Class 11 Commerce Economics - Correlation

Define the positive co-relation.

A positive correlation is a relationship between two variables such that their values increase or decrease together .

Examples of positively correlated variables include:

Hours spent studying and grade point averages.
Education and income levels.
Poverty and crime levels.
Evaluated stress levels and blood pressure readings.
Smoking and lung disease.

For the given lines of regression, $3x - 2y = 5$ and $x - 4y = 7$ , find:
coefficient of correlation $r(x, y)$

The two regression lines are

$3x - 2y = 5 ....(1)$

$x - 4y = 7 .... (2)$
From equation

$(1)\ 3x - 2y = 5$

$3x = 2y + 5$

$x = \dfrac {2}{3}y + \dfrac {5}{3}$

$\therefore b_{xy} = \dfrac {2}{3}$ (Regression of

$x$ on

$y$ )
From equation

$(2)\ x - 4y = 7$

$4y = x - 7$

$y = \dfrac {x}{4} - \dfrac {7}{4}$

$\therefore b_{yx} = \dfrac {1}{4}$ (Regression of

$y$ and

$x$ )

The two regression lines are
Coefficient of correlation

$= |r|$

$= \sqrt {b_{xy}\cdot b_{yx}}$

$= \sqrt {\dfrac {1}{4}\times \dfrac {2}{3}}$

$= \sqrt {\dfrac {1}{6}} = \dfrac {\sqrt {6}}{6}$

From the equation of the two regression lines, $4x + 3y + 7 = 0$ and $3x + 4y + 8 = 0$ , find: Coefficient of correlation.

Coefficient of correlation

$|r| = \sqrt{b_{yx} . b_{xy}} = \sqrt{\dfrac{-3}{4} \times \dfrac{-3}{4}}$

$\Rightarrow r = \pm \dfrac{3}{4}$
But r has the same sign as regression coefficients.

$\therefore r = \dfrac{-3}{4}$

Prove that coefficient of correlation lies between $-1$ and $1$ .

$p=\cfrac { \sum { \left( x.\overline { x } \right) \left( y-\overline { y } \right) } }{ \sqrt { \sum { { \left( x.\overline { x } \right) }^{ 2 } } \sum { { \left( y-\overline { y } \right) }^{ 2 } } } }$

$=\cfrac { \sum { XY } }{ \sqrt { \sum { { X }^{ 2 } } \sum { { Y }^{ 2 } } } }$
Where

$(X=\left( x.\overline { x } \right) ;Y=\left( y-\overline { y } \right)$

$\therefore$ by Swchwarz's inequality

$(S{X}^{2}{Y}^{2})\le \sum { { X }^{ 2 } } \sum { { Y }^{ 2 } }$

$\cfrac { { \left( \sum { XY } \right) }^{ 2 } }{ \sum { { X }^{ 2 } } \sum { { Y }^{ 2 } } } \le 1$

${ p }^{ 2 }\le 1$

$\Rightarrow$

$-1\le p\le 1$
Hence value of correlation coefficient lies between

$-1$ and

$1$

Find the coefficient of correlation from the following data.
$x$ $2$ $3$ $5$ $7$ $3$
$y$ $15$ $17$ $4$ $5$ $4$

$n=5$

$x$	$y$	$x-\bar{x}$	$y-\bar{y}$	$(x-\bar{x})(y-\bar{y})$	$(x-\bar{x})^2$	$(y-\bar{y})^2$
$2$	$15$	$-2$	$6$	$-12$	$4$	$36$
$3$	$17$	$-1$	$8$	$-8$	$1$	$64$
$5$	$4$	$1$	$-5$	$-5$	$1$	$25$
$7$	$5$	$3$	$-4$	$-12$	$9$	$16$
$3$	$4$	$-1$	$-5$	$5$	$1$	$25$
$\sum x=20$	$\sum y=45$			$\sum(x-\bar{x})(y-\bar{y})=-32$	$\sum(x-\bar{x})^2=16$	$\sum(y-\bar{y})^2=166$

$\bar{x}=\displaystyle\frac{\sum x}{n}=\frac{20}{5}=4$

$\bar{y}=\displaystyle\frac{\sum y}{n}=\frac{45}{5}=9$
Coefficient of correlation

$r=\displaystyle\frac{\sum (x-\bar{x})(y-\bar{y})}{\sqrt{\sum(x-\bar{x})^2}\sqrt{\sum(y-\bar{y})^2}}$

$=\displaystyle\frac{-32}{\sqrt{16}\sqrt{166}}$

$=\displaystyle\frac{-32}{4\times 12.8841}$

$=\displaystyle\frac{-32}{51.5364}$

$=-0.64$ .

Calculate coefficient of correlation from the following data:
$x$ $y$
$3$ $15$
$10$ $17$
$8$ $4$
$6$ $5$
$8$ $4$

$x$	$y$	$x-\bar{x}$	$(y-\bar{y})$	$(x-\bar{x})(y-\bar{y})$	$(x-\bar{x})^2$	$(y-\bar{y})^2$
$3$	$15$	$-4$	$6$	$-24$	$16$	$36$
$10$	$17$	$3$	$8$	$24$	$9$	$64$
$8$	$4$	$1$	$-5$	$-5$	$1$	$25$
$6$	$5$	$-1$	$-4$	$4$	$1$	$16$
$8$	$4$	$1$	$-5$	$-5$	$1$	$25$

$\sum (x-\bar{x})(y-\bar{y})=-6, \sum(x-\bar{x})^2=28$

$\sum x=35, \sum y=45$

$\sum(y-\bar{y})^2=166$

$\bar{x}=7, \bar{y}=9$

$r_{xy}=\displaystyle\frac{\sum(x-\bar{x})(y-\bar{y})}{\sqrt{\sum (x-\bar{x})^2}\sqrt{\sum (y-\bar{y})^2}}$

$=\displaystyle\frac{-6}{\sqrt{28}\cdot\sqrt{166}}$

$=\displaystyle\frac{-6}{5.29\times 12.88}$

$=\displaystyle\frac{-6}{68.13}$

$=-0.088$ Approx.

Calculate the correlation coefficient between $x$ and $y$ for the following data:
$x$ $5$ $9$ $13$ $17$ $21$
$y$ $12$ $20$ $25$ $33$ $35$

${ x }_{ i }$	${ y }_{ i }$	${ u }_{ i }$	${ v }_{ i }$	${ u }_{ i }^{ 2 }$	${ v }_{ i }^{ 2 }$	${ u }_{ i }{ v }_{ i }$
5	12	-8	-13	64	169	104
9	20	-4	-5	16	25	20
13	25	0	0	00	00	00
17	33	4	8	16	64	32
21	35	8	10	64	100	80
$\sum { { x }_{ i } } =65$	$\sum { { y }_{ i } } =125$	$\sum { { u }_{ i } } =0$	$\sum { { v }_{ i } } =0$	$\sum { { u }_{ i }^{ 2 } } =160$	$\sum { { v }_{ i }^{ 2 } } =358$	$\sum { { u }_{ i }{ v }_{ i } } =236$

$\overline { x } =\cfrac { 1 }{ n } \sum { { x }_{ i } } =\cfrac { 1 }{ 5 } \times 65=13$

$\overline { y } =\cfrac { 1 }{ n } \sum { { y }_{ i } } =\cfrac { 1 }{ 5 } \times 125=25$

$\rho (x,y)=\rho (u,v)=\cfrac { n\sum { { u }_{ i }{ v }_{ i } } -\sum { { u }_{ i } } \sum { { v }_{ i } } }{ \sqrt { n\sum { { u }_{ i }^{ 2 } } -{ (\sum { { u }_{ i }^{ 2 } } ) }^{ 2 } } \sqrt { n\sum { { v }_{ i }^{ 2 } } -{ (\sum { { v }_{ i }^{ 2 } } ) }^{ 2 } } }$

$=\cfrac { 5\times 236-0\times 0 }{ \sqrt { 5\times 160-{ 0 }^{ 2 } } \sqrt { 5\times 358-{ 0 }^{ 2 } } } =\cfrac { 236 }{ 239.33 }$

$=0.98$

If " $r$ " is a coefficient of correlation of two variables $x$ and $y$ , then prove that:
$r = \dfrac {\sigma_{x}^{2} + \sigma_{y}^{2} - \sigma_{x - y}^{2}}{2\sigma_{x}.\sigma_{y}}$ Where $\sigma_{x}^{2}, \sigma_{y}^{2}$ and $\sigma_{x - y}^{2}$ are the variance of $x, y$ and $x - y$ respectively.

Coefficient of correlation of

$x$ and

$y$ is

$r$ and

$\sigma_{x}^{2}, \sigma_{y}^{2}$ and

$\sigma_{x - y}^{2}$ are varience of

$x, y$ and

$x - y$

$\sigma_{x - y}^{2} = \dfrac {1}{n}[(x - y) - (\overline {x} - \overline {y})]^{2}$

$\sigma_{x - y}^{2} = \dfrac {1}{n}[(x - y) - (\overline {x}) + (\overline {y})]^{2}$

$\sigma_{x - y}^{2} = \dfrac {1}{n} [(x - \overline {x}) - (y - \overline {y})]^{2}$

$\sigma_{x - y}^{2} = \dfrac {1}{n} [(x - \overline {x})^{2} + (y - \overline {y})^{2} - 2(x - \overline {x})(y - \overline {y})]$

$\sigma_{x - y}^{2} = \dfrac {1}{n} [(x - \overline {x})^{2}] + \dfrac {1}{n} [(y - \overline {y})^{2}] - 2\dfrac {1}{n}(x - \overline {x})(y - \overline {y})$

$\sigma_{(x - y)}^{2} = \sigma_{x}^{2} + \sigma_{y}^{2} - 2r\sigma_{x}\sigma_{y}$

$2r\sigma_{x}\sigma_{y} = \sigma_{x}^{2} + \sigma_{y}^{2} - \sigma_{(x - y)}^{2}$

$\therefore r = \dfrac {\sigma_{x}^{2} + \sigma_{y}^{2} - \sigma_{(x - y)}^{2}}{2\sigma_{x}\sigma_{y}}$ .

Find the coefficient of correlation from the following data.
$x$ $-10$ $-5$ $0$ $5$ $10$
$y$ $5$ $9$ $7$ $11$ $13$

$n=5$

$x$	$y$	$x-\bar{x}$	$y-\bar{y}$	$(x-\bar{x})(y-\bar{y})$	$(x-\bar{x})^2$	$(y-\bar{y})^2$
$-10$	$5$	$-10$	$-4$	$40$	$100$	$16$
$-5$	$9$	$-5$	$0$	$0$	$25$	$0$
$0$	$7$	$0$	$-2$	$0$	$0$	$4$
$5$	$11$	$5$	$2$	$10$	$25$	$4$
$10$	$13$	$10$	$4$	$40$	$100$	$16$
$\sum x=0$	$\sum y=45$			$\sum(x-\bar{x})(y-\bar{y})=90$	$\sum(x-\bar{x})^2=250$	$\sum(y-\bar{y})^2=40$

$\bar{x}=\displaystyle\frac{\sum x}{n}=\frac{0}{5}=0$

$\bar{y}=\displaystyle\frac{\sum y}{n}=\frac{45}{5}=9$
Coefficient of correlation

$r=\displaystyle\frac{\sum(x-\bar{x})(y-\bar{y})}{\sqrt{\sum(x-\bar{x})^2}\sqrt{\sum(y-\bar{y})^2}}$

$=\displaystyle\frac{\sqrt{90}}{\sqrt{250}\sqrt{40}}$

$=\sqrt{\displaystyle\frac{90}{10000}}$

$=\displaystyle\frac{90}{100}$

$=0.9$

Calculate the Spearman's ranks correlation coefficient for the following data and interpret the result:
$X$ 35 54 80 95 73 73 35 91 83 81
$Y$ 40 60 75 90 70 75 38 95 75 70

$X$	$Y$	Rank $X$	Rank $Y$	$d$	${ d }^{ 2 }$
$35$	$40$	$1$	$2$	$1$	$1$
$54$	$60$	$2$	$3$	$1$	$1$
$80$	$75$	$4$	$5$	$1$	$1$
$95$	$90$	$8$	$6$	$2$	$4$
$73$	$70$	$3$	$4$	$1$	$1$
$73$	$75$	$3$	$5$	$2$	$4$
$35$	$38$	$1$	$1$	$0$	$0$
$91$	$95$	$7$	$7$	$0$	$0$
$83$	$75$	$6$	$5$	$1$	$1$
$81$	$70$	$5$	$4$	$1$	$1$

$\sum { { d }^{ 2 } } =14$

Sperman rank correlation coefficient

$=r=1-\dfrac { 6\sum { { d }^{ 2 } } }{ n({ n }^{ 2 }-1) }$

$r=1-\dfrac { 6(14) }{ 10(100-1) } \\ r=1-\dfrac { 84 }{ 990 } =1-.084=.915$

For a bivariate data $b_{YX} = -1.2$ and $b_{XY} = -0.3$ , find the correlation coefficient between $x$ and $y$ .

We know, correlation co efficient

$r=\sqrt { { b }_{ XY }\times { b }_{ YX } }$

$\therefore r=\sqrt { -1.2\times -0.3 }$

$r=\sqrt { 0.36 }$

$r=-0.6$ (as

${ b }_{ XY }, { b }_{ YX }$ negative)

$\therefore$ X and Y are negatively correlated with

$r=-0.6$

If Cov(x,y)= 1125, Ox= 47.5, and Oy= 39.Find r.

Cov / Ox * Oy = 1125 / (47.8*39.6)

= 0.59

Calculate the coefficient of correlation between $X$ and $Y$ series from the following data:
$n = 15, \overline {x} = 25, \overline {y} = 18, \sigma_{X} = 3.01, \sigma_{Y} = 3.03, \sum (x_{i} - \overline {x})(y_{i} - \overline {y}) = 122$ .

Correlation

$= \rho = \dfrac{Cov(X, Y)}{\sigma_x \sigma_y}$

$\displaystyle Cov(x, y) = \sum^{n}_{i = 1} \frac{(X_i - \bar{X})(Y_i - \bar{Y})}{n - 1}$

$Cov (X, Y) = \dfrac{122}{14} = \dfrac{61}{7}$

$\rho = \dfrac{61}{7(3.01)(3.01)} = 0.955$

If $\displaystyle \sum d_i^2 = 25$ , $n = 6$ find rank correlation coefficient where $d_i$ is the difference between the ranks of $i^{th}$ values.

Given that:

$\sum di^2=25$ ,

$n=6$

The rank coefficient is given by

$r=1-\dfrac{6\sum di^2}{n(n^2-1)}$

$r=1-\dfrac{6\times 25}{6(6^2-1)}$

$r=1-\dfrac{5}{7}$

$r=\dfrac{2}{7}.$

This is the required solution.

If the rank correlation coefficient is $0.6$ and the sum of squares of difference of ranks is $66$ , then find the number pairs of observations.

$P = 1 - 6 \dfrac{\sum di^2}{n(n^2 - 1)}$

$0.6 = 1 - \dfrac{6 \times 66}{n(n^2 - 1)}$

$\dfrac{6 \times 66}{n(n^2 - 1)} = 0.4$

$n(n^2 -1) = \dfrac{3}{2} \times 66 \times 10 = 990$

$n = 10$

$10(10^2 - 1) = 10 (99) = 990$

Find out Spearman's rank correlation:

X

Y

20

60

11

63

72

26

65

35

43

43

29

51

50

37

Interpretate the value of coefficient of correlation (r) lies between -1 to 1.

Interpretation of degree of correlation:
1. Perfect: If the value is near ± 1, then it said to be a perfect correlation: as one variable increases, the other variable tends to also increase (if positive) or decrease (if negative).
2. High degree: If the coefficient value lies between ± 0.50 and ± 1, then it is said to be a strong correlation.
3. Moderate degree: If the value lies between ± 0.30 and ± 0.49, then it is said to be a medium correlation.
4. Low degree: When the value lies below + .29, then it is said to be a small correlation.
5. No correlation: When the value is zero.

Find the value of correlation from the following information:

Price (Rs.)

PPE kit (supply)

500

25

700

35

800

40

1000

50

1100

55

1200

60

1400

70

1500

75

1800

90

2000

100

The two lines of regressions are $x+2y-5=0$ and $2x+3y-8=0$ and the variance of $x$ is $12$ . Find the variance of $y$ and the coefficient of correlation.

The given equation of the lines of regression are

$x+2y-5=0.......(i)$
and

$2x+3y-8=0.....(ii)$

Rewriting the equations

$(i)$ and

$(ii)$ , we have
From equation

$(i)$

$y=\cfrac{-x}{2}+\cfrac{5}{2}$

$y=-0.5x+2.5$ *regression line of

$y$ on

$x$ )

${ b }_{ yx }=r\cfrac { { \sigma }_{ y } }{ { \sigma }_{ x } } =-0.5....(iii)$

From eqution

$(ii)$ ,

$x=\cfrac{-3}{2}y+\cfrac{8}{2}$

$x=-1.5y+4$ ( regression line of

$x$ on

$y$ )

${ b }_{ xy }=r\cfrac { { \sigma }_{ x } }{ { \sigma }_{ y } }$

$\therefore { r }^{ 2 }={ b }_{ yx }\times { b }_{ xy }=(-0.5)\times (-1.5)=0.75$

$\therefore$

$r=\sqrt{0.75}=\pm 0.866$

But

${b}_{xy}$ and

${b}_{yx}$ being both

$-$ ve therefore,

$r$ is also

$-$ ve.

Correlation coefficient

$(r)=-0.866$

Varianceof

$x$ i.e.,

${ \sigma }_{ x }^{ 2 }=12$

$\therefore$

${\sigma}_{x}=\sqrt{12}$

From equation

$(iii)$

$r\cfrac { { \sigma }_{ y } }{ { \sigma }_{ x } } =-0.5$

$-0.866.\cfrac{{\sigma}_{x}}{\sqrt { 12 } }=-0.5$

${ \sigma }_{ y }=\cfrac { 0.5\times \sqrt { 12 } }{ 0.866 } =2$

$\therefore$ Variance of

$y$ i.e.,

${ \sigma }_{ y }^{ 2 }=4$

In a contest the competitors are awarded marks out of $20$ by two judges. The scores of the $10$ competitors are given below. Calculate Spearman's rank correlation.
Competitors $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ $I$ $J$
Judge $A$ $2$ $11$ $11$ $18$ $6$ $5$ $8$ $16$ $13$ $15$
Judge $B$ $6$ $11$ $16$ $9$ $14$ $2$ $4$ $3$ $13$ $17$

Competitors	Judge $A$ Score	$R_1$	Judge $B$ Score	$R_2$	Difference $d=R_1-R_2$	$d^2$
$A$	$2$	$10$	$6$	$8$	$2$	$4$
$B$	$11$	$5.5$	$11$	$6$	$-0.5$	$0.25$
$C$	$11$	$5.5$	$16$	$3$	$2.5$	$6.25$
$D$	$18$	$1$	$9$	$7$	$-6$	$36$
$E$	$6$	$8$	$14$	$4$	$4$	$16$
$F$	$5$	$9$	$20$	$1$	$8$	$64$
$G$	$8$	$7$	$4$	$9$	$-2$	$4$
$H$	$16$	$2$	$3$	$10$	$-8$	$64$
$I$	$13$	$4$	$13$	$5$	$-1$	$1$
$J$	$15$	$3$	$17$	$2$	$1$	$1$
						$\sum d^2=196.5$

$B$ and

$C$ has same job rating of

$5$ , their average rank is

$\dfrac{5+6}{2}=5.5$ .
As there is one tie in ranks

$R_1$ of 2 items, there is one correction in Spearman's rank correlation.

$r=1-\dfrac{6\begin{bmatrix}\sum d^2+\dfrac{1}{12}(m^3-m)\end{bmatrix}}{n(n^2-1)}$

$=1-\dfrac{6[196.5+\dfrac{1}{12}(2^3-2)]}{10(10^2-1)}$

$=1-\dfrac{6[196.5+\dfrac{1}{12}(8-2)]}{10(100-1)}$

$=1-\dfrac{6(196.5+0.5)}{990}$

$=1-\dfrac{6(197)}{990}=1-\dfrac{1182}{990}$

$=1-1.193=-0.1939$

$= -0.194$

The two lines of regressions are $4x+2y-3 = 0$ and $3x+6y+5 = 0$ . Find the correlation coefficient between $x$ and $y$ .

Let

$4x+2y-3=0$ be the regression equation of

$y$ on

$x$ and

$3x+6y+5=0$ the regression equation of

$x$ on

$y$ then

$2y=-4x+3$

$y=-2x+\dfrac{3}{2}$

$\therefore b_{yx}=-2$
and,

$3x=-6y-5$

$x =-\dfrac{6}{3}y-\dfrac{5}{3}$

$x =-2y-\dfrac{5}{3}$

$\therefore b_{xy} = -2$

$r^2 = b_{yx}\times b_{xy}$

$=-2\times -2=4>1$ which is impossible.

So regression equation of

$y$ on

$x$ is

$3x+6y+5=0$ ..........(i)

and regression equation of

$x$ on

$y$ is

$4x+2y-3=0$ ...........(ii)

From (i), we get

$6y=-3x-5$

$y=-\dfrac{1}{2}x-\dfrac{5}{6}$

$\therefore b_{yx}=-\dfrac{1}{2}$

From (ii), we get

$4x=-2y+3$

$x=-\dfrac{1}{2}y+\dfrac{3}{4}$

$\therefore b_{xy}=-\dfrac{1}{2}$

$\therefore$ Correlation coefficient is given by

$|r|=\sqrt{b_{yx}.b_{xy}}$

$=\sqrt{-\dfrac{1}{2}\times -\dfrac{1}{2}} = \dfrac{1}{2}$

$\Rightarrow r = \pm \dfrac{1}{2}$

But

$r$ has the same sign as regression coefficient

$\therefore r =-\dfrac{1}{2}$ .

A psychologist selected a random sample of $22$ students. He grouped them in $11$ pairs so that the students in each pair have nearly equal scores in an intelligence test. In each pair, one student was taught by method $A$ and the other by method $B$ and examined after the course. The marks obtained by them after the course are as follows:
Pairs $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$
Method $A$ $24$ $29$ $19$ $14$ $30$ $19$ $27$ $30$ $20$ $28$ $11$
Method $B$ $37$ $35$ $16$ $26$ $23$ $27$ $19$ $20$ $16$ $11$ $21$
Calculate Spearman's Rank correlation

Pairs	Method $A (X)$	$R_{X}$	Method $B(Y)$	$R_{Y}$	$(R_{X} - R_{Y})^{2} = d^{2}$
$1$	$24$	$6$	$37$	$1$	$(5)^{2} = 25$
$2$	$29$	$3$	$35$	$2$	$(1)^{2} = 1$
$3$	$19$	$8.5$	$16$	$9.5$	$(-1)^{2} = 1$
$4$	$14$	$10$	$26$	$4$	$(6)^{2} = 36$
$5$	$30$	$1.5$	$23$	$5$	$(-3.5)^{2} = 12.25$
$6$	$19$	$8.5$	$27$	$3$	$(5.5)^{2} = 30.25$
$7$	$27$	$5$	$19$	$8$	$(-3)^{2} = 9$
$8$	$30$	$1.5$	$20$	$7$	$(-5.5)^{2} = 30.25$
$9$	$20$	$7$	$16$	$9.5$	$(-2.5)^{2} = 6.25$
$10$	$28$	$4$	$11$	$11$	$(-7)^{2} = 49$
$11$	$11$	$11$	$21$	$6$	$(5)^{2} = 25$
$N = 11$					$\sum d^{2} = 225$

Spearman's Rank correlation:

$R = 1 - \dfrac {6\left [\sum d^{2} + \dfrac {1}{12} ((2^{3} - 2) + (2^{3} - 2) + (2^{3} - 2))\right ]}{n^{3} - n}$

$= 1 - \dfrac {6\left [225 + \dfrac {1}{12} (18)\right ]}{(11)^{3} - 11}$

$= 1 - \dfrac {6[225 + 1.5}{1331 - 11} = 1 - \dfrac {6[226.5]}{1320}$

$= 1 - \dfrac {1359}{1320} = 1 - 1.0296 = -0.0296$

The following results were obtained with respect to two variable x and y:
$\sum x = 30, \sum y = 42, \sum xy = 199, \sum x^2 = 184, \sum y^2 = 318, n = 6.$ Find the correlation coefficient between x and y.

$b_{yx} = \dfrac{\sum xy - \dfrac{1}{n} \sum x. \sum y}{\sum x^2 - \dfrac{1}{n} (\sum x)^2}$

$= \dfrac{199 - \dfrac{1}{6} (30)(42)}{184 - \dfrac{1}{6} (30)^2}$

$= \dfrac{199 - 210}{184 - 150} = \dfrac{-11}{34} = - 0.323$

$b_{xy} = \dfrac{\sum x y - \dfrac{1}{n} \sum x \sum y}{\sum y^2 - \dfrac{1}{n} (\sum y)^2}$

$= \dfrac{199 - \dfrac{1}{6} (30)(42)}{318 - \dfrac{1}{6}(42)^2}$

$= \dfrac{199 - 210}{318 - 294} = \dfrac{-11}{24} = -0.458$

Correlation coefficient between x and y.

$r^2 = b_{yx} . b_{xy}$

$= (-0.323) \times (- 0.458)$

$= 0.1479$

$\therefore r = \sqrt{0.1479} = -0.384$
Since

$b_{xy}$ and

$b_{yx}$ have the same sign, that is why we have used negative sign with r.

For $10$ pairs of observation on two variables $x$ and $y$ , the following data are available.
$\sum (x - 2) = 30, \sum (y - 5) = 40, \sum (x - 2)^{2} = 900$ .
$\sum (y - 5)^{2} = 800, \sum (x - 2)(y - 5) = 480$ .
Find the correlation coefficient between $x$ and $y$

Let

$x = x_{i}$ and

$y = y_{i}$
Let

$u_{i} = x_{i} - 2$ and

$v_{i} = y_{i} - 5$
Then it is given that,

$n= 10, \sum u_{i} = 30, \sum v_{i} = 40, \sum u_{i}^{2} = 900, \sum v_{i}^{2} = 800, \sum u_{i}v_{i} = 480$

$\therefore \overline {u} = \dfrac {\sum u_{i}}{n} = \dfrac {30}{10} = 3$

$\overline {v} = \dfrac {\sum vi}{n} = \dfrac {40}{10} = 4$
Since the correlation coefficient is independent of change of origin and scale, correlation coefficient between

$x$ and

$y$ is

$r(x, y) = r(u, v)$

$= \dfrac {\dfrac {1}{n}\sum u_{i} v_{i} - \overline {u}\overline {v}}{\sqrt {\dfrac {1}{n}\sum u_{i}^{2} - (\overline {u})^{2}} \cdot \sqrt {\dfrac {1}{n} \sum v_{i}^{2} - (\overline {v^{2}})}}$

$= \dfrac {\dfrac {480}{10} - 3\times 4}{\sqrt {\dfrac {900}{10} - (3)^{2}}\cdot \sqrt {\dfrac {800}{10} - (4)^{2}}}$

$= \dfrac {48 - 12}{\sqrt {81}\cdot \sqrt {64}} = \dfrac {36}{9\times 8} = \dfrac {4}{8} = 0.5$

The following table shows the mean and standard deviation of the marks of Mathematics and Physics scored by the students in a school
Mathematics Physics
Mean $84$ $81$
Standard Deviation $7$ $4$
The correlation coefficient between the given marks is $0.86$ . Estimate the likely marks in Physics if the marks in Mathematics are $92$ .

$\overline x = 84, \overline y = 81$

$\sigma_x = 7, \sigma_y = 4$

$r = 0.86$

$\therefore b_{yx} = r . \dfrac{\sigma_y}{\sigma_x}$

$= 0.86 \times \dfrac{4}{7} = 0.49$

$\therefore$ Regression equation of y on x

$y - \overline y = b_{yx} (x - \overline x)$

$y - 81 = 0.49 (x - 84)$

$y - 81 = 0.49 x - 41.16$

$\Rightarrow y = 0.49 x + 39.84$

Putting

$x = 92,$

$y = 45.08 + 39.84 = 84.92$

Hence, the likely marks in Physics are

$84.92.$

For $10$ points of observations on two variables $X$ and $Y$ , the following data are available:
$\sum (x - 2) = 30, \sum y - 5 = 40, \sum (x - 2)^{2} = 900$
$\sum (y - 5)^{2} = 800, \sum (x - 2)(y - 5) = 480$ .
Find the correlation coefficient between $X$ and $Y$ .

The basic formula for calculating correlation coefficient

$\text{corr}(X,Y)=\dfrac{n\sum xy - \sum x\sum y }{\sqrt{(n\sum x^2 - (\sum x)^2)(n\sum y^2-(\sum y)^2)}}$ .....(1)

where

$n$

$=$ no. of observations and

$n=10$ in our case.

We need to find out each of the terms in equation (1).

From the given data,

$\sum (x-2)=30$

$\implies \sum x - 2\sum 1 = 30$

$\implies \sum x -2n = 30$

$\implies \sum x -20 = 30$

$\therefore \sum x =10$ ....(2)

$\sum y - 5 = 40$

$\therefore \sum y =45$ .....(3)

$\sum(x-2)^2 = 900$

$\implies \sum (x^2-4x-4) = 900$

$\implies \sum x^2 -4\sum x - 4\sum 1 = 900$

$\implies \sum x^2 -4\times 10 + 4\times 10 = 900$ .... from

$(2)$

$\therefore \sum x^2 =900$ ------

$(4)$

$\sum (y-5)^2 = 800$

$\implies \sum (y^2-10y+25) = 800$

$\implies \sum y^2 -10\sum y+25\sum 1 =800$

$\implies \sum y^2 -10\times 45+25\times 10 = 800$ ..... from

$(3)$

$\therefore \sum y^2 = 1000$ --------

$(5)$

$\sum (x-2)(y-5) = 480$

$\implies \sum (xy - 5x -2y+10) =480$

$\implies \sum xy - 5\sum x -2\sum y+10\sum 1=480$

$\implies \sum xy-5\times 10-2\times 45+10\times 10=480$ .....from

$(2) \& (3)$

$\therefore \sum xy = 520$ ---------

$(6)$

Substituting

$(2),(3),(4),(5),(6)$ in

$(1)$ gives us

$corr(X,Y) = \dfrac{10\times 520 - 10\times 45}{\sqrt{(10\times 900-10^2)(10\times 1000-45^2)}}$

$corr(X,Y)=0.564$

The equations of the two regression line are $2x+3y-6=0$ and $5x+7y-12=0$
Find
$(a)$ Correlation coefficient
$(b)$ $\dfrac{\sigma_x}{\sigma_y}$

Let

$2{x}+3{y}-6=0$ be

$x$ on

$y$ and

$5{x}+7{y}-12=0$ be

$y$ on

$x$

$2{x}+3{y}-6=0\implies x=-\dfrac{3}{2}y+3\implies bxy=-\dfrac{3}{2}$

$5{x}+7{y}-12=0\implies y=-\dfrac{5}{7}x+\dfrac{12}{7}\implies byx=-\dfrac{5}{7}$

$r=\sqrt{bxy byx}=\sqrt{\dfrac{15}{14}}$

$byx=\dfrac{r \sigma _y}{\sigma_x}=-\dfrac{5}{7},bxy=\dfrac{r\sigma_x}{\sigma_y}=-\dfrac{3}{2}\implies \dfrac{\sigma_x^2}{\sigma_y^2}=\dfrac{\dfrac{3}{2}}{\dfrac{5}{7}}=\dfrac{21}{10}\implies \dfrac{\sigma_x}{\sigma_y}=\sqrt{\dfrac{21}{10}}$

Calculate coefficient of correlation from the following data and interpret the result.
Number of Years of Schooling of Farmers 0 3 6 9 12 15
Annual Yield per Acre (in '000) 6 6 8 12 12 10

The solution is as follows:
Interpretation: Since r = 0.81, there is fairly high degree of positive correlation between between number of years of schooling of farmers and annual yield per acre.
2149574_2027378_ans_51f4bab311264a8b90e7db617257e412.PNG

2149574_2027378_ans_51f4bab311264a8b90e7db617257e412.PNG

Calculate coefficient of correlation for the ages of husband and wife.
Age of Husband (X) 24 25 22 30 34 37
Age of Wife (Y) 20 21 18 26 28 30

Solution is as follows:
2150268_2027698_ans_eca7f6388d5f4245a41b5fc930fb8fdb.PNG

Correlation - Class 11 Commerce Economics - Extra Questions

Define the positive co-relation.

For the given lines of regression, $3x - 2y = 5$ and $x - 4y = 7$ , find:
coefficient of correlation $r(x, y)$

From the equation of the two regression lines, $4x + 3y + 7 = 0$ and $3x + 4y + 8 = 0$ , find: Coefficient of correlation.

Prove that coefficient of correlation lies between $-1$ and $1$ .

Find the coefficient of correlation from the following data.
$x$ $2$ $3$ $5$ $7$ $3$
$y$ $15$ $17$ $4$ $5$ $4$

Calculate coefficient of correlation from the following data:
$x$ $y$
$3$ $15$
$10$ $17$
$8$ $4$
$6$ $5$
$8$ $4$

Calculate the correlation coefficient between $x$ and $y$ for the following data:
$x$ $5$ $9$ $13$ $17$ $21$
$y$ $12$ $20$ $25$ $33$ $35$

If " $r$ " is a coefficient of correlation of two variables $x$ and $y$ , then prove that:
$r = \dfrac {\sigma_{x}^{2} + \sigma_{y}^{2} - \sigma_{x - y}^{2}}{2\sigma_{x}.\sigma_{y}}$ Where $\sigma_{x}^{2}, \sigma_{y}^{2}$ and $\sigma_{x - y}^{2}$ are the variance of $x, y$ and $x - y$ respectively.

Find the coefficient of correlation from the following data.
$x$ $-10$ $-5$ $0$ $5$ $10$
$y$ $5$ $9$ $7$ $11$ $13$

Calculate the Spearman's ranks correlation coefficient for the following data and interpret the result:
$X$ 35 54 80 95 73 73 35 91 83 81
$Y$ 40 60 75 90 70 75 38 95 75 70

For a bivariate data $b_{YX} = -1.2$ and $b_{XY} = -0.3$ , find the correlation coefficient between $x$ and $y$ .

If Cov(x,y)= 1125, Ox= 47.5, and Oy= 39.Find r.

Calculate the coefficient of correlation between $X$ and $Y$ series from the following data:
$n = 15, \overline {x} = 25, \overline {y} = 18, \sigma_{X} = 3.01, \sigma_{Y} = 3.03, \sum (x_{i} - \overline {x})(y_{i} - \overline {y}) = 122$ .

If $\displaystyle \sum d_i^2 = 25$ , $n = 6$ find rank correlation coefficient where $d_i$ is the difference between the ranks of $i^{th}$ values.

If the rank correlation coefficient is $0.6$ and the sum of squares of difference of ranks is $66$ , then find the number pairs of observations.

Find out Spearman's rank correlation:

X

Y

20

60

11

63

72

26

65

35

43

43

29

51

50

37

Interpretate the value of coefficient of correlation (r) lies between -1 to 1.

Find the value of correlation from the following information:

Price (Rs.)

PPE kit (supply)

500

25

700

35

800

40

1000

50

1100

55

1200

60

1400

70

1500

75

1800

90

2000

100

The two lines of regressions are $x+2y-5=0$ and $2x+3y-8=0$ and the variance of $x$ is $12$ . Find the variance of $y$ and the coefficient of correlation.

The two lines of regressions are $4x+2y-3 = 0$ and $3x+6y+5 = 0$ . Find the correlation coefficient between $x$ and $y$ .

The following results were obtained with respect to two variable x and y:
$\sum x = 30, \sum y = 42, \sum xy = 199, \sum x^2 = 184, \sum y^2 = 318, n = 6.$ Find the correlation coefficient between x and y.

For $10$ pairs of observation on two variables $x$ and $y$ , the following data are available.
$\sum (x - 2) = 30, \sum (y - 5) = 40, \sum (x - 2)^{2} = 900$ .
$\sum (y - 5)^{2} = 800, \sum (x - 2)(y - 5) = 480$ .
Find the correlation coefficient between $x$ and $y$

For $10$ points of observations on two variables $X$ and $Y$ , the following data are available:
$\sum (x - 2) = 30, \sum y - 5 = 40, \sum (x - 2)^{2} = 900$
$\sum (y - 5)^{2} = 800, \sum (x - 2)(y - 5) = 480$ .
Find the correlation coefficient between $X$ and $Y$ .

The equations of the two regression line are $2x+3y-6=0$ and $5x+7y-12=0$
Find
$(a)$ Correlation coefficient
$(b)$ $\dfrac{\sigma_x}{\sigma_y}$

Calculate coefficient of correlation from the following data and interpret the result.
Number of Years of Schooling of Farmers 0 3 6 9 12 15
Annual Yield per Acre (in '000) 6 6 8 12 12 10

Calculate coefficient of correlation for the ages of husband and wife.
Age of Husband (X) 24 25 22 30 34 37
Age of Wife (Y) 20 21 18 26 28 30

Class 11 Commerce Economics Extra Questions

Price (Rs.)	PPE kit (supply)
500	25
700	35
800	40
1000	50
1100	55
1200	60
1400	70
1500	75
1800	90
2000	100

Competitors	$A$	$B$	$C$	$D$	$E$	$F$	$G$	$H$	$I$	$J$
Judge $A$	$2$	$11$	$11$	$18$	$6$	$5$	$8$	$16$	$13$	$15$
Judge $B$	$6$	$11$	$16$	$9$	$14$	$2$	$4$	$3$	$13$	$17$

Number of Years of Schooling of Farmers	0	3	6	9	12	15
Annual Yield per Acre (in '000)	6	6	8	12	12	10

Correlation - Class 11 Commerce Economics - Extra Questions

Define the positive co-relation.

For the given lines of regression, 3x−2y=53x - 2y = 5 and x−4y=7x - 4y = 7, find:coefficient of correlation r(x,y)r(x, y)

From the equation of the two regression lines, 4x+3y+7=04x + 3y + 7 = 0 and 3x+4y+8=03x + 4y + 8 = 0, find: Coefficient of correlation.

Prove that coefficient of correlation lies between −1-1 and 11.

Find the coefficient of correlation from the following data.xx2233557733yy15151717445544

Calculate coefficient of correlation from the following data: xxyy33151510101717884466558844

Calculate the correlation coefficient between xx and yy for the following data:xx5599131317172121yy12122020252533333535

Find the coefficient of correlation from the following data.xx−10-10−5-500551010yy55997711111313

Calculate the Spearman's ranks correlation coefficient for the following data and interpret the result:XX35548095737335918381YY40607590707538957570

For a bivariate data bYX=−1.2b_{YX} = -1.2 and bXY=−0.3b_{XY} = -0.3, find the correlation coefficient between xx and yy.

If Cov(x,y)= 1125, Ox= 47.5, and Oy= 39.Find r.

Calculate the coefficient of correlation between XX and YY series from the following data:n=15,¯x=25,¯y=18,σX=3.01,σY=3.03,∑(xi−¯x)(yi−¯y)=122n = 15, \overline {x} = 25, \overline {y} = 18, \sigma_{X} = 3.01, \sigma_{Y} = 3.03, \sum (x_{i} - \overline {x})(y_{i} - \overline {y}) = 122.

If ∑d2i=25\displaystyle \sum d_i^2 = 25, n=6n = 6 find rank correlation coefficient where did_i is the difference between the ranks of ithi^{th} values.

If the rank correlation coefficient is 0.60.6 and the sum of squares of difference of ranks is 6666, then find the number pairs of observations.

Find out Spearman's rank correlation: X Y 20 60 11 63 72 26 65 35 43 43 29 51 50 37

Interpretate the value of coefficient of correlation (r) lies between -1 to 1.

Find the value of correlation from the following information: Price (Rs.) PPE kit (supply) 500 25 700 35 800 40 1000 50 1100 55 1200 60 1400 70 1500 75 1800 90 2000 100

The two lines of regressions are x+2y−5=0x+2y-5=0 and 2x+3y−8=02x+3y-8=0 and the variance of xx is 1212. Find the variance of yy and the coefficient of correlation.

In a contest the competitors are awarded marks out of 2020 by two judges. The scores of the 1010 competitors are given below. Calculate Spearman's rank correlation.CompetitorsAABBCCDDEEFFGGHHIIJJJudge AA22111111111818665588161613131515Judge BB661111161699141422443313131717

The two lines of regressions are 4x+2y−3=04x+2y-3 = 0 and 3x+6y+5=03x+6y+5 = 0. Find the correlation coefficient between xx and yy.

The following results were obtained with respect to two variable x and y:∑x=30,∑y=42,∑xy=199,∑x2=184,∑y2=318,n=6.\sum x = 30, \sum y = 42, \sum xy = 199, \sum x^2 = 184, \sum y^2 = 318, n = 6. Find the correlation coefficient between x and y.

The equations of the two regression line are 2x+3y−6=02x+3y-6=0 and 5x+7y−12=05x+7y-12=0Find (a)(a) Correlation coefficient(b)(b) σxσy\dfrac{\sigma_x}{\sigma_y}

Calculate coefficient of correlation from the following data and interpret the result.Number of Years of Schooling of Farmers03691215Annual Yield per Acre (in '000)668121210

Calculate coefficient of correlation for the ages of husband and wife.Age of Husband (X)242522303437Age of Wife (Y)202118262830

Class 11 Commerce Economics Extra Questions

Support mcqexams.com by disabling your adblocker.

For the given lines of regression, $3x - 2y = 5$ and $x - 4y = 7$ , find:
coefficient of correlation $r(x, y)$

From the equation of the two regression lines, $4x + 3y + 7 = 0$ and $3x + 4y + 8 = 0$ , find: Coefficient of correlation.

Prove that coefficient of correlation lies between $-1$ and $1$ .

Find the coefficient of correlation from the following data.
$x$ $2$ $3$ $5$ $7$ $3$
$y$ $15$ $17$ $4$ $5$ $4$

Calculate coefficient of correlation from the following data:
$x$ $y$
$3$ $15$
$10$ $17$
$8$ $4$
$6$ $5$
$8$ $4$

Calculate the correlation coefficient between $x$ and $y$ for the following data:
$x$ $5$ $9$ $13$ $17$ $21$
$y$ $12$ $20$ $25$ $33$ $35$

Find the coefficient of correlation from the following data.
$x$ $-10$ $-5$ $0$ $5$ $10$
$y$ $5$ $9$ $7$ $11$ $13$

Calculate the Spearman's ranks correlation coefficient for the following data and interpret the result:
$X$ 35 54 80 95 73 73 35 91 83 81
$Y$ 40 60 75 90 70 75 38 95 75 70

For a bivariate data $b_{YX} = -1.2$ and $b_{XY} = -0.3$ , find the correlation coefficient between $x$ and $y$ .

Calculate the coefficient of correlation between $X$ and $Y$ series from the following data:
$n = 15, \overline {x} = 25, \overline {y} = 18, \sigma_{X} = 3.01, \sigma_{Y} = 3.03, \sum (x_{i} - \overline {x})(y_{i} - \overline {y}) = 122$ .

If $\displaystyle \sum d_i^2 = 25$ , $n = 6$ find rank correlation coefficient where $d_i$ is the difference between the ranks of $i^{th}$ values.

If the rank correlation coefficient is $0.6$ and the sum of squares of difference of ranks is $66$ , then find the number pairs of observations.

Find out Spearman's rank correlation:

X

Y

20

60

11

63

72

26

65

35

43

43

29

51

50

37

Find the value of correlation from the following information:

Price (Rs.)

PPE kit (supply)

500

25

700

35

800

40

1000

50

1100

55

1200

60

1400

70

1500

75

1800

90

2000

100

The two lines of regressions are $x+2y-5=0$ and $2x+3y-8=0$ and the variance of $x$ is $12$ . Find the variance of $y$ and the coefficient of correlation.

The two lines of regressions are $4x+2y-3 = 0$ and $3x+6y+5 = 0$ . Find the correlation coefficient between $x$ and $y$ .

The following results were obtained with respect to two variable x and y:
$\sum x = 30, \sum y = 42, \sum xy = 199, \sum x^2 = 184, \sum y^2 = 318, n = 6.$ Find the correlation coefficient between x and y.

The equations of the two regression line are $2x+3y-6=0$ and $5x+7y-12=0$
Find
$(a)$ Correlation coefficient
$(b)$ $\dfrac{\sigma_x}{\sigma_y}$

Calculate coefficient of correlation from the following data and interpret the result.
Number of Years of Schooling of Farmers 0 3 6 9 12 15
Annual Yield per Acre (in '000) 6 6 8 12 12 10

Calculate coefficient of correlation for the ages of husband and wife.
Age of Husband (X) 24 25 22 30 34 37
Age of Wife (Y) 20 21 18 26 28 30