correlation. Forexample : T.V. registrations and cinema attendance.

1. The nature of the graph gives us the idea of the linear type of correlationbetween two variables. If the graph is in a straight line, the correlation is called

a "linear correlation" and if the graph is not in a straight line, the correlation

is non-linear orcurvi-linear.


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For example, if variable x changes by a constant quantity, say 20 then y also changes

by a constant quantity, say 4. The ratio between the two always remains the same (1/5

in this case). In case of a curvi-linear correlation this ratio does not remain constant.

6.4 Degrees of Correlation

Through the coefficient of correlation, we can measure the degree or extent of the

correlation between two variables. On the basis of the coefficient of correlation we

can also determine whether the correlation is positive or negative and also its degree

or extent.

1. Perfect correlation: If two variables changes in the same direction and in thesame proportion, the correlation between the two is perfect positive.

According to Karl Pearson the coefficient of correlation in this case is +1. On

the other hand if the variables change in the opposite direction and in the same

proportion, the correlation is perfect negative. its coefficient of correlation is -1. In practice we rarely come across these types of correlations.

2. Absence of correlation: If two series of two variables exhibit no relationsbetween them or change in variable does not lead to a change in the other

variable, then we can firmly say that there isno correlation or absurd

correlation between the two variables. In such a case the coefficient of

correlation is 0.

3. Limited degrees of correlation: If two variables are not perfectly correlated or is there aperfect absence of correlation, then we term the correlation as Limited correlation. It maybe positive, negative or zero but lies with the limits 1.

4. High degree, moderate degree or low degree are the three categories of thiskind of correlation. The following table reveals the effect ( or degree ) ofcoefficient or correlation.

Degrees Positive Negative

Absence of correlation Zero 0

Perfect correlation + 1 -1

High degree + 0.75 to +1

- 0.75 to -1

Moderate degree + 0.25 to +

0.75

- 0.25 to -

0.75


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Low degree 0 to 0.25 0 to - 0.25

6.5 Methods Of Determining Correlation

We shall consider the following most commonly used methods.(1) Scatter Plot (2)Kar Pearsons coefficient of correlation (3) Spearmans Rank-correlation coefficient.

1) Scatter Plot ( Scatterdiagramor dot diagram ): In this method the values of the

two variables are plotted on a graph paper. One is taken along the horizontal ( (x-axis)

and the other along the vertical (y-axis). By plotting the data, we get points (dots) on

the graph which are generally scattered and hence the name Scatter Plot.

The manner in which these points are scattered, suggest the degree and the direction

of correlation. The degree of correlation is denoted by r and its direction is given

by the signs positive and negative.

i) If all points lie on a rising straight line the correlation is

perfectly positive and r = +1 (see fig.1 )

ii) If all points lie on a falling straight line the correlation is

perfectly negative and r = -1 (see fig.2)

iii) If the points lie in narrow strip, rising upwards, the

correlation is high degree of positive (see fig.3)

iv) If the points lie in a narrow strip, falling downwards, the

correlation is high degree of negative (see fig.4)

v) If the points are spread widely over a broad strip, rising

upwards, the correlation is low degree positive (see fig.5)

vi) If the points are spread widely over a broad strip, falling

downward, the correlation is low degree negative (see

fig.6)

vii) If the points are spread (scattered) without any specific

pattern, the correlation is absent. i.e. r = 0. (see fig.7)
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Though this method is simple and is a rough idea about the existence and the degree of

correlation, it is not reliable. As it is not a mathematical method, it cannot measure the degree ofcorrelation.

2)Karl Pearsons coefficient of correlation: It gives the numerical expression for the measure

of correlation. it is noted by r . The value of r gives the magnitude of correlation and signdenotes its direction. It is defined as

r =

where

N = Number of pairs of observation

Note : r is also known as product-moment coefficient of correlation.

OR r =

OR r =

Now covariance of x and y is defined as

Example Calculate the coefficient of correlation between the heights of father and his

son for the following data.


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Heightof

father

(cm):165 166 167 168 167 169 170 172

Heightof son

(cm):

167 168 165 172 168 172 169 171

Solution: n = 8 ( pairs of observations )

Height of

father

xi

Height of

son

yi

x

=

xi-

x

y =

yi-yxy x2 y2

165 167 -3 -2 6 9 4

166 168 -2 -1 2 4 1

167 165 -1 -4 4 1 16

167 168 -1 -1 1 1 1

168 172 0 3 0 0 9

169 172 1 3 3 1 9

170 169 2 0 0 4 0

172 171 4 2 8 16 4

xi=1344 yi=1352 0 0 xy=24 x2=36 y2=44

Calculation:

Now,


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Since r is positive and 0.6. This shows that the correlation is positive and moderate

(i.e. direct and reasonably good).

Example From the following data compute the coefficient of correlation between x

and y.

Example If covariance between x and y is 12.3 and the variance of x and y are 16.4

and 13.8 respectively. Find the coefficient of correlation between them.

Solution: Given - Covariance = cov ( x, y ) = 12.3

Variance of x ( x2 )= 16.4


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Variance of y (y2 ) = 13.8

Now,

Example Find the number of pair of observations from the following data.

r = 0.25, (xi - x ) ( yi - y ) = 60, y = 4, ( xi - x )2 = 90.

Solution: Given - r = 0.25


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If the values of x and y are very big, the calculation becomes very tedious and if we

change the variable x to u = and y to where x0 and y0 are the

assumed means for variable x and y respectively, then rxy= ruv

The formula for r can be simplified as

Example Marks obtained by two brothers FRED and TED in 10 tests are as follows:

Find the coefficient of correlation between the two.

Solution: Here x0 = 60, c = 4, y0 = 60 and d = 3


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Calculation:

6.6 Coefficient Of Correlation For Bivariate Grouped Data

When the number of observations is very large, we need to arrange the data into

different classes, which are either discrete or continuous. Items having values falling

in a particular class are placed together and those having values falling in another

class are placed together. Due to this the whole data is divided into horizontal rows

and vertical columns, with one variable placed horizontally and the other placedvertically. The table so obtained is a two-way frequency distribution table and is

called the correlation table or Bi-variate frequency distribution table. The formula for

calculating and for bi-variate distribution is given by

STEPS:

1. First write down the mid-points of x along a horizontal raw and those of yalong a vertical column.

2. Find


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3. Multiply each frequency by the corresponding value of u then by correspondingvalue of v to get fuv. Write these numbers in the same box at the top.

4. Add the frequencies horizontally, and write down the total. Similarly add thefrequencies vertically and write down its total.

5. Multiply this additions of x by u to get f u.6. Multiply this addition of y by v to get f v.7. Multiply these frequencies by the square of u to get f u2.8. Multiply these frequencies by the square of v to get f v2.9. Add horizontally ( or vertically ) the top numbers denoting f u v written in each

box ( or cell )

10.Write down f u, f u2, f v, f v2 and f u v and then use the aboveformula.

Example Calculate the coefficient of correlation for the following data.

Age

(years)

of

Husband

Age (years) of wife

Total10 -20 20 -30 30 -40 40 -50 50 -60

10 - 25

25 - 35

35 - 45

45 - 55

55 - 65

5

3

3

15

11

11

14

7

7

12

3

3

6

8

29

32

22

9

Total 8 29 32 22 9 100


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Inserting, fuv = 94, n = 100, fu = -5, fv = -5, fu2 = 119 and fv2 = 119 in

6.7 ProbableError


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It is used to help in the determination of the Karl Pearsons coefficient of correlation r . Due to this r is corrected to a great extent but note that r depends on the

random sampling and its conditions. it is given by

P. E. = 0.6745

i. If the value of r is less than P. E., then there is no evidence of correlation i.e. ris not significant.

ii. If r is more than 6 times the P. E. r is practically certain .i.e. significant.iii. By adding or subtracting P. E. to r , we get the upper and Lower limits

within which r of the population can be expected to lie.

Symbolically e = r P. E.

P = Correlation ( coefficient ) of the population.

Example If r = 0.6 and n = 64 find out the probable error of the coefficient of correlation.

Solution: P. E. = 0.6745

= 0.6745

=

= 0.57

6.8 Spearmans Rank Correlation Coefficient

This method is based on the ranks of the items rather than on their actual values. Theadvantage of this method over the others in that it can be used even when the actual

values of items are unknown. For example if you want to know the correlation

between honesty and wisdom of the boys of your class, you can use this method by

giving ranks to the boys. It can also be used to find the degree of agreements between

the judgements of two examiners or two judges. The formula is :
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R =

where R = Rank correlation coefficient

D = Difference between the ranks of two items

N = The number of observations.

Note: -1 R 1.

i) When R = +1 Perfect positive correlation or complete

agreement in the same direction

ii) When R = -1

Perfect negative correlation or completeagreement in the opposite direction.

iii) When R = 0 No Correlation.

Computation:

i. Give ranks to the values of items. Generally the item with the highest value isranked 1 and then the others are given ranks 2, 3, 4, .... according to their

values in the decreasing order.

ii.

Find the difference D = R1 - R2where R1 = Rank of x and R2 = Rank of y

Note that D = 0 (always)

iii. Calculate D2 and then find D2iv. Apply the formula.Note :

In some cases, there is a tie between two or more items. in such a case each items

have ranks 4th and 5th respectively then they are given = 4.5th rank. If three

items are of equal rank say 4th then they are given = 5th rank each. If m be

the number of items of equal ranks, the factor is added to S D2. If there

are more than one of such cases then this factor added as many times as the number of

such cases, then
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Example Calculate R from the following data.

Student No.: 1 2 3 4 5 6 7 8 9 10

Rank in

Maths :

1 3 7 5 4 6 2 10 9 8

Rank in

Stats:

3 1 4 5 6 9 7 8 10 2

Solution :

Student

No.

Rank

inMaths

(R1)

Rank

inStats

(R2)

R1 - R2D

(R1 - R2 )2

D2

1 1 3 -2 4

2 3 1 2 4

3 7 4 3 9

4 5 5 0 0

5 4 6 -2 4

6 6 9 -3 9

7 2 7 -5 25

8 10 8 2 4

9 9 10 -1 1

10 8 2 6 36

N = 10 S D = 0 S D2 = 96

Calculation of R :
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Example Calculate R of 6 students from the following data.

Marks

in Stats :40 42 45 35 36 39

Marks

inEnglish

:

46 43 44 39 40 43

Solution:

Marks

in

Stats

R1

Marks

in

English

R2 R1 - R2 (R1 -R2)2=D

2

40 3 46 1 2 4

42 2 43 3.5 -1.5 2.25

45 1 44 2 -1 1

35 6 39 6 0 0

36 5 40 5 0 0

39 4 43 3.5 0.5 0.25

N = 6 S D = 0 S D2

= 7.50

Here m = 2 since in series of marks in English of items of values 43 repeated twice.
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Example The value of Spearmans rank correlation coefficient for a certain number ofpairs of observations was found to be 2/3. The sum of the squares of difference

between the corresponding rnks was 55. Find the number of pairs.

Solution: We have

6.9Linear Regression

Correlation gives us the idea of the measure of magnitude and direction betweencorrelated variables. Now it is natural to think of a method that helps us in estimating

the value of one variable when the other is known. Also correlation does not imply

causation. The fact that the variables x and y are correlated does not necessarily mean

that x causes y or vice versa. For example, you would find that the number

ofschoolsin a town is correlated to the number of accidents in the town. The reason

for these accidents is not the school attendance; but these two increases what is known

as population. A statistical procedure called regression is concerned with causation

in a relationship among variables. It assesses the contribution of one or more variable

calledcausing variable or independent variable or one which is

beingcaused(dependent variable). When there is only one independent variable thenthe relationship is expressed by a straight line. This procedure is called simple linear

regression.

Regression can be defined as a method that estimates the value of one variable when

that of other variable is known, provided the variables are correlated. The dictionary

meaning of regression is "to go backward." It was used for the first time by Sir
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Francis Galton in hisresearchpaper "Regression towards mediocrity in hereditary

stature."

Lines of Regression: Inscatter plot, we have seen that if the variables are highly

correlated then the points (dots) lie in a narrow strip. if the strip is nearly straight, we

can draw a straight line, such that all points are close to it from both sides. such a linecan be taken as an ideal representation of variation. This line is called the line of best

fit if it minimizes the distances of all data points from it.

This line is called the line of regression. Now prediction is easy because now all we

need to do is to extend the line and read the value. Thus to obtain a line of regression,

we need to have a line of best fit. But statisticians dont measure the distances bydropping perpendiculars from points on to the line. They measure deviations (

orerrorsor residuals as they are called) (i) vertically and (ii) horizontally. Thus we

get two lines of regressions as shown in the figure (1) and (2).

(1) Line of regression of y on x

Its form is y = a + b x

It is used to estimate y when x is given

(2) Line of regression of x on y

Its form is x = a + b y

It is used to estimate x when y is given.

They are obtained by (1) graphically - by Scatter plot (ii)

Mathematically - by the method of least squares.

ii. Let y = a + b y ..... (1) where a and b are given by the normal equations y = n a + b x ..... (2)

xy = a x + b x2

.... (3)

where n be the number of pairs of values of x and y.
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Equation (6) is the equation of the line of regression of y on x.


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is called the coefficient of regression of y on x which is obviously the

slope of this line. Interchanging x and y in equation (6), the equation of the line of

regression of x and y is given by

Naturally bxy is the slope of this line which is equal to


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Example A panel of two judges A and B graded dramatic performance by

independently awarding marks as follows:

Solution:

The equation of the line of regression of y on x

Inserting x = 38, we get


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y - 33 = 0.74 ( 38 - 33 )

y - 33 = 0.74 5

y - 33 = 3.7

y = 3.7 + 33

y = 36.7 = 37 ( approximately )

Therefore, the Judge B would have given 37 marks to 8th performance.

Example The tworegression equationsof the variables x an y are

x = 19.13 - 0.87 y and y = 11.64 - 0.50 x

Find (1) Mean of xs

(2) Mean of ys

(3) Correlation coefficient between x and y

Solution:

1. Calculation of Mean

\Mean of xs = 15.94 and Mean of ys = 3.67

2.Calculation of r


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x = 19.93 - 0.87 y

Therefore,

and y = 11.64 - 0.50 x

Therefore,

From (3) and (4)

r = 0.66

But regression coefficient are negative

r = - 0.66

Example In a partially destroyed laboratory record of an analysis of correlation data,

the following results are legible:

Variance of x = 9Regression equations : 8 x - 10 y + 66 = 6

40 x - 18 y = 214

What are (1) Means of xs and ys (2) the coefficient of correlation between x and y(3) the standard deviation of y ?

Solution:

1. Means:8 x - 10 y = -66 ----- (1)

40 x - 18 y = 214 ----- (2)

Solving (1) and (2) as


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40 x - 50 y = -330 ----- (1)

40 x - 18 y = 214 ----- (2)

-32 y = -544

y = 17

Mean of ys 17

Substituting y = 17 in (1) we get 8x - 10 17 = -66

or 8x = 104 x = 13

Mean of xs = 13

2.Coefficient of correlation between x and y

40 x = 18 y + 214

Also -10 y = - 8 x - 66

Therefore,

3. Standard deviation of y Variance of x i.e. x2 = 9 x = 3

Now byx =


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y = 0.4

Example From 10 observations of price x and supply y of a commodity the results

obtained x = 130, y = 220, x2 = 2288, xy = 3467

Compute the regression of y on x and interpret the result. Estimate the supply whenthe price of 16units.

Solution: The equation of the line of regression of y on x

y = a + b x

Also from normal equations

y = n a + b x and xy = a x + b x2

we get

220 = 10 a + 130 b (1)

3467 = 130 a + 2288 (2)

Solving (1) and (2) as

2860 = 130 a + 1690 b

3467 = 130 a + 2288 b

On subtraction

607 = 598 b b = 1.002

Putting b = 1.002 in 220 = 10 a + 130 b, we get a = 8.974.

Hence the 3 equation of the line of regression of y on x is

y = 8.974 + 1.002 x

When x = 16, we get

y = 8.974 + 1.002 ( 16 )

y = 25.006

Example If is the acute angle between the two regression lines in the case of two

variables x and y show that


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P 25 f 25

with usual meanings. Explain the significance when r = 0 and r = 1.

Solution: The slopes of the two regression lines are

If r = then tan = or = /2 i.e. there is no relationship between two variables i.e.

independent or uncorrelated.

If r = 1 then = 0. The two regression lines are coincident or parallel and the

correlation is perfect.

**********

CHAPTER 6_correlation and Regression

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