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Questions:

Generic 5

4 x 5 = 20

For the following observations, find the regression co-efficient b_yx and b_xy and hence find the correlation co-efficient (4,2)(2,3)(3,2)(4,4)(2,4).

Obtain the two regression lines from the following

X	6	2	10	4	8
Y	9	11	5	8	7

An examination of 11 applicants for a accountant post was taken by a finance company. The marks obtained by the applicants in the reasoning and aptitude tests are given below.

Applicant	A	B	C	D	E	F	G	H	I	J	K
Reasoning test	20	50	28	25	70	90	76	45	30	19	26
Aptitude test	30	60	50	40	85	90	56	82	42	31	49

Calculate Spearman’s rank correlation coefficient from the data given above.

Calculate correlation coefficient for the following data.

X	25	18	21	24	27	30	36	39	42	48
Y	26	35	48	28	20	36	25	40	43	39

Generic 2

1 x 2 = 2

Obtain the two regression lines from the following data N = 20, ΣX = 80, ΣY = 40, ΣX²= 1680, ΣY²= 320 and ΣXY = 480

Generic 3

4 x 3 = 12

If two lines of regression are 3X-2Y+1=0, 2X-Y-2=0, find \(\bar {X}\)and \(\bar{Y}\).

Explain the concept of correlation and co-efficient of correlation.

X and Y are a pair of correlated variables. Ten observations of their values (X, Y) have the following results. ΣX = 55, ΣXY = 350, ΣX² = 385, ΣY = 55, Predict the value of Y when the value of X is 6.

A survey was conducted to study the relationship between expenditure on accommodation (X) and expenditure on Food and Entertainment (Y) and the following results were obtained:

Details	Mean	SD
Expenditure on Accommodation (Rs)	Rs. 178	63.15
Expenditure on Food and Entertainment (Rs)	Rs 47.8	22.98
Coefficient of Correlation	0.43

Write down the regression equation and estimate the expenditure on Food and Entertainment, if the expenditure on accommodation is Rs. 200.

Multiple Choice Question

5 x 1 = 5

10.

If Cov(x, y) = –16.5, \({ \sigma }_{ x }^{ 2 }=2.89,{ \sigma }_{ y }^{ 2 }\) = 100. Find correlation coefficient ________.

(a)

-0.12

(b)

0.001

(c)

-1

(d)

-0.97

11.

The lines of regression intersect at the point ________.

(a)

(X,Y)

(b)

\(\left( \bar { X } ,\bar { Y } \right) \)

(c)

(0,0)

(d)

(σ_x, σ_y)

12.

When one regression coefficient is negative, the other would be ________.

(a)

Negative

(b)

Positive

(c)

Zero

(d)

None of them

13.

The variable whose value is influenced (or) is to be predicted is called ________.

(a)

dependent variable

(b)

independent variable

(c)

regressor

(d)

explanatory variable

14.

Correlation co-efficient lies between ______.

(a)

0 to ∞

(b)

-1 to +1

(c)

-1 to 0

(d)

-1 to ∞

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Answers Key:

Generic 5

4 x 5 = 20

x	y	x²	y²	xy
4	2	16	4	8
2	3	4	9	6
3	2	9	4	6
4	4	16	16	16
2	4	4	16	8
15	15	49	49	44

Here n=5,
b_yx=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ n\sum { { x }^{ 2 }-{ (\sum { x } ) }^{ 2 } } } \)
=\(\frac { 5(44)-(15)(15) }{ 5(49)-{ (15) }^{ 2 } } \)
=\(\frac { 220-225 }{ 245-225 } =\frac { -5 }{ 20 } =\frac { -1 }{ 4 } \)
b_xy=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ n\sum { { y }^{ 2 }-{ (\sum { y } ) }^{ 2 } } } \)
=\(\frac { 5(44)-15(15) }{ 5(49)-{ (15) }^{ 2 } } \)
\(\frac { 220-225 }{ 245-225 } =\frac { -5 }{ 20 } =\frac { -1 }{ 4 } \)
b_xy and b_yx are negative, r is also negative.
\(\therefore\)Correlation co-efficient
r(x,y)=-\(\sqrt { { b }_{ xy }.{ b }_{ yx } } \)
=-\(\sqrt { \left( \frac { -1 }{ 4 } \right) \left( \frac { -1 }{ 4 } \right) } =\frac { -1 }{ 4 } \)
r=-0.25

Here N=5

X	Y	X²	Y²	XY
6	9	36	81	54
2	11	4	121	22
10	5	100	25	50
4	8	16	64	32
8	7	64	49	56
30	40	220	340	214

\(\overline { X } =\frac { \sum { X } }{ N } =\frac { 30 }{ 5 } \)=6
\(\overline { Y } =\frac { \sum { Y } }{ N } =\frac { 40 }{ 5 } \)=8
\({ b }_{ xy }=\frac { N\sum { XY-(\sum { X } )(Y) } }{ N\sum { { X }^{ 2 }-{ (\sum { Y } ) }^{ 2 } } } =\frac { 5(214)-(30)(40) }{ 5(340)-{ (40) }^{ 2 } } \)
=\(\frac { 1070-1200 }{ 1700-1600 } =-\frac { 130 }{ 100 } \)=-1.3
b_yx=\({ b }_{ xy }=\frac { N\sum { XY-(\sum { X } )(Y) } }{ N\sum { { X }^{ 2 }-{ (\sum { X } ) }^{ 2 } } } =\frac { 5(214)-(30)(40) }{ 5(2200)-{ (30) }^{ 2 } } \)
\(=\frac { 1070-1200 }{ 1700-900 } =\frac { -130 }{ 200 } \)=-0.65
Regression equation of X on Y is
\(\Rightarrow\)X-\(\bar{X}\)=b_xy(Y-\(\bar{Y}\))
X-6=-1.3(Y-8)
X=-1.3Y+10.4+6
X=-1.3Y+16.40
Regression equation of Y on X is
Y-\(\bar{Y}\) =b_yx(X-\(\bar{X}\))
Y-8=-0.65(X-6)
Y=-0.065X+3.9+8
Y=-0.65X+11.90

Applicant	X	Y	R_x	R_y	d=R_x-R_y	d²
A	20	30	2	1	1	1
B	50	60	8	8	0	0
C	28	50	5	6	-1	1
D	25	40	3	3	0	0
E	70	85	9	10	-1	1
F	90	90	11	11	0	0
G	76	56	10	7	3	9
H	45	82	7	9	-2	4
I	30	42	6	4	2	4
J	19	31	1	2	-1	1
K	26	49	4	5	-1	1
						\(\sum\)d²= 22

\(\rho =1-\frac{6 \sum d^2}{n\left(n^2-1\right)} \)
\(=1-\frac{6(22)}{11(120)}=1-\frac{1}{10}=1-0.1=0.9\)

X	Y	x²	y²	xy
25	26	625	676	650
18	35	324	1225	630
21	48	441	2304	1008
24	28	576	784	672
27	20	729	400	540
30	36	900	1296	1080
36	25	1296	625	900
39	40	1521	1600	1560
42	43	1764	1849	1806
48	39	2304	1521	1872
\(\sum\)X = 310	\(\sum\)Y = 340	\(\sum\)X² = 10480	\(\sum\)Y² = 12280	\(\sum\)XY = 10718

\(r(x, y)=\frac{N \Sigma X Y-\left(\sum X\right)\left(\sum Y\right)}{\sqrt{N \Sigma X^2-(\Sigma Y)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(10718)-(310)(340)}{\sqrt{10(10480)-(310)^2} \sqrt{10(12280)-(340)^2}} \)
\(=\frac{107180-105400}{\sqrt{104800-96100} \sqrt{122800-115600}} \)
\(=\frac{1780}{\sqrt{8700 \times 7200}}=\frac{1780}{7914.54}=0.2249\)

Generic 2

1 x 2 = 2

\(\bar{X} =\frac{\Sigma X}{N}=\frac{80}{20}=4 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{40}{20}=2 \)
\(b_{x y} =\frac{N \sum X Y-(\Sigma X)(\Sigma Y)}{N \Sigma Y^2-(\Sigma Y)^2} \)
\(=\frac{20(480)-(80)(40)}{20(320)-(40)^2} \)
\(=\frac{9600-3200}{6400-1600}=\frac{6400}{4800}=1.33 \)
\(b_{y x} =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{N \Sigma X^2-(\Sigma X)^2}=\frac{6400}{20(1680)-(80)^2} \)
\(= \frac{6400}{33600-6400}=\frac{6400}{27200}=0.235\)
Regression equation of X on Y
\(X-\bar{X} =b_{x y}(Y-\bar{Y}) \)
X - 4 = 1.33(Y - 2)
X = 1.33 Y - 2.66 + 4
X = 1.33 Y + 1.34
Regression equation of Y on X is
\(Y-\bar{Y} =b_{y x}(X-\bar{X}) \)
Y - 2 = 0.235(X - 4)
Y = 0.235 X - 0.94 + 2
Y = 0.235X + 1.06

Generic 3

4 x 3 = 12

The two lines of regression are
3X-2Y =-1 ..(1)
2X-Y =2 ..(2)
Since these lines meet in (\(\bar {X}\), \(\bar{Y}\)), let us solve (1) and (2)
(1) 3X-2Y =-1
(-) (+) (-)
(2)x2 4X-2Y=4
_________________
Subtracting, -X=-5 \(\Rightarrow\)\(\bar{X}\)=5
Substitutuing \(\bar{X}\)=5 in (1) we gwt,
\(\Rightarrow\)3(5)-2Y=-1 15-2Y=-1
\(\Rightarrow\)16=2Y
\(\Rightarrow\)Y=8
\(\bar{Y}\)=8
Hence, \(\bar{X}\)=5 and \(\bar{Y}\)=8

The relationship between two variables such that a change in one results in a corresponding change in the other is known as correlation.
The degree to which the variables are interrelated is measured by a co-efficient known as the co-efficient of correlation.

\(\bar{X} =\frac{\Sigma X}{N}=\frac{55}{10}=5.5 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{55}{10}=5.5 \)
\(b_{y x} =\frac{N \Sigma X Y-\Sigma X \Sigma Y}{N \Sigma X^2-(\Sigma X)^2} \)
\(=\frac{10(350)-(55)(55)}{10(385)-55^2}=\frac{3500-3025}{3850-3025} \)
\(=\frac{475}{825}=0.576\)
Regression line of Y on X is.
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 5.5 = 0.576(X - 5.5)
Y = 0.576 X - 3.168 + 5.5
Y = 0.576 X + 2.332
At X = 6, Y = 0.576(6) + 2.332
= 3.456 + 2.332
= 5.788

Let X, Y represent the expenditure on accommodation and food.
\(\bar{X} =178, \bar{Y}=47.8 \)
\(\sigma_x =63.15, \sigma_y=22.98, r=0.43 \)
\(b_{y x} =r \frac{\sigma_y}{\sigma_x}=0.43\left(\frac{22.98}{63.15}\right)=0.1565\)
Regression equation of Y on X is
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 47.8 = 0.1565(X - 178)
Y - 47.8 = 0.1565 X - 27.86
Y = 0.1565 X + 19.94
If X = 200
Y = 31.3 + 19.94 = 51.24