What is the regression coefficient bxy from the following details x 7 3y 28.10 y 1.5 x 10

Paper 4 Fundamentals Of Business Mathematics Statistics

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PAPER 4: FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS (SYLLABUS 2012) _ MCQ 35. If 0.5 of A = 0.6 of B = 0.75 of C and A+B+C = 60, then the number which is to be added to A so that the result of this addition and B, C will be in continued proportion, is: (a) 1 (b) 2 (c) 3 (d) 4 36.

today we are going to discuss one more question the question is that find the regression Coefficient b y x sin BX why when the lines of regression R 6 minus 5 Y + 10 = 20 and 10 x minus 3 Y - 28 equal to zero line of regulations are 6 x minus 5 Y + 10 equal to zero and x minus 3 Y - 28 equal to zero so here is our equation for and here is our equation so regression equation first from equation fast the regression the recreation line of Y on X is

minus 5 y equals to - 10 - 6 x 10 can be written as 5 Y is equals to 10 + 6x here so why can be written as 10 by 5 + 6 by 5x so we can write it as y = 26 by 5 x + 2 here so so b y x = 26 by 5 here again we have to find b x on Y from equation second from equation second the regression degree creation line of x on Y

is 10 x minus 3 Y - 28 = 20 can be written as X equals to 10 X equals to C white + 28 x can be written as 3 B N Y + 28 Y plan so So B to b x y equals to three by ten year I hope you understand the explanation thanks for watching

Correlation and Association 6.59

2. If σ 2 = 6.25, σ2y = 4 and cov(x,y) = 0.9, then the coefficient of correlation
x

will be [C.U. B.Com. 2000, 2012, 2015 (G)] [Ans. (a)]

(a) 0.18 (b) 0.17 (c) 0.19 (d) 0.20

3. Given rxy = 0.8, if u = x + 5 and v = y − 5, then the value of ruv will be

(a) 0.85 (b) 0.7 (c) 0.75 (d) 0.8

[C.U. B.Com. 2011, 2014 (G)] [Ans. (d)]

4. The correlation coefficient of two variables x and y is 0.6 and their covariance
is 12. If the standard deviation of x be 5, the standard deviation of y will be

(a) 2 (b) 5 (c) 4 (d) 3

[C.U. B.Com. 2011] [Ans. (c)]

5. If rxy = 0.6, 4 = 2x − 3 and v = −3y + 2, then the value of ruv will be

(a) −0.55 (b) −0.6 (c) −0.5 (d) −0.45

[C.U. B.Com. 2014 (H)] [Ans. (b)]

6. If r = 0.4, cov(x, y) = 10 and σy = 5, then the value of σx will be

(a) 5 (b) 4 (c) 6 (d) 3

[C.U. B.Com. 2014 (G)] [Ans. (a)]

7. Karl Pearson’s coefficient of correlation between two variables x and y is 0.46,
their covariance is 3.68; if the variance of x is 16, then the standard deviation
of y will be

(a) 3 (b) 4 (c) 2 (d) 5

[C.U. B.Com. 2013 (H)] [Ans. (c)]

8. If rxy = 0.6, σx= 4 and σy = 5, then the value of cov(x,y) will be

(a) 9 (b) 10 (c) 11 (d) 12

[C.U. B.Com. 2015 (H)] [Ans. (d)]

9. If x = X − X and y = Y − Y, and ∑x2 = 10, ∑y2 = 24, ∑xy = 12, then the coefficient
of correlation between the two variables x and y will be

(a) 0.77 (b) 0.82 (c) 0.85 (d) 0.71

[C.U. B.Com. 1992] [Ans. (a)]

10. The correlation coefficient between x and y is 0.5 and u = 2x + 11 and v = 3y + 7,
then the correlation coefficient between u and v will be

(a) 0.3 (b) 0.5 (c) 0.6 (d) 0.4

[C.U. B.Com. 2008] [Ans. (b)]

11. If x = X − X, y = Y − Y, where X, Y being respectively arithmetic means of X

and Y and if ∑x2 = 60, ∑xy = 57, r = 0.95 and variance of y = 6 2 , then the value
3
of n will be

(a) 6 (b) 7 (c) 9 (d) 8

[C.U. B.Com. 2009] [Ans. (c)]

6.60 Business Mathematics and Statistics

12. From the following data the Karl Pearson coefficient of correlation is

x : 6 8 10 7 10 7 [Ans. (c)]
y : 12 10 8 12 8 10
(a) 0.97 (c) −0.93
(b) 0.85 (d) 0.65

13. From the following data the Karl Pearson coefficient of correlation is

x : 11 15 15 12 15 10

y : 18 13 11 15 11 16 [Ans. (b)]
(a) −0.99 (c) 0.80
(b) −0.89 (d) −0.50

14. Number of observations n = 10, mean of x =22, mean of y = 15, sum of squared
deviations of x from mean value = 148, sum of squared deviations of y from
mean value = 168, sum of multiplication of deviation of x and y = 124.

From the above details the coefficient of correlation will be

(a) 0.79 (c) 0.65

(b) 0.87 (d) 0.43 [Ans. A]

15. Sum of deviations of x from mean value = 8, sum of squared deviation of y
from mean value = 54. Sum of multiplication of deviation of x and y = 32. Sum
of squared deviations of x from mean value = 60.

From the above details the coefficient of correlation will be

(a) 0.58 (c) 0.61

(b) 0.56 (d) 0.47 [Ans. (b)]

16. If the coefficient of correlation is 0.8, the coefficient of determination

will be

(a) 0.98 (c) 0.66

(b) 0.64 (d) 0.54 [Ans. (b)]

[Hints: coefficient of determination = (coefficient of correlation)2]

17. If the coefficient of determination is 0.49, what is the coefficient of correlation.

(a) 0.7 (c) 0.9

(b) 0.8 (d) 0.6 [Ans. (a)]

18. If the coefficient of correlation between x and y is 2 and the standard deviations
3

of x is 3 and standard deviation of y is 4, the covariance between x and y will be

(a) 3 (b) 6 (c) 7 (d) 8 [Ans. (d)]

19. If the correlation is perfect then what is the value of r?

(a) 1 (b) 2 (c) 3 (d) 4 [Ans. (a)]

20. If there is no relation between two variables, then what will be the value of
coefficient of correlation of these two variables?

(a) 1 (b) 0 (c) −1 (d) 2 [Ans. (b)]

Correlation and Association 6.61

21. Given r = 0.8, ∑xy = 80, σx= 2, ∑y2 = 100, where x = X − X and y = Y − Y; the

number of items will be

(a) 12 (c) 20

(b) 15 (d) 25 [Ans. (d)]

22. If the two variables are independent of each other, then value of ‘r’ is

(a) 1 (b) 2 (c) 0 (d) 0.5 [Ans. (c)]

23. The value of the coefficient of correlation lies between

(a) ±1 (b) ±2 (c) ±3 (d) ±1.5 [Ans. (a)]

24. What is the covariance if the coefficient of correlation between x and y is 0.87
and the variance of x and y are 36 and 25 respectively

(a) 18.25 (b) 26.10 (c) 19.25 (d) 21.6 [Ans. (b)]

25. When we conduct a study that examines the relationship between two variables,
then we are working with

(a) Univariate data (c) Bivariate data

(b) Multivariate data (d) None of there [Ans. (c)]

Rank Correlation

26. If ∑d2 = 33 and n = 10, where d represents the difference between the rank of

two series and n is the number of pairs of observations, then the value of the

coefficient of rank correlations will be [C.U. B.Com. 1985]

(a) 0.68 (c) 0.75

(b) 0.7 (d) 0.8 [Ans. (d)]

27. The value of R (Spearman Rank correction coefficient) when ∑d2 = 30 and n =

10, is

(a) 0.75 (c) 0.82

(b) 0.65 (d) 0.9

[C.U. B.Com. 2001, 2012, 2014 (G), 2016 (G), 2016 (H)] [Ans. (c)]

28. The following are the ranks of 10 students in English and Maths

Sr. no. : 1 2 3 4 5 6 7 8 9 10
9 6 4 5 10 3 1 7 2 8
Rank in Math : 8 9 3 6 7 1 2 5 5 10

Rank in English :

The coefficient of rank correlation between the marks in Maths and English is

(a) 0.61 (c) 0.59

(b) 0.769 (d) 0.79 [Ans. (b)]

29. In case of tie in ranks, the rank of the tied scores are calculated by using

(a) Mean (c) Mode

(b) Median (d) S.D [Ans. (a)]

30. Differences of the ranks (d) are squared to remove

(a) deviations (c) positive values

(b) negative values (d) none of there [Ans. (b)]

6.62 Business Mathematics and Statistics

Association of Attributes

31. A qualitative characteristic is called (c) attribute [Ans. (c))
(a) constant (d) association
(b) variable

32. If an attribute has two classes, it is called [Ans. (b)]
(a) Trichotomy
(b) Dichotomy
(c) Simple classification
(d) Manifold classification

33. With two attributes A and B, the total number of ultimate frequencies is

(a) Two (c) Six

(b) Four (d) Nine [Ans. (b)]

34. If (AB) = (A)(B), the two attributes A and B are
N

(a) Independent (c) Correlated

(b) Dependent (d) Quantitative [Ans. (a)]

35. If the class frequency (AB) = 0, the value of Q is equal to

(a) 0 (c) −1

(b) 1 (d) 0 to 1 [Ans. (c)]

36. If for two attributes the class frequencies are (AB) (αβ) = (Aβ) (αB), then Q is

equal to:

(a) 0 (c) +1

(b) −1 (d) α [Ans. (a)]

37. If two attributes A and B are independent, then the coefficient of association

is:

(a) −1 (c) 0

(b) +1 (d) 0.5 [Ans. (c)]

38. If (AB) < (A)(B), the association between two attributes A and B is;
N
(a) Positive (c) Symmetrical

(b) Zero (d) Negative [Ans. (d)]

39. If (AB) (αβ) > (Aβ) (αB), then A and B are said to be [Ans. (b)]
(a) Negatively associated
(b) Positively associated
(c) Independent
(d) Difficult to tell

40. If two attributes A and B have perfect positive association, the value of coeffi-

cient of association is equal to

(a) 0 (c) (r − 1) (c − 1)

(b) −1 (d) +1 [Ans. (d)]

Correlation and Association 6.63

(ii) Short Essay Type
Correlation

1. No. of study hours: 2 4 6 8 10

No. of sleeping hours: 10 9 8 7 6

The correlation coefficient between the number of study hours and the number

of sleeping hours of different students is

(a) 0.97 (c) –1

(b) –0.89 (d) +1 [Ans. (c)]

2. X series Y series
15 15
Number of Items 25 18
Arithmetic Mean 136 138
Sum of square deviations

The summation of the products of the deviations of X and Y series from their
arithmetic means = 122.

From the above data, the coefficient of correlation between X and Y will be:

(a) 0.97 (c) –0.89

(b) 0.89 (d) –0.97 [Ans. (b)]

3. X: 86434
Y: 97446

From the above data the Karl Pearson’s coefficient of correlation is

(a) –0.87 (c) –0.92

(b) 0.89 (d) 0.94 [Ans. (d)]

4. Height (in meter): 1.60 1.64 1.71

Weight (in kg): 53 57 60

Covariance for the above data is (c) 0.141 [Ans. (a)]
(a) 0.126 (d) 0.153
(b) 0.135

5. N = 25, ∑x = 125, ∑y = 100, ∑x2 = 650, ∑y2 = 436, ∑xy = 520

Correlation coefficient from the above data is

(a) 0.75 (c) –0.667

(b) 0.72 (d) –0.59 [Ans. (c)]

6. If n = 10, ∑(x – x)2 = 144, ∑(y – y)2 = 49 and ∑(x – x) (y – y) = 77, then the

value of r is (c) 0.92 [Ans. (c)]
(a) –0.91 (d) 0.95
(b) –0.95

6.64 Business Mathematics and Statistics

7. If r = 0.8, ∑xy = 80, σx = σy = 2, then the number of pairs of n (where x = X –
X– and y = Y – Y–) is

(a) 24 (c) 26 [Ans. (b)]
(b) 25 (d) 27

8. If 2u + 5x = 17, 5v – 2y = 11 and cov (x, y) = 3, the cov(u, v) is

(a) –3 (c) –2 [Ans. (a)]
(b) –4 (d) –5

9. If r = 0.8, ∑xy = 60, σy = 2.5, and ∑x2 = 9 then the number of items (x and y are
deviations from the arithmetic means) is

(a) 7 (c) 9 [Ans. (d)]
(b) 8 (d) 10

10. If rxy = 0.6, find ruv where (ii) u = 3x + 5, v = –4y + 3
(i) u = 3x + 5, v = 4y – 3
[C.U. B. Com. 1987]
(a) 0.7, –0.7
(b) 0.5, –0.5 (c) 0.6, –0.6

(d) 0.8, –0.8 [Ans. (c)]

11. For 10 pairs of observations of x and y, the correlation coefficient is 0.7. Here,

x = 15, y = 18, σx = 4, σy = 5. Later it is found that the pair (x = 10, y = 8) is

wrongly copied. If it is omitted, the correlation coefficient of the remaining 9

pairs of observations is

(a) –0.71 (c) 0.69

(b) –0.63 (d) 0.62 [Ans. (d)]

12. The correlation coefficient and covariance of two variables x and y are respec-

tively 0.28 and 7.6. If the variance of x is 9, the standard deviation of y is

[V.U. B. Com. ’94]

(a) 7 (c) 9

(b) 8 (d) 10 [Ans. (c)]

13. In a question on correlation, the value of r is 0.917 and its probable error is

0.034. Then the value of N is

(a) 10 (c) 12

(b) 11 (d) 13 [Ans. (a)]
[Hints: P.E. = 0.6745 × 1 - r2 ]

N

14. In a question on correlation the value of r is 0.64 and its P.E. = 0.1312. The

value of N is

(a) 7 (c) 9

(b) 8 (d) 10 [Ans. (c)]

15. r = 0.5, ∑xy = 120, σy = 8, ∑x2 = 90

From the data given above, the number of items, i.e., n is

(a) 9 (c) 11

(b) 10 (d) 12 [Ans. (b)]

Correlation and Association 6.65

Rank Correlation

16. The rankings of ten students in Statistics and Economics are as follows:

Statistics: 3 5 8 4 7 10 2 1 6 9

Economics: 6 4 9 8 1 2 3 10 5 7

The coefficient of rank correlation is (c) +0.3 [Ans. (d)]
(a) –0.2 (d) –0.3
(b) +0.2

17. The coefficient of rank correlation between marks in Statistics and marks

in Accountancy obtained by a certain group of students is 0.8. If the sum of

the squares of the differences in ranks is given to be 33, then the number of

students in the group is

(a) 9 (c) 11

(b) 10 (d) 12 [Ans. (b)]

18. The coefficient of rank correlation of the marks obtained by 10 students in

Statistics and Accountancy was found to be 0.8. It was later discovered that

the difference in ranks in the two subjects obtained by one of the students was

wrongly taken as 7 instead of 9. The correct coefficient of rank correlation is

(a) 0.509 (c) 0.606

(b) 0.512 (d) 0.612 [Ans. (c)]

19. The coefficient of the rank correlation between debenture prices and share

prices is found to be 0.143. If the sum of squares of the differences in ranks is

given to be 48, then the value of N is

(a) 6 (c) 8

(b) 7 (d) 9 [Ans. (b)]

20. From the following data, Spearman’s rank correlation is

X: 10 12 8 15 20 25 40
Y: 15 10 6 25 16 12 8

(a) 0.14 (c) 0.12 [Ans. (a)]
(b) 0.13 (d) 0.15

Association of Attributes

21. Given N = 2,000, (A) = 1,500, (B) = 100, (AB) = 350. Are the data consistent?

(a) Yes (b) No [Ans. (b)]

22. Given N = 280, (A) = 250, (B) = 85, (AB) = 35. Are the data consistent?

(a) Yes (c) Cannot say [Ans. (b)]

(b) No

23. In a sample of 1,000 individuals, 100 possess the attribute A and 300 possess

attribute B. If A and B are independent, how many individual possess A and B?

(a) 28 (c) 30

(b) 25 (d) 35 [Ans. (c)]

6.66 Business Mathematics and Statistics

24. In a report of consumer preference, it was given that out of 500 persons

surveyed, 400 preferred variety A, 380 preferred variety B and 270 liked both

A and B. Are the data consistent?

(a) Yes (c) Cannot say [Ans. (b)]

(b) No

25. Given N = 1,482, (A) = 368, (B) = 343 and (AB) = 35, Then Yule’s coefficient

of association is

(a) 0.58 (c) 0.63

(b) –0.55 (d) –0.57 [Ans. (d)]

26. Total adults = 10,000, Literates = 1,290, Unemployed = 1,390, Literate

unemployed = 820. The association between literacy and unemployment from

the above figures is

(a) 0.923 (c) 0.792

(b) 0.891 (d) 0.956 [Ans. (a)]

27. From the following data find out the nature of (αβ). [Ans. (b)]
N = 100, (A) = 40, (B) = 80, (AB) = 30
(a) Independent
(b) Disassociated
(c) Positively associated
(d) Negatively associated

28. Find if, A and B are independent, dependent, positively associated or negatively

associated from the data given below:

(A) = 470, (B) = 620, (AB) = 320, N = 1,000

(a) Independent

(b) Disassociated

(c) Positively associated

(d) Negatively associated [Ans. (c)]

29. Given N = 800, (A) = 470, (β) = 450 and (AB) = 230. Then Yule’s coefficient

of association is

(a) 0.215 (c) 0.273

(b) 0.253 (d) 0.262 [Ans. (b)]

30. Given (AB) = 100, (αB) = 80, (Aβ) = 50, (αβ) = 40. The total number of obser-

vations are

(a) 320 (c) 300

(b) 250 (d) 270 [Ans. (d)]

Regression CHAPTER
Analysis
7

SYLLABUS

Least Squares Method, Simple Regression Lines, Properties of Regression,
Identification of Regression Lines

THEMATIC FOCUS

7.1. Regression
7.2. Regression Analysis
7.3. Types of Regression Techniques

7.3.1. Linear Regression
7.3.2. Least Squares Method
7.3.3. Non-linear Regression
7.4. Regression Line
7.5. Derivation of the Regression Equations
7.6. Regression Coefficients
7.7. Properties of Regression Lines
7.8. Identification of regression Lines
7.9. Uses of Regression
7.10. Difference between Correlation and Regression
7.11. Explanation of having Two Regression Lines
7.12. Illustrative Examples

7.1 REGRESSION

Regression is a statistical measure used in finance, investment and other disci-
plines to determine the strength of the relationship between one dependent
variable (usually denoted by y) and one or more independent variables. It also

7.2 Business Mathematics and Statistics

can be used to predict the future value of one variable based on the values of
others. It helps investment and financial managers to value assets, understand the
relationship between variables, such as commodity prices, stocks of businesses
dealing in those commodities and to predict sales based on weather, previews
sales, GDP growth or other conditions.

7.2 REGRESSION ANALYSIS

Regression analysis is a statistical tool which investigates the relationship between
variables in the form of mathematical equations. It is the most useful form of
analysis as it studies the variables individually and determines their significance
with greater accuracy. Say, for example, you want to find out the effect of price-
increase on the customer’s demand for fixing selling price of a product or you want
to study the relationship between salaries and qualifications on the job perfor-
mance of an employee. These studies will naturally involve a lot of co-related
variables that will individually have an effect on the dependent variable. These
complex questions can be easily answered with the help of regression analysis. In
short, a good regression analysis needs sound reasoning and proper interpretation
of data for highly accurate predictions, forecast and solutions.

7.3 TYPES OF REGRESSION TECHNIQUES

Regression techniques basically involve the assembling of data on the variables
under study and estimating the quantitative effect of a variable on another. There
are different kinds of regression techniques such as:

7.3.1 Linear Regression

This is a simple and easy to use method that models the relationship between a
scalar dependent variable y and one or more explanatory (independent) variables
denoted as x. Usually, more than one independent variable influences the
dependent variable. When one independent variable is used in a regression, it is
called a Simple Linear Regression; when two or more independent variables are
used, it is called a Multiple Linear Regression. This method uses linear predictor
functions for data modelling wherein unknown parameters are estimated from the
data. Since linear models are linearly dependent on unknown parameters, they are
easier to fit than non-linear models and lead to easier determination of statistical
parameters. Multiple regression analysis introduces several additional complex-
ities but may produce more realistic results then simple regression analysis.

7.3.2 Least Squares Method

The typical procedure for finding the line of best fit is called the least-squares
method. This method is a commonly used method to solve for linear regression

Regression Analysis 7.3

equations. It’s best suited for data fitting applications such as fitting a straight
line on to the points i.e. closeness to all the points in a scatter diagram etc. The
criteria for what we mean by ‘closeness’ is called the least square principle.
Recall the discussion on variance, where we learnt how the variance squares the
deviations around the mean. In regression we will square the deviations around
the regression line instead of around the mean. The best fit regression line that
has the smallest value for the squared deviations around it, the least squared
deviations. That’s essentially the whole idea of least squares. Therefore, the least
square method is based upon the principle that the sum of the squared residuals
(residual is nothing but the difference between the actual or observed y value and
y value calculated by the linear regression line equation) should be made as small
as possible so the regression line has the least error. This method can be used for
linear as well as non-linear regression depending on the nature of the residuals
and equations.

7.3.3 Non-linear Regression

When the relationship between variable are represented by curved lines, then it is
called non-linear regression. The non-linear regression analysis uses the method
of successive approximations. Here, the data are modeled by a function, which
is a non-linear combination of model parameters and depends on one or more
explanatory (independent) variables. Therefore, in non-linear regression too, the
models could be based on simple or multiple regressions. This method takes into
account the nature of relationship between the variables and tries to find some
kind of transformation in them so that the relationship can be expressed easily
as a straight line.

7.4 REGRESSION LINE

In statistics, a regression line is a line that best describes the behavior of a set
of data. It is the best fit straight line that we can draw through or between the
points on the scatter diagram. Obviously a straight line cannot connect all the
dots because then you would have to bounce up and down and up and down from
one dot to the next, and it would not be a straight line. So we want to be able
to draw a single line that comes as close to all the dots as possible. Regression
line is the line that best fits the data, such that the overall distance from the line
to the points (variable values) plotted on a graph is the smallest. In other words,
the line which is used to minimize the squared deviations of predictions is called
Regression line. It is also known as the line of ‘best fit’.

Consider the scatter diagram given below. One possible line of best fit has been
drawn on the diagram. Some of the points lie above the line and some lie below it.
The vertical distance each point is above or below the line has been added to the

7.4 Business Mathematics and Statistics

diagram. These distances are called deviations or errors – they are symbolized as
d1, d2,… , dn. When drawing in a regression line, the aim is to make the line fit the
points as closely as possible. We do this by making the total of the squares of the
deviations as small as possible, i.e. we minimize Σdi2 . If a line of best fit is found
using this principle, it is called the least-squares regression line.

Y
b = Slope of regression line

Distance from the line to a typical data point
(= “error” between the line and this y value)

a

OX

Figure 7.1 Regression line

The regression line formula is like the following:

y = a + bx + e

The multiple regression formula look like this:
y = a + b1x1 + b2x2 + b3x3 + . . . + bnxn + e

where,
y = the value of the dependent variable (y), what is being predicted or explained
a or alpha = a constant; intercept of regression line; equals the value of y when
the value of x = 0
b or beta = the coefficient of x; the slope of the regression line; how much y
changes for each one-unit change in x.
x = the value of the independent variable (x), what is predicting or explaining
the value of y
e = the error term; the error in predicting the value of y, given the value of x (it
is not displayed in most regression equations).
A simple equation for the regression line is

y = a + bx
There are as many number of regression lines as variables. Suppose we take
two variables, say x and y, then there will be two regression lines:
(1) Regression line of y on x: This give the most probable values of y from the

given values of x.

Regression Analysis 7.5

(2) Regression line of x on y: This gives the most probable values of x from
the given values of y.
The algebraic expression of these regression lines is called as Regression
Equations. There will be two regression equations for the two regression
lines.
(i) Regression equation of y on x : y = a + bx
(ii) Regression equation of x on y : x = a + by

7.5 DERIVATION OF THE REGRESSION EQUATIONS

(1) Regression Equation of y on x

We know that the regression equation of y on x by method of least squares

from observation is y = a + bx …(1)

Let (x1, y1), (x2, y2),. . .,(xn, yn ) be n pair of observations of the two variables
x and y.

Now to find the values of a and b, we apply the method of least squares and

solve the following two normal equations:

Σy = n a + b Σx …(2)
and Σxy = aΣx + bΣx2 …(3)
Now, dividing (2) by n, we get
…(4)
Σy = a + b Σx or y = a + bx
nn

Subtracting (4) from (1) we get y – y = b (x – x) …(5)

Again multiplying (2) by Sx and (3) by n we get

and, (Σx) (Σy) = n a (Σx) + (Σx)2
Now, n Σxy = n a (Σx) + n b(Σx2)

( )(Σx)(Σy) − n Σxy = b(Σx)2 − n b Σx2 (by subtracting)

( )or 
b (Σx)2 −n Σx2  = (Σx)(Σy) − n Σxy

(Σx)(Σy) − nΣxy nΣxy − (Σx)(Σy) [Changing sign]
( ) ( )or
b = (Σ x)2 − n Σ x2 = − (Σ x)2
n Σ x2

Sxy - Sx . Sy cov(x, y)
n n n
= = …(6)
ö2 s 2
Sx2 - æ Sx ÷ø x
n çè n

7.6 Business Mathematics and Statistics

Replacing b by byx in (5), the regression equation of y on x is
_ _
y – y = byx (x – x)

where, byx = regression coefficient of y on x

= cov(x, y) …(7)
s 2
x

Again we know r = cov(x, y) or cov(x, y) = r.s x .s y
s x .s y

Then from (7), byx = r.s x .s y = r . s y
s x
s 2
x

(2) Regression Equation of x on y

The regression equation of x on y by method of least squares from
observation is

x = a + by …(8)

Now to find the values of a and b, we are to apply the method of least
squares and solve the following two normal equations:

Σx = n a + b Σy …(9)

and Σxy = a Σy = b Σy2 …(10)

solving (9) and (10) for a and b and proceeding in the same way as before,

we get the regression equation of x on y as x – x = b(y – y). …(11)

where, b = Sxy - Sx . Sy = cov(x, y)
n nn
Sy2
n - æ Sy ö2 s 2
çè n ÷ø y

Replacing b by bxy in (ii), the regression equation of x on y is
x − x = bxy (y − y)

where, bxy = regression coefficient of x on y

= cov(x, y) = r. s x
s
s 2 y
y

Theorem 1 Show that the correlation coefficient is the geometric mean of

regression coefficients

Proof: We know byx = r. s y and bxy = r. s x
s x s y

Now multiplying we get s s
s s
byx ´ bxy = r. y ´r. x
x y

Regression Analysis 7.7

= r2

or r = ± byx × bxy
It is clear from the relationship that r is the geometric mean between the two

regression coefficients.
This relation clearly shows that byx and bxy must have the same sign otherwise

r2 will be negative which is impossible.
(i) If byx and bxy are both positive, then r = + byx × bxy
(ii) If byx and bxy are both negative, then r = − byx × bxy

7.6 REGRESSION COEFFICIENTS

There are two regression coefficient:

(i) Regression coefficient of y on x

Σxy − Σx . Σy
= n n n
byx Σx2 2
 Σx
n − n

(ii) Regression coefficient of x on y

Σxy − Σx . Σy
= n nn
bxy Σy2
 Σy 2
n − n

Since sx, s and r are independent of the change of origin, byx and bxy

y

are also independent of the change of origin. Thus the formulae for actual

computation of byx and bxy are given below.
(a) When deviations are taken from assumed mean of x and y.

That is, U = x – A and V = y – B, then

Σuv − Σu . Σv Σuv − Σu . Σv
= n nn = n nn
byx Σu2 and bxy Σv2
 Σu 2  Σv 2
n − n n − n

(b) When deviations _are taken from_arithmetic mean of x and y.
That is, U = x – x and V = y – y , then ΣU = 0 and ΣV = 0

Hence, byx = Σuv and bxy = Σuv
Σu2 Σv2

7.8 Business Mathematics and Statistics

7.7 PROPERTIES OF REGRESSION LINES (INCLUDING

CORRELATION COEFFICIENT AND REGRESSION

COEFFICIENTS)

_ __ _
(1) The two regression_lines y – _y = byx (x – x ) and x – x = bxy (y – y ) are

satisfied when x = x a_nd_y = y , it implies that the two lines of regression
intersect at the point (x , y).

(2) Putting byx = r. s y in the regression equation of y on x
sx

We get, y - y = r. s y (x - x )
sx

or y - y = r. x - x
sy sx

Again, putting bxy = r. s x in the regression equation of x on y we get
sy

x - x = s x (y - y)
r.

sy

or x - x =r. y - y
sx sy

(a) Two lines of regression are different. They coincide, i.e. become

identical when r = –1 or 1 or in other words, there is a perfect negative
or positive correlation between the two variables under discussion.

(b) The two lines of regression are perpendicular to each other when r = 0

(i.e. parallel to the x-axis and y-axis)
(3) The slope of the regression line of y on x is byx and slope of the regression

1
line of x on y is .

bxy
(4) The correlation between the variables depend on the distance between two

regression lines, such as the nearer the regression lines to each other the

higher is the degree of correlation, and the further the regression lines to
each other the lesser is the degree of correlation.
(5) The correlation coefficient is the geometric mean of two regression coeffi-

cients. Symbolically , it can be expressed as: r = byx × bxy .
(6) The value of the coefficient of correlation cannot exceed unity, i.e. 1.

Therefore, if one of the regression coefficients is greater than unity, the

other must be less than unity.
(7) The coefficient of correlation will have the same sign as that of the

regression coefficients, such as if the regression coefficients have a positive
sign, then ‘r’ will be positive or vice versa.

Regression Analysis 7.9

(8) The sign of both the regression coefficients will be same, i.e. they will
be either positive or negative. Thus, it is not possible that one regression
coefficient is negative while the other is positive.

(9) Arithmetic mean of the regression coefficients is greater than the corre-
lation coefficient r, provided r > 0. Symbolically, it can be represented as
bxy + byx > r.
2

(10) The regression coefficients are independent of the change of origin, but
not of the scale. By origin, we mean that there will be no effect on the
regression coefficients if any constant is subtracted from the value of x
and y. By scale, we mean that if the value of x and y is either multiplied
or divided by some constant, then the regression coefficients will also
change.

Proof: Let u= x−a and v= y − c , i.e. x = a + bu
b d

and y = c + dv or x = a + bu and y = c + dv

We know, cov(x, y) = 1 Σ(x − x )(y − y)
n
= 1 Σ(a + bu − a − bu)(c + dv − c − dv )
n
= 1 Σb(u − u). d(v − v )
n
= bd. 1 Σ(u − u)(v − v )
n
= b.d.cov(u, v)

s x2 = 1 S(x - x )2 = 1 S(a + bu - a - bu )2
n n

= 1 Σ(bu − bu )2 = b2. 1 Σ(u − u )2
nn

= b2 .s 2
u

and s 2 = 1 S(y - y )2 = 1 S(c + dv -c - d v )2
y n n

= 1 Σ(dv − dv )2 = d2 1 Σ(v − v )2
nn

= d 2 .s 2
v

7.10 Business Mathematics and Statistics

Therefore, byx = cov(x, y) = bd cov(u, v)

s 2 b2 .s 2
x u

= d . cov(u, v) = d bvu
b b
s 2
u

Similarly, bxy = b . buv
d

Thus the regression coefficients are independent of a and c but dependent of
b and d, i.e. independent of the change of origin but not of the scale.

ILLUSTRATION 1

If u = –5x + 3 and v = 7y + 2 and the regression coefficient of x on y is 0.7, find
the regression coefficient of u on v.

Solution: u = −5x + 3

or 5x = −4 + 3

or x = − 1 u + 3
55

It is in the form x = a + bu, where b = − 1
5

Again, v = 7y + 2 or 7y = v − 2

or y = 1 v − 2
77

It is in the form y = c + dv, where d = 1
7

Therefore, regression coefficient of u on v

1

buv = d . bxy = 7 × 0.75
b 1
−

5

= − 5 × 0.7 = − 0.5
7

7.8 IDENTIFICATION OF REGRESSION LINES

Suppose the two regression equations are
(i) a1x + b1y + c1 = 0 and (ii) a2x + b2y + c2 = 0

Regression Analysis 7.11

Out of these two equations which of the two is the equation of x on y; is not

given. In such a case the following steps should be followed:

(1) Assume any one of the equations [suppose equation (i)] as the regression

equations of x on y and calculate slope, i.e. bxy  bxy = − a1 
 here b1 .
 

(2) Calculate slope of another equation [here equation (ii)]

That is, byx  byx = − b2 
 here a2 

(3) Multiply bxy and byx, i.e. find r 2 .  here a1b2 
 a2b1 
 

(i) If r2 <  a1b2 <  then our assumption is correct, i.e. a1x +
1 here a2b1 1

b1y + c1 = 0 in the regression equation of x on y and a2x + b2y + c2 = 0
is the regression equation of y on x.

(ii) If r2 >  here a1b2 >  then our assumption is wrong. Therefore,
1 a2b1 1

a1x + b1y + c1 = 0 is equation of y on x and a2x + b2y + c2 = 0 is equation
of x on y. Calculate regression coefficients of the changed equations

following the same procedure.

NOTE

If one of the regression coefficient is greater than unity, the other must be less
than unity. [Property No. 6]

ILLUSTRATION 2

Two lines of regression are 4x – 5y + 30 = 0 and 20x – 9y = 56. Calculate
coefficient of correlation between x and y.

Solution: For calculating the value of coefficient of correlation (r), we will
have to find out the regression coefficients. But we do not know which of the
two regression equations is the equation of x on y.

We assume first equation as the regression equation of x on y

4x − 5y + 30 = 0

or 4x = −30 + 5y

or x = − 30 + 5 y or bxy = 5
4 4 4

7.12 Business Mathematics and Statistics

From 2nd equation we get

20x − 9y = 56

or −9y = 56 − 20x

or y = 56 − 20 x
−9 −9

= − 56 + 20 x
9 9

or byx = 20
9

Now r2 = bxy × byx = 5 20 100 = 2.78 > 1
4× 9 = 36

Since r2 >1, our assumption is wrong.

Hence, the first equation is equation of y on x

then, 4x − 5y + 30 = 0

or −5y = −30 − 4x

or y = 30 + 4 x
5 5

or byx = 4
5

Second equation in equation of x on y
20x – 9y = 56

or 20x = 56 + 9y

or x = 56 + 9 y
20 20

9
or bxy = 20

then, r2 = byx × bxy = 4 ´ 9 = 36
5 20 100

= 0.36

or r = 0.36 = 0.6

[As bxy and byx both are positive, then r will also be positive]
Therefore, the required value of coefficient of correlation is 0.6

Regression Analysis 7.13

7.9 USES OF REGRESSION

1. Regression is used in Economics, Commerce and other disciplines to
determine the strength of relationship between one dependent variable and
one or more independent variables.

2. Regression is used to estimate the future value of one variable based on the
values of others.

3. Regression lines are used to study the relative variation of the two variables.
4. Regression lines are very useful for forecasting procedures. By using the

equation obtained from the regression line an analyst can forecast future
behaviour of the dependent variable by inputting different values for the
independent ones.
5. Regression analysis helps investment and financial manager to perform
valuations for many different securities.
6. We calculate coefficient of correlation with the help of regression coeffi-
cients. The square of correlation coefficient (r2) is called the coefficient of
determination which measure the degree of association of correlation that
exists between the two variables.

7.10 DIFFERENCE BETWEEN CORRELATION
AND REGRESSION

Both correlation and regression can be said as the tools used in statistics that
actually deals through two or more than two variables. Even though both identify
with some topic, there exist contrasts between these two methods. The main
difference is that correlation finds out the degree of relationship between two
variables, while regression explains the nature of relationship. Other differences
between these methods are given below:

1. The correlation term is used when (i) both variables are random variables,
and (ii) the end goal is simply to find a number that expresses the relation
between the variables.
The regression term is used when (i) one of the variables is a fixed variable,
and (ii) the end goal is to use the measure of relation to predict or estimate
values of the random variable based on values of the fixed variable.

2. In correlation, there exists no distinction amongst explanatory (independent)
and dependent variable that shows correlation amongst x and y is as like as
y and x.
In regression, there exists the distinction amongst explanatory and dependent
variable that shows the regression of y on x is is not quite the same as x on y.

3. Correlation coefficient is independent of the change of scale and origin.
Regression coefficients are independent of the change of origin, but not of
the scale.

7.14 Business Mathematics and Statistics

4. Correlation never fit in a line which passes through the points of data.
Regression finds the finest line which estimates the behavior of y from x.

5. Correlation does not help us in ascertaining whether one variable is the
cause and the other the effect. But regression helps us in studying the cause
and effect relationship between the two variables.

7.11 EXPLANATION OF HAVING TWO REGRESSION LINES

Regression line is obtained by minimizing the sum of squares of overall distances
(i.e. vertical and horizontal distances) from the line to the points (variable values)
plotted on a graph. One line cannot serve both the purposes, i.e. minimize the sum
of squares of vertical distances as well as horizonal distances. So it is essential to
have two regression lines. The line which is used to minimize the sum of squares
of the vertical, distances is known as the regression line of x on y. In this case it
is assumed that the values of x are exactly known but the values of y are subject
to error. The line which is used to minimize the sum of squares of horizontal
distances is known as the regression line of x on y. In this case it is assumed that
the values of x are exactly known but the values of x are subject to error. The first
line is used to obtain best estimates of y for given values of x, and the second line
is used to obtain best estimates of x for given values of y.

ILLUSTRATIVE EXAMPLES

A. SHORT TYPE

EXAMPLE 1

Find the regression equation of y on x from the following values:
__
x = 10, y = 15 and byx = 2.50 [C.U. B.Com. 2016]

Solution: We know that the regression equation of y on x is

y – y = byx (x – x)
Putting the values of x, y and byx we get

y – 15 = 2.50 (x – 10)

or y – 15 = 2.50x – 25
or y = 2.50x – 25 + 15
or y = 2.50x – 10

EXAMPLE 2

If two regression equations are 8x – 10y + 66 = 0 and 40x – 18y = 214, find the

average values of x and y. [C.U. B.Com. 2010, 2016 (H)]

Regression Analysis 7.15

Solution: We know that two regression equations intersect at a point (x, y).

Hence, for finding the average values of x and y, we are to solve two regression

equations.

8x – 10y + 66 = 0 …(i)

40x – 18y – 214 = 0 …(ii)

Multiplying equation (i) by 5 and then subtracting from equation (ii) we get

40x − 18y − 214 = 0

40x − 50y + 330 = 0
−+ −

32y − 544 = 0
or 32y = 544 or y = 544 = 17

32
Putting the value of y in equation (i) we get

or 8x − 170 + 66 = 0
Hence, 8x − 104 = 0 or 8x = 104 or x = 104 = 13

8

x = 13 and y = 17.

EXAMPLE 3

Using the following regression coefficient, find the value of correlation coefficient

r where byx = –0.6 and bxy = –1.35. [C.U. B.Com. 2011]

Solution: We know that, r = byx × bxy

= −0.6 × −1.35

= 0.81 = 0.9

As bxy and byx both are negative, therefore, value of r will be –0.9.

EXAMPLE 4

__
If x = 36, y = 85, s x = 11, s y = 8 and correlation coefficient between x and y

= 0.66, then find the two regression equations. [C.U. B.Com. 2013 (G)]

__
Solution: Given that, x = 36, y = 85, sx = 11, sy = 8 and rxy = 0.66
Regression equation of x on y:

or x − x = bxy (y − y) [bxy = rxy sx = 0.66 ´ 11 = 0.9075]
x − 36 = 0.9075(y − 85) sy 8

or x − 36 = 0.9075y − 77.1375

7.16 Business Mathematics and Statistics

or x = 0.9075y − 77.1375 + 36
or x = 0.9075y − 41.1375

Regression equation of y on x:

y − y = byx (x − x )

or y - 85 = 0.48(x - 36) [byx = rxy × s y = 0.66 ´ 8 = 0.48]
s x 11

or y − 85 = 0.48x − 17.28

or y = 0.48x − 17.28 + 85

or y = 0.48x + 67.72

EXAMPLE 5

If x + 2y = 5 and 2x + 3y = 8 be two regression lines then find rxy.
[C.U. B.Com. 2014 (H)]

Solution: Assume the regression equation of y on x is

x + 2y = 5

or 2y = 5 − x

or y = 5 − 1 . x
22

Therefore, byx = 1
−

2

The regression equation of x on y is

2x + 3y = 8

or 2x = 8 − 3y

or x = 4 − 3 . y
2

Therefore, bxy = − 3
2

Now, r2 = byx × bxy = − 1 ×− 3 = 3 <1
2 2 4

Therefore, our assumption is correct.

Hence, r2 = 3
4

or r= 3 0.75 = 0.87
=
4

As bxy and byx both are negative, therefore, the value of r will be –0.87.

Regression Analysis 7.17

EXAMPLE 6

If s x = 10, s y = 12, bxy = –0.8, find the value of rxy . [C.U. B.Com. 2015 (G)]

Solution: We know, bxy = r. s x
s y

or −0.8 = rxy . 10 or rxy = −0.8 × 12 = −0.96
12 10

EXAMPLE 7

The regression equation of y on x is 3x – 5y = 13 and the regression equation of
x on y is 2x – y = 7. Estimate the value of x when y = 10. [C.U. B.Com. 1996]

Solution: We are to estimate the value of x for a given value of y
So, regression equation of x on y is to be selected
That is 2x − y = 7

or 2x − 10 = 7 [Putting the value of y]
or 2x = 17
or x = 17 = 8.5.

2
EXAMPLE 8

If rxy = 0.6, s y = 4 and byx = 0.48, find the value of sx. [C.U. B.Com. 2003]

Solution: We know that, s
s
byx = rxy . y
x

or 0.48 = 0.6 ´ 4
sx

or 0.48 s x = 2.4

or s x = 2.4 = 5
0.48

Therefore, required value of s is 5.

x

B. SHORT ESSAY TYPE

EXAMPLE 9

Find the two regression equations from the following table:

X12345
Y23456

If X = 2.5, what will be the value of Y? [B.U. B.Com (H), 1994]

7.18 Business Mathematics and Statistics

Solution: Calculation of regression equations

X Y x=X–X y=Y–Y x2 y2 xy

1 2 –2 –2 4 4 4
2 3 –1 –1 1 1 1
3 5 0 1 0 1 0
4 4 1 0 1 0 0
5 6 2 2 4 4 4

SX = 15 SY = 20 Sx = 0 Sy = 0 Sx2 = 10 Sy2 = 10 Sxy = 9

X = ΣX 15 = 3, Y = ΣY = 20 =4 (both are integers).
=
n5 n5

Regression coefficient of Y on X (bYX) = Σ(X − X)(Y − Y )
Σ(X − X)2

= Σ xy = 9 = 0.9
Σx2 10

Regression coefficient of X on Y (bXY) = Σ (X − X)(Y − Y ) = Σ xy
Σ (Y − Y )2 Σy2

= 9 = 0.9
10

Therefore,

Regression equation of Y on X is

Y − Y = bYX (X − X)
or Y − 4 = 0.9(X − 3) or Y − 4 = 0.9X − 2.7
or Y = 0.9X − 2.7 + 4 or Y = 0.9X + 1.3

If X = 2.5, thenY = 0.9 × 2.5 + 1.3 = 2.25 + 1.3 = 3.55
Regression equation of x on y is

X − X = bXY (Y − Y )
or X − 3 = 0.9(Y − 4) or X − 3 = 0.9Y − 3.6
or X = 0.94 − 3.6 + 3 or X = 0.9Y − 0.6

EXAMPLE 10
Find the two linear regression equations from the following:

X 12 23 37 46 57 76 82
Y 36 42 57 64 68 82 95

[C.U. B.Com 2016 (G)]

Regression Analysis 7.19

Solution: Calculation of regression equations

X Y x=X–a y=Y–b x2 y2 xy

12 36 –35 –27 1225 729 945

23 42 –24 –21 576 441 504

37 57 –10 –6 100 36 60

46 64 –1 1 1 1 –1

57 68 10 5 100 25 50

76 82 29 19 841 361 551

80 95 33 32 1089 1024 1056

SX = 331 SY = 444 Sx = 2 Sy = 3 Sx2 = 3932 Sy2 = 2617 Sxy = 3165

Actual mean: X = ΣX = 331 = 47.29
n7

Y = ΣY = 444 = 63.43
n7

X and Y are not integers, hence deviations are to be taken from assumed mean
or calculation may be done directly (without taking any deviation).

Assumed mean: A = 47, B = 63
Since the regression coefficients are independent of origin,
Hence,

Regression coefficient of X on Y (bXY) = bxy = nΣxy − Σx.Σy
n.Σy2 − (Σy)2

= 7 × 3165 − 2 × 3 = 22155 − 6 = 22149 = 1.21
7 × 2617 − 9 18319 − 9 18310

and regression coefficient of Y on X (bYX ) = byx = nΣxy − Σx.Σ y
n.Σx2 − (Σx)2

= 7 × 3165 − 2 × 3 = 22155 − 6 − 22149 = 0.8
7 × 3932 − 4 27524 − 4 27520

Therefore,
Regression equation of X on Y is

X − X = bXY (Y − Y )
or X − 47.29 = 1.21(Y − 63.43)

or X − 47.29 = 1.21Y − 76.75
or X = 1.21Y − 76.75 + 47.29
or X = 1.21Y − 29.46

7.20 Business Mathematics and Statistics

Regression equation of Y on X is
Y − Y = bYX (X − X)

or Y − 63.43 = 0.8(X − 47.29)
or Y − 63.43 = 0.8X − 37.83
or Y = 0.8X − 37.83 + 63.43
or Y = 0.8X + 25.60

EXAMPLE 11

With the help of a suitable regression line estimate the value of x when y = 22 by
using the following data:

x: 4 5 8 9 11
y: 16 10 8
76
[C.U. B.Com. 2011]

Solution: We are to estimate the value of x for a given value of y; therefore, we
are to find the regression equation of x on y.

Calculation of regression equation of x on y

x y y2 xy

4 16 256 64
5 10 100 50
8 8 64 64
9 7 49 63
11 6 36 66

Sx = 37 Sy = 47 Sy2 = 505 Sxy = 307

Here, x = Σx = 37 = 7.4 and y = Σy = 47 = 9.4
n5 n5

and bxy = nΣxy − Σx.Σy = 5 × 307 − 37 × 47
n.Σy2 − (Σy)2 5 × 505 − (47)2

= 1535 − 1739 = −204 = −0.646 (Approx.)
2525 − 2209 316

Therefore, the regression equation of x on y is.

x − x = bxy (y − y) or x − 7.4 = −0.646 (y − 9.4)

or x − 7.4 = −0.646y + 6.0724
or x = −0.646y + 6.0724 + 7.4
or x = −0.646y + 13.4724

Regression Analysis 7.21

Putting y = 22 in the above equation we get
x = −0.646 × 22 + 13.4724
= −14.212 + 13.4724
= −0.7396
= −0.74 (Approx)

Therefore, the required estimated value of x is –0.74 when y = 22.

EXAMPLE 12

You are given the following data:

xy
A.M 36 85
S.D. 11 8

and correlation coefficient between x and y is 0.66 [C.U. B.Com. 2000]
(i) Find the two regression equations
(ii) Estimate the value of x when y = 33.5
(iii) Estimate the value of y when x = 73.3

Solution: Given that, x = 36, y = 85, s x = 11, s y = 8 and r = 0.66

byx = r. s y = 0.66 ´ 8 = 0.48
s x 11

bxy = r. s x = 0.66 ´ 11 = 0.91
s y 8

(i) Therefore, the regression equation of y on x is,

y − y = byx (x − x )

or y − 85 = 0.48(x − 36)

or y − 85 = 0.48x − 17.28

or y = 0.48x − 17.28 + 85

or y = 0.48x + 67.72

and the regression equation of x on y is
x − x = bxy (y = y)

or x − 36 = 0.91(y − 85)

or x − 36 = 0.91y − 77.35

or x = 0.91y − 77.35 + 36
or x = 0.91y − 41.35

7.22 Business Mathematics and Statistics

(ii) To estimate the value of x for a given value of y we are to use regression
equation of x on y.
That is, x = 0.91y − 41.35

or x = 0.91× (−33.5) − 41.35 (putting y = 33.5)

or x = 30.485 − 41.35
= −10.865

Therefore, the estimated value of x is –10.865 when y = 33.5
(iii) To estimate the value of y for a given value of x we are to use regression

equation of y on x
That is, y = 0.48 + 67.72
or y = 0.48 × 73.3 + 67.72 [puting x = 73.3]

= 35.184 + 67.72 = 102.904

Therefore, the estimated value of y is 102.904 when x = 73.3.

EXAMPLE 13

The lines of regression of y on x and x on y are respectively y = x + 5 and 16x =
9y – 94. Find the variance of x if the variance of y is 16. Also, find the covariance
of x and y.

Solution: The regression equation of y on x is y = x + 5
Therefore, the regression coefficient of y on x (byx) = 1
Again, the regression equation of x on y is

16x = 9y − 94

or 9 94
x= y−
16 16

9
Therefore, the regression coefficient of x on y (bxy) = 16

We know that, r = byx × bxy

= 1× 9 = 93
=

16 16 4

Given that, variance of y = 16

That is, s 2 = 16 or s = 4
Now, bxy
y y s
s
= r. x
y

or 9 = 3 . s x
16 4 4

Regression Analysis 7.23

or 9 ´ 16 = sx or sx =3
16 3

Therefore, variance of x = s2 = (3)2 = 9

x

Again, bxy = cov(x, y)

s 2
y

or 9 cov(x, y)
=
16 16

or cov (x, y) = 9

EXAMPLE 14

Find the regression equation of x on y from the following data and find the
estimated value of x, when y = 6.

Sx = 24, Sy = 44, Sxy = 306, Sx2 = 164, Sy2 = 574, n = 4
[C.U. B.Com. 2008]

Solution: Given: Sx = 24, Sy = 44, Sxy = 306, Sx2 = 164, Sy2 = 574, n = 4

Now, x = Σx = 24 = 6, y = Σy = 44 = 11
n4 n4

Regression coefficient of x on y (bxy) = nΣxy − Σx.Σy
n.Σy2 − (Σy)2

4 × 306 − 24 × 44 1224 − 1056
==

4 × 574 − (44)2 2296 − 1936

= 168 = 0.47
360

Therefore, the regression equation of x on y is
x − x = bxy (y − y) or x − 6 = 0.47(y − 11)

or x − 6 = 0.47y − 5.17 or x = 0.47y − 5.17 + 6 or x = 0.47y + 0.83

when y=6

Then, the estimated value of x is

x = 0.47 × 6 + 0.83
= 2.82 + 0.83 = 3.65

EXAMPLE 15

If u = 2x – 3 and v = 1 y + 1.5 , find the values of regression coefficients buv and
3

bvu when bxy = 0.5 and byx = 1.4

7.24 Business Mathematics and Statistics

Solution: u = 2x – 3
or 2x = u + 3

or x= 3 + 1u
2 2

It is in the form x = a + bu  when u = x − a 
b

where a = 3 and b = 1
22

again, v= 1 y + 1.5
3

or 1 y = −1.5 + v
3

or y = – 4.5 + 3v

It is in the form y = c + dv  when v = y − c 
d

where, c = – 4.5 and d = 3

Now, bxy = cov(x, y) = bd cov(u, v) = b × cov(u, v) = b × buv
d
s 2 d 2 .s 2 d s 2
y v v

1

or 0.5 = 2 buv or buv = 1.5 × 2 = 3
Similarly, 3

or d
byx = b .bvu

1

bvu = b = 2 × 1.4 = 1.4 = 0.23
d .byx 3 6

Therefore, the required values of buv and bvu are 3 and 0.23 respectively.

EXAMPLE 16

For the variables x and y, variance of x = 12, regression equations are: x + 2y = 5
and 2x + 3y = 8

Find the following:

(i) The average values of x and y [C.U. B.Com. 2004]
(ii) Correlation coefficient between x and y
(iii) Standard deviation of y.

Regression Analysis 7.25

Solution:
(i) We know that two regression equations intersect at a point (x, y). Hence, for
finding x and y, we are to solve two regression equations.

x + 2y = 5 …(i)
2x + 3y = 8 …(ii)

Multiplying equation (i) by 2 and then subtracting from equation (ii) we get

2x + 3y = 8
2x + 4y = 10
–– –

–y = –2
or y = 2

Putting the value of y in equation (i) we get

or x+2×2=5
Hence, x + 4 = 5 or x = 5 – 4 = 1
x = 1 and y = 2

(ii) For finding correlation coefficient (r) we are to find the values of byx and bxy.
Out of the two equations, which one is meant for y on x is not given. Let us

assume that equation (i) is meant for y on x and equation (ii) is for x on y.

From (i), 2y = −x + 5 o r y = − 1 x + 5
22

or byx = − 1
2

and from (ii), 2x = −3y + 8

3
or x = − 2 y + 4

3
or bxy = − 2

Now, 1 33
byx × bxy = − 2 × − 2 = 4 < 1.

So our assumption is correct.

Therefore, r2 = byx × bxy = 3
4

or r= 3 3 = 1.732 = 0.866
=
42 2

As bxy and byx both are negative, then value of r will be –0.866.
(iii) Given, variance of x = 12

That is, s2 = 12

x

7.26 Business Mathematics and Statistics

or s x = 12 = 3.46
Now,
byx = r. s y
or s x
or
- 1 = -0.866. s y
2 3.46

sy = 2 3.46 = 3.46
´ 0.866 1.732

= 1.99 = 2 (approx.)

Therefore, standard deviation of y is 2.

EXAMPLE 17

In order to find regression coefficients between two variables x and y from 5 pairs
of observations the following results are given:

Σ x = 30, Σ y = 40, Σ x2 = 220, Σ y2 = 340, Σ xy = 214. Later it was found that
one particular set of observations namely x = 4, y = 8 was wrongly taken. The
correct value being x = 2, y = 6, Find the corrected values of the regression coeffi-
cients and hence find the equations.

Solution: Here n = 5
For determining correct value we have to subtract the wrong observation and

then add the correct observations.

Corrected Σ x = 30 − 4 + 2 = 28
Corrected
Corrected Σ y = 40 − 8 + 6 = 38
Corrected
Corrected Σ x2 = 220 − 42 + 22 = 220 − 16 + 4 = 208
Corrected
Σ y2 = 340 − 82 + 62 = 340 − 64 + 36 = 312
Corrected
Σ xy = 214 − (4 × 8) + (2 × 6) = 214 − 32 + 12 = 194

5 × 194 − 28 × 38 970 − 1064 −94
byx = = = = −0.37
5 × 208 − (28)2 1040 − 784 256

5 × 194 − 28 × 38 970 − 1064 −94
bxy = = = = −0.81
5 × 312 − (38)2 1560 − 1444 116

x = Σ x = 28 = 5.6, y = Σ y = 38 = 7.6
n5 n5

Regression equation of y on x:

y − y = byx (x − x )

Regression Analysis 7.27

or y − 7.6 = −0.37(x − 5.6)
or y = −0.37x + 2.072 + 7.6
or y = −0.37x + 9.672

Regression equation of x on y:

x – x = bxy (y – y)
or x − 5.6 = −0.81(y − 7.6)
or x − 5.6 = −0.81y + 6.156
or x = −0.81y + 6.156 + 5.6
or x = −0.81y − 11.756

EXAMPLE 18

The correlation coefficient (r) = 0.60, variance of x and y are respectively 2.25

and 4.00; x = 10, y = 20. From the above data find the regression equations. Find

the estimated value of y when x = 25. [C.U. B.Com 2006(old)]

Solution: Given that r = 0.60;s 2 = 2.25 or sx = 2.25 = 1.5;
x

s 2 = 4 or sy = 4 = 2; x = 10, y = 20
y

Regression equation of x on y:

x - x = bxy (y - y) or x - x = r × sx (y - y)
sy

or x − 10 = 0.6.1.5 (y − 20) or x − 10 = 0.45 (y − 20)
2

or x − 10 = 0.45y − 9 or x = 0.45y − 9 + 10

or x = 0.45y + 1

Regression equation of y on x:

y- y = byx (x - x) or y - y = r × s y (x - x)
s x

or y − 20 = 0.6. 2 (x − 10) or y − 20 = 0.8(x − 10)
1.5

or y = 0.8x − 8 + 20 or y = 0.8x + 12

when x = 25

Then y = 0.8 × 25 + 12 = 20 + 12 = 32

7.28 Business Mathematics and Statistics

EXERCISE

A. THEORY

1. Briefly explain the concept of regression.

2. What are the regression coefficients?

3. Why are there two regression lines?

4. Show that correlation coefficient is a geometric mean of regression coeffi-

cients. [C.U. B.Com. 1991]

5. If two regression lines coincide, determine the value of the correlation coefficient
between x and y and mention when they will be perpendicular to each other.

6. Obtain the equations of the two lines of regression for a bivariate distribution.
[C.U.B.Com. 1985, 1987]

7. What are the differences between correlation and regression?

8. Mention the important properties of regression lines.

9. What are the uses of regression?

10. Write short notes on: [C.U. B.Com. 1982]
(i) Regression [C.U. B.Com. 1983]
(ii) Regression analysis

(iii) Regression coefficients
(iv) Regression equations

B. SHORT TYPE

1. Find the regression equation of y on x from the following values:

(i) x = 15, y = 20,byx = 3.5 [Ans. y = 3.5x – 32.5]

(ii) x = 10, y = 15,byx = 2.5 [Ans. 2y = 5x –20]

2. Find the regression equation of x on y from the following values:

(i) x = 15, y = 10 and bxy = 2.5 [Ans. x = 2.5y –10]

(ii) x = 6, y = 1 and bxy = −0.4 [Ans. x = –0.4y + 6.4]

3. Find the equations of the lines of regression using the following data:

x = 4, y = 5,bxy = 0.39 and bxy = 0.69

[C.U. B.com. 1982] [Ans. y = 0.39x + 3.44; x = 0.69y + 0.55]

4. Find the regression equation of y on x from the following values:
__
x = 10, y = 15 and bxy = 2.50 [C.U. B.Com. 1988] [Ans. y = 2.5x – 10]

Regression Analysis 7.29

5. If x = 4, y = 5,byx = 0.35,bxy = 0.65 then find the equation of the regression lines.

[C.U. B.Com. 1991] [Ans. x = 0.65y + 0.75; y = 0.35x + 3.6]

6. If x = 3, y = 4,bxy = byx = 0.9 , then find the equation of the regression lines.

[Ans. y = 0.9x + 1.3; x = 0.9y − 0.6]

7. Estimate the correlation coefficient between x and y from the following two

regression lines: y = 1.5x + 2.3, x = 0.4y + 1.8 [Ans. 0.77]

8. Find the value of the correlation coefficient r when byx = –0.4 and bxy = –0.9.
(Here byx and bxy are the regression coefficients of the two regression lines).
[C.U. B.Com 1983, 86, 89] [Ans. –0.6]

9. The regression coefficient of y on x and x on y are 1.2 and 0.3 respectively. Find

the coefficient of correlation. [C.U. B.Com. 1994] [Ans. +0.6]

10. The regression coefficient of y on x and x on y are –1.2 and –0.3 respectively.
Find the coefficient of correlation.
[C.U. B.Com. 1997] [Ans. –0.6]

11 If the regression coefficient are 0.8 and 0.6, what would be the value of the

coefficient of correlation. [Ans. 0.69]

12. Find the mean values of x and y when the regression equations are 3x – 2y = 4.5

and 2x – y = 3.5 [Ans. x = 2.5, y = 1.5]
__
13. Find x and y if the regression equations are 5x – 7y = 15 and 4x – 15y = –35.
_
[Ans. x = 10, y = 5]
__
14. Find x and y; if the regression equations are 5x – 2y – 4 = 0 and 4x – 7y + 13 = 0

[Ans. x = 2, y = 3]

15. If the two regression equations are 8x – 10y + 66 = 0 and 40x – 18y = 214, find

the average values of x and y. __

[C.U. B.Com. 2010, 2016(H)] [Ans. x = 13, y = 17]

16. The regression equation of y on 3x – 5y = 13 and the regression equation of x
on y is 2x – y = 7. Estimate the value of x when y = 10.
[C.U. B.Com. 1996] [Ans. 8.5]

17. The regression equation of y on x is 4x – 3y = 11 and the regression equation of
x on y is 5x – 2y = 9
Estimate (i) the value of x when y = 3 (ii) the value of y when x = 5
[Ans. (i) 3, (ii) 3]

18. The regression equation of y on x is 15x – 4y = 14 and the regression equation

of x on y is 7x + 2y = 11. Estimate (i) the value of x when y = 2 (ii) the value of

y when x = 4. [Ans. (i) x = 1 (ii) y = 11.5]

19. If s = 10, s = 12, bxy = –0.8 find the value of r.

x y

. [C.U. B.Com. 1998] [Ans. –0.96]

7.30 Business Mathematics and Statistics

20. If s = 4.5, s = 13 and byx = 1.04, then find the value of r. [Ans. 0.36]

x y

21. If r = ± 1, will the two regression lines be perpendicular to each other?

[V.U. B.Com. 1988] [Ans. No]

22. Find the regression coefficients for the following data:

n = 10,s x = 12,s y = 8 and S (x - x) (y - y) = 250 [Ans. 0.26]

C. SHORT ESSAY / PROBLEM TYPE

1. Obtain two lines of regression from the following data:

x: 4 5 6 8 11

y: 12 10 8 7 5

[Ans. y = –0.93x + 14.72; x = –0.98y + 15.03]

2. From the following data find the two regression equations:

x: 1 2 3 4 5
y: 2 3 5 4 6

Predict the value of y when x = 2.5.
[C.U. B.Com. 1983, 89] [Ans. y = 0.9x + 1.3; x = .9y – .6; 3.55]

3. Find the linear regression equation of Y on X for the data:

x: 1 2 3 4 5
y: 3 2 5 4 6

[C.U. B.Com. 1985] [Ans. y = 0.8x + 1.6]
4. Find the linear regression equation of Y on X for the following data:

X: 1 2 3 4 5

Y: 6 8 11 8 12

Find also the most probable value of Y when X = 2.5.
[C.U. B.Com. 1992] [K.U. B.Com. ’97, ‘99] [Ans. Y = 1.2X + 5.4;8.4]

5. Find the linear regression equation of y on x from the following data:

x: 1 2 3 4 5
y: 3 3 5 4 6

[C.U. B.Com. 1994] [Ans. y = 0.8x + 1.6]
6. From the following data find the two regression equations:

Age (Years): 1 3 4 5 7
Weight (Kg.) 3 5 8 12 17

What will be the most probable weight of a baby at the age of 8 years?

[C.U. B.Com. 1996] [Ans. Y = 2.4X − 0.8; X = 0.39Y + 0.49;18.8 kg.]

Regression Analysis 7.31

7. (i) What do you mean by Regression ?
(ii) Find the two regression equations from the following data:

x: 6 2 10 4 8 12 14 16
y: 9 11 5 8 7 11 16 18

[C.U. B.Com. 1998] [Ans. x = 0.67y + 1.88; y = 0.55x + 5.675.]

8. From the following data obtain the two regression equations and calculate the
correlation coefficient:

x: 1 2 3 4 5 6 7 8 9
y: 9 8 10 12 11 13 14 16 15

Estimate the value of y which should correspond on an average to x = 6.2

[Ans. x = 0.95y − 6.4; y = 0.95x + 7.25; y = 13.14,r = 0.95]

9. Find the equation of the line of regression of Y on X from the following data:

Husband’s age (X) 25 27 29 33 26 32 35 31 35 30
24 24 29 21 26 28 30 29 28
Wife’s age (Y) 20
[C.U. B.Com. 1988] [Ans. 0.319]
10. Given the bivariate data:
321173
X: 1 5 001215
Y: 6 1

(a) Fit the regression line of Y on X and hence predict Y if X = 10.
(b) Fit the regression line of X on Y and hence predict X, if Y = 2.5.

[Ans. (a) Y = 2.86 − 0.30X; − 0.14 (b)X = 3.43 − 0.284Y;2.73]

11. Write the equation of the line of regression of Y on X.

X: 78 89 97 69 59 79 68 61
Y: 125 137 156 112 107 136 123 108

[Ans. Y = 1.22X + 34]
12. The grade of 9 students at the college test (X) and at the University examination

(Y) are as follows:

X: 77 50 71 72 81 94 96 99 67
Y: 82 66 78 84 47 85 99 99 68

Find a linear regression equation for the data and then estimate the grade of
the University examination for a student who received 85 in the college test but
was sick at the time of University examination.

[C.U. B.Com. (Hons) ’83] [Ans. Y = 0.615X + 30.36; 82.635]

7.32 Business Mathematics and Statistics

13. By using the following data, find out the two lines of regression and from them
compute the Karl Pearson’s coefficient of correlation:
Σx = 250;Σy = 300;Σxy = 7900; Σx2 = 6500; Σy2 = 10,000 and N = 10

[Ans. x = 0.4y + 13; y = 1.6x − 10;r = 0.8]

14. Find the regression line of y on x from the following results:

N = 10, Σx = 350, Σ y = 310, Σ (x − 35)2 = 162

Σ (y − 31)2 = 222, Σ (x − 35)(y − 31) = 92

[Ans. y = 0.568x – 19.88]
15. Given the following data:

x = 36, y = 85, s x = 11, s y = 8, r = 0.66
Find the two regression equations and estimate the value of x when y = 75.

[Ans. x = 0.9075y − 41.1375; y = 0.48x + 67.72; x = 26.925]

16. Find the regression equation of x on y from the following data :

n = 10,Σx = 30,Σ y = 90, Σx2 = 110, Σy2 = 858,Σ xy = 294 .

Find the estimated value of x, when y = 8. [Ans. x = 0.5y – 1.5; x = 2.5]

17. You are given the following data:

xy

Arithmetic Mean: 20 25

Standard Deviation: 5 4

Correlation coefficient between X and Y is 0.5.

Find the two regression equations.

é byx = r. s y and bxy = r sx ù
êGiven: s x sy ú
ëê ûú

[C.U. B.Com. 1986] [Ans. y = 0.48x + 15.4; x = 0.75y + 1.25]

18. The following data are given for marks in English (x) and Mathematics (y) in a
certain examination:

Mean Marks English Mathematics
S. D. Marks 39.5 47.5
10.8 16.8

Coefficient of correlation between marks in English and Mathematics = +0.40.
Find the two regression equations.

[C.U. B.Com. 1990] [Ans. x = 0.25y + 27.63; y = 0.62x + 23.01]

Regression Analysis 7.33

19. The following results were obtained from the record of age (x) and blood
pressure (y) of a group of 10 women:

Mean x y
Variance 53 142
130 165

Σ (x − x ) (y − y) = 1,220

Find the regression equation of y on x and use it to estimate the blood pressure
of a woman of age 45.

[C.U. B.Com. 1993] [Ans. y = 0.94x + 92.18;134.48]

20. Given

Mean x series y series
Standard Deviation 18 100
14 20

Coefficient of correlation between X and Y is + 0.8. Find out the most probable
value of y if x is 70 and most probable value of x if y is 90.

[Ans. y = 79.48 + 1.14x, y = 159.28; x = 0.56y − 38, x = 12.4]

21. Given the following data find what will be the probable yield when the rainfall
is 29''.

Rainfall Production
_
x 25'' 40 units per acre

s 3'' 6 units

r between rainfall and production = 0.8 [Ans. 50 units per acre]

22. In a correlation study the following values are obtained:

Mean x y
Standard Deviation 65 67
2.5 3.5

Coefficient of correlation 0.8.
Find the two regression equations that are associated with the above values.

[Ans. x = 0.5714y + 26.72; y = 1.12x − 5.8]

23. For the variables x and y the equations of regression lines of y on x and x on y

respectively are 4x – 5y + 33 = 0 and 20x – 9y = 107. What is the correlation

coeff_ici_ent. If variance of x is 9 find the standard deviati_on of y_. Also
find x , y [C.U. B.Com. 1999] [Ans. + 0.6; 4; x = 13, y = 17]

24. Two random variables have the least square regression lines with equations
3x + 2y = 26 and 6x + y = 31.

7.34 Business Mathematics and Statistics

Find the mean values and the coefficient of correlation be_tween_x and y.
[Ans. x = 4, y = 7, r = –0.5]

25. For certain x and y series which are correlated, the two lines of regression are

5x – 6y + 90 = 0 and 15x – 8y – 130 = 0 . Find the me_ans of t_he two series and
the correlation coefficient. [Ans. x = 30, y = 40, r = 0.667]

26. Find out sy and r from the following data: 3x = y, 4y = 3x and sx = 2.

[Ans. sy = 3; r = 0.5]

27. The equations of two lines of regression obtained in a correlation analysis are

as following:

2x = 8 –3y and 2y = 5 – x

obtain the value of the correlation coefficient. [Ans. – 0.866]

28. Two lines of regression are given by x + 2y = 5 and x+ 3y = 9 and s2x = 12.
__ _ _
Calculate the values of x , y, s2y and r. [Ans. x = 1, y = 2, sy2 = 4, r = – 0.87]

29. The lines of regression of y on x and x on y are respectively y = x + 5 and

16x = 9y – 94. Find the variance of x if the variance of y is 16. Also find the

covariance of x and y. [Ans. sx2 = 9; cov (x, y) = 9]

30. From the data given below find:

(a) The two regression equations.

(b) The coefficient of correlation between the marks Economics and Statistics.

(c) The most likely marks in Statistics when marks in Economics are 30.

Marks in Economies 25 28 35 32 31 36 29 38 34 42
Marks in Statistics 43 46 49 41 36 32 31 30 33 39

[D.U. ’82] [Ans. (a) y = 59.1 − 0.66x; x = 40.7 − 0.23y (b)r = 0.389 (c) y = 39.3]

31. Calculate the (i) two regression coefficient, (ii) coefficients of correlation, and
(iii) the two regression equations from the following informations:

N = 10,Σ x = 350,Σ y = 310,Σ (x − 35)2 = 162,Σ (y − 31)2 = 222,
Σ (x − 35)(y − 31) = 92

[D.U. B.Com. ’92]
Ans. bxy = 0.41, byx = 0.57, r = 0.24, x = 22.15 + 0.41y, y = 11.12 + 0.57x
32. In studying a set of pairs of related variates a statistician has completed the
preliminary calculations. The results are as follows:

n = 16,Σ x = 749,Σ y = 77.90,Σx2 = 42.177,Σy2 = 454.81,Σxy = 3,156.80

Compute the linear regression equation of x on y.
[C.U. B.Com. (Hons)’81] [Ans. x = 78.4 – 6.49y]

33. Find the regression equation of x on y from the following data:

Regression Analysis 7.35

s x = 24,Sy = 44,Sxy = 306,Sx2 = 164,Sy2 = 574, N = 4
[Ans. x = 0.467y + 0.863, x = 3.7]

34. Find the two regression equations from the following data:
n = 20,Σx = 80,Σy = 40,Σxy = 480,Σx2 = 1680,Σy2 = 320

[Ans. y = 0.235x + 1.06, x = 1.33y + 1.334]

35. Calculate the regression coefficients from the following informations:

Σx = 50,Σy = 30,Σxy = 1000,Σx2 = 3000,Σy2 = 1800, N = 10

[C.A. (F) Nov.’95] [Ans. byx = 0.31; bxy = 0.50]
36. To study the relationship between expenditure on accommodation, (` x) and

Expenditure on Food and Entertainment, (` y), an enquiry into 50 families gave
the following results.

å x = 8500, å y = 9600, s x = 60, s y = 20, r = 0.6.

Estimate the Expenditure on Food and Entertainment when Expenditure on

accommodation is ` 200. [C.A. (Inter) Nov. ’79] [Ans. ` 198]

37. For a bivariate data, you are given the following information: Σ (x – 58) = 46,
Σ (x – 58)2 = 3086, Σ (y – 58) = 9, Σ(y – 58)2 = 483, Σ (x – 58) (y – 58) = 1905

Number of pairs of observations = 7.

You are required to determine (i) the two regression equations and (ii) the
coefficient of correlation between x and y series.
[D.U. B.Com. (Hons.) ’88] [Ans. y = 0.37x + 35.399; x = 2.20y – 65.87; 0.902]

D. MISCELLANEOUS

38. You are given the following data:

A.M. X Y
S.D. 36 85
11 8

If correlation coefficient between X and Y is 0.66 find the two regression
equations. Also find the value of X when Y = 33.5 and that of Y when X = 73.3
from the two regression lines.

[Ans. X = 0.9075Y – 41.1375; Y = 0.48X + 67.72; – 10.73625; 102.904]

39. Find the means of X and Y variables and the coefficient of correlation between
them from the following two regression equations:
2y – x – 50 = 0; 3y – 2x – 10 = 0

[D.U. B.Com. (Hons.) ’83] [Ans. x = 130, y = 90, 0.866]

7.36 Business Mathematics and Statistics

40. The following data about the sales and advertisement expenditure of a firm is
given below:

Mean Sales Advertisement expenditure
Standard deviations (in crores of `) (in crores of `)
6
40 1.5
10

Coefficient of correlation = r = 0.9

(i) Estimate the likely sales for a proposed advertisement expenditure of

` 10 crores.

(ii) What should be the advertisement expenditure if the firm proposes a sales

target of 60 crores of rupees? [D.U.B.Com. (Hons.) ’85]

[Ans. (i) X (sales) = 6Y (Adv.exp) + 4, ` 64 crores
(ii) Y = 0.135 X + 0.60, `8.70 crores]

41. The following data relate to marks in Advanced Accounts and Business
Statistics in B.Com.(Hons.) 1st year examination of a particular year in Delhi

University:

Mean marks in Advanced Accounts = 30

Mean Marks in Business Statistics = 35

Standard Deviation of Marks in Advanced Accounts = 10

Standard Deviation of Marks in Business Statistics = 7

Coefficient of correlation between the marks of Advanced Accounts and
Business Statistics = 0.8

Form the two regression lines calculate the expected marks in Advanced
Accounts if the marks secured by a student in Business Statistics are 40.

[D.U. B.Com. (Hons.) ’87] [Ans. Marks in Advanced Account = X, Marks in
8

Business Statistics = Y; Y = 0.5X + 18.2; X = Y – 10; 36 Marks]
7

42. In order to find the correlation coefficient between two variables x and y from
12 pairs of observations, the following calculations were made: Sx = 30; Sx2 =
670; Sy = 5; Sy2 = 285; Sxy = 344 on subsequent verification it was discovered
that the pair (x = 11, y = 4) was copied wrongly, the correct values being x =
10, y = 14. After making necessary corrections, find: (a) the two regression
coefficients. (b) the two regression equations (c) the correlation coefficient.

[D.U. B.Com. (Hons.) ’90]
[Ans. (a) 0.898, 0.694 (b) x = 0.898y + 1.294, y = 0.694x – 0.427 (c) 0.79]

43. A panel of judges A and B graded seven debators and independently awarded
the following marks:

Regression Analysis 7.37

Debator: 1234567

Marks by A: 40 34 28 30 44 38 31

Marks by B: 32 39 26 30 38 34 28

An eighth debtor was awarded 36 marks by judge A while judge B was not
present.

If judge B was also present, how many marks would you expect him to award
to eighth debator assuming same degree of relationship exists in judgment?

[D.U. B.Com. (H) ’93] [Ans. 33 (Approx).]

[Hints: y = 0.587x + 11.885. Putting x = 36 we get y = 33.017 or 33 Approx.]

44. The line of regression of marks in Statistics (x) on marks in Accountancy (y)
for a class of 50 students is 3y – 5x + 180 = 0. Average marks in Accountancy is

44 and variance of marks in Statistics is 9 th of variance of marks in Accoun-
16

tancy. Find: (i) the average marks in statistics; and (ii) coefficient of correlation
between marks in Statistics and marks in Accountancy.

[D. U.B.Com (H) ’94] [Ans. (i) 62.4 marks (ii) 0.8]

E. MULTIPLE CHOICE QUESTIONS (MCQs)

(i) Short Type

Mark the correct alterative in each of the following:

1. If 33 then the value of rxy is [C.U. B.Com. 2013 (G)]
bxy = − ,byx = − ,

20 5

(a) –0.25 (c) –0.4
(b) –0.3
(d) 4 [Ans. (b)]
−5

2. If regression coefficient of y on x is –0.4 and regression coefficient of x on y is

–0.9, the correlation coefficient between x and y is

[C.U. B.Com. 2013 (H)]

(a) –0.6 (c) –0.45

(b) –0.5 (d) –0.7 [Ans. (a)]

3. Two lines of regression are given by 3x – 2y = 5, 2x – y = 4, the value of x and

y are [C.U. B.Com. 2012]

(a) (3 and 2) (c) (4 and 3)

(b) (5 and 4) (d) (3 and 1) [Ans. (a)]

4. If s x = 10, s y = 12, bxy = –0.8, the value of rxy is

(a) 0.75 (c) –0.96

(b) 0.89 (d) –0.98 [Ans. (c)]

7.38 Business Mathematics and Statistics

__
5. s x = 36, y = 30, x = 36, r = 0.8, sy = 32

The coefficient of regression of x and y from the above information is

(a) 0.48 (c) 0.40

(b) 0.55 (d) 0.90 [Ans. (d)]

6. In simple linear regression, the numbers of unknown constants are:

(a) one (c) three

(b) two (d) four [Ans. (b)]

7. If the value of any regression coefficient is zero, then two variables are.

(a) dependent (c) qualitative

(b) independent (d) correlated [Ans. (b)]

8. In the regression equation y = a + bx, the y is called

(a) dependent variable (c) qualitative variable

(b) independent variable (d) none of the above [Ans. (a)]

9. When bxy is positive, then byx will be: (c) negative [Ans. (d)]
(a) zero (d) positive

(b) one

10. Regression coefficient is independent of

(a) units of measurement (c) both (a) and (b)

(b) scale and origin (d) none of these [Ans. (c)]

11. When the two regression lines are parallel to each other, then their slopes are:

(a) zero (c) same

(b) different (d) positive [Ans. (c)]

12. In the regression equation y = a + bx, a is called

(a) x-intercept (c) dependent variable

(b) y-intercept (d) none of the above [Ans. (b)]

13. The regression equations always passes through:

(a) (x, y) (c) (x, y)

(b) (a, b) (d) (x, y) [Ans. (c)]

14. The straight line graph of the linear equation y = a + bx, slope will be

downward if:

(a) b < 0 (c) b = 0

(b) b > 0 (d) b ≠ 0 [Ans. (a)]

15. If y = –10x and x = – 0.1y, then r is equal to

(a) 10 (c) 1

(b) 0.1 (d) –1 [Ans. (d)]

16. If rxy = 1, then (c) bbyyxx.<bxbyx=y 1 [Ans. (d)]
(a) byx = bxy (d) [Ans. (c)]
(b) byx > bxy
(c) > 0
17. If rxy > 0, then byx and bxy are both (d) < 1
(a) 0

(b) < 0

Regression Analysis 7.39

18. If bxy = 0.20 and rxy = 0.50, then byx is equal to

(a) 0.20 (c) 0.50

(b) 0.25 (d) 0.125 [Ans. (d)]

19. If bxy = –0.5 and rxy = –1, then byx is equal to

(a) –1 (c) –0.5

(b) –2 (d) 0.5 [Ans. (b)]

20. If sx = 10, sy = 12 and bxy = – 0.8, then the value of r is

(a) 0.92 (c) –0.96

(b) –0.86 (d) 0.89

[V.U. B.Com 1992] [Ans. (c)]

(ii) Short Essay Type

21. x = 0.64y + 19.10 ; y = x + 5.25

The regression coefficient bxy from the above details is

(a) 0.85 (c) 0.98

(b) 0.64 (d) 1 [Ans. (b)]

22. 9x = 5y + 9.10 ; y = 3x – 7

The regression coefficient bxy from the above details is

(a) 5 (c) 1.08
9

(b) 9 (d) 2.3 [Ans. (a)]
5

7
23. x = 3 y + 28.10 ; y = 1.5x + 10

The regression coefficient byx from the above details is

(a) 2.9 (c) 3/77

(b) 1.5 (d) 7/3 [Ans. (b)]

24. If the regression coefficient bxy is 2.5, the value of a in the given equation
2x = ay + 12.6 is

(a) 4 (c) 3.32

(b) 2.5 (d) 5.0 [Ans. (d)]

25. If the regression coefficient byx is 3.0, the value of b in given equation 2y = bx +
18 is

(a) 2.5 (c) 6.0

(b) 1.5 (d) 4.0 [Ans. (c)]

26. From the regression equations 2x – 8y + 60 = 0 and 40x – 18y – 220 = 0, the

value of bxy and byx are

(a)  9 , 14  (b)  19 , 2 
20 20 5

7.40 Business Mathematics and Statistics

(c)  8 , 45  (d)  15 , 15 [Ans. (a)]
20 20

27. From the regression equations 8x – 10y + 66 = 0 and 40x – 18y – 214 = 0, the

value of mean x and mean y are

(a) (19,21) (c) (11,15)

(b) (13,17) (d) (16,19) [Ans. (b)]
__
28. If rxy = 0.5, s x = 1.5, s y = 3.0, x = 10 and y = 8, the regression line of x on y is

(a) x = 0.25y + 8 (c) x = 1.54y – 10.6

(b) x = 0.56y + 9 (d) x = 0.34y – 8 [Ans. (a)]
__
29. If rxy = 0.6, s x = 4, s y = 1.33, x = 15, and y =10, the regression line y on x is

(a) y = 2.6x –14 (c) y =1.5x – 10

(b) y = 1.88x – 15 (d) y = 0.2x+7 [Ans. (d)]

30. Average rainfall in W.B. = 40 cm, standard deviation of rainfall = 3 cm,

mean of paddy yield = 800 quintal, standard deviation of paddy production

= 10 quintal, correlation = 0.6, the estimate of production of paddy in 2017

corresponding to the estimate of 72 cm rainfall is

(a) 978 quintal (c) 753.84 quintal

(b) 640.9 quintal (d) 773 quintal [Ans. (c)]

31. If n = 10, sx = 15, sy = 10 and ∑(x – x)(y – y) = 980, then byx is

(a) 0.435 (c) 0.413

(b) 0.529 (d) 0.517 [Ans. (a)]

32. The regression equation of y on x and x on y are y = x + 5 and 16x = 9y – 94
respectively. If the variance of y is 16, the standard deviation of x is

[V.U. B.Com 1990]

(a) 2 (c) 4

(b) 3 (d) 5 [Ans. (b)]

33. If x = 5, y = 12, s x = 3, s = 4 and r = 0.75, the regression equation of x

y

on y is

(a) x = 0.42y + 1.15 (c) x = 0.45y + 2.32

(b) x = 0.5y – 2.32 (d) x = 0.56y – 1.75 [Ans. (d)]

34. The regression equation of x on y from the following data byx = –1.2,

bxy = –0.3, x = 10, y = 12 is

(a) x = –0.3y + 13.6 (c) x = 0.29y + 11.25

(b) x = 0.42y – 12.3 (d) x = –0.23y + 12.7 [Ans. (a)]

35. Two lines of regression are given by x + 2y = 5 and 2x + 3y = 8. The values of

x and y are [V.U. B.Com.’97]

(a) 3, 4 (c) 1, 2

(b) 2, 3 (d) 2, 5 [Ans. (c)]

36. The regression coefficient of y on x for the following information:
Sx = 50, Sy = 30, Sxy = 1000, Sx2 = 3000, Sy2 = 1800, N = 10 is

Regression Analysis 7.41

(a) 0.35 (c) 0.41 [Ans. (b)]
(b) 0.31 (d) 0.45

37. The regression coefficient of y on x is 4 and that of x on y is 1/9. The value of r

is 2/3. If standard deviation of y is 12, then the standard deviation of x is

(a) 2 (c) 4

(b) 3 (d) 5 [Ans. (a)]

38. The regression equation of x and y from the following data is:

Sx = 24, Sy = 44, Sxy = 306, Sx2 = 164, Sy2 = 574, n = 4

(a) x = 0.52y + 0.29 (c) x = 0.467y + 0.86

(b) x = 0.37y + 1.73 (d) x = 0.49y – 3.23 [Ans. (c)]

39. The value of sy for the following data 3x – y, 4y = 3x and sx = 2 is

(a) 6 (c) 4

(b) 5 (d) 3 [Ans. (d)]

40. Given bxy = 0.85, byx = 0.89 and the standard deviation of x = 6. Then the value

of sy is

(a) 5.69 (c) 6.25

(b) 6.14 (d) 5.78 [Ans. (b)]

How do you find the regression coefficient of BXY?

What is the regression coefficient BXY from the following details?

What is regression coefficient formula?

What is BXY and Byx?