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Correlation coefficient - class-XII

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Write True/False in the following statement:
If regression coefficient are $0.8$ and $0.2$ then the value of corelation coefficient is $+0.4$

  1. True

  2. False


Correct Option: A
Explanation:

Given:-

Coefficient of regression are $0.8$ and $0.2$.
Therefore,
Coefficient of correlation $= \sqrt{0.8 \cdot 0.2} = \sqrt{0.16} = 0.4$
Hence the given statement is true.

It is perfect correlaion if

  1. $0.7\lt r\lt 0.99$

  2. $-0.7\gt r\gt -0.99$

  3. $r=1$

  4. $0.5\lt r \lt0.699$


Correct Option: C

The coefficient of correlation is always between

  1. $0\ and \ 1$

  2. $-1\ and \ 1$

  3. $-\infty \ and \ \infty$

  4. $-10\ and \ 10$


Correct Option: B
Explanation:

$\Rightarrow$  Correlation coefficients are expressed as values between $-1$ and $+1$. 

$\Rightarrow$  A coefficient of $+1$ indicates a perfect positive correlation: A change in the value of one variable will predict a change in the same direction in the second variable. 
$\Rightarrow$  A coefficient of $-1$ indicates a perfect negative correlation: A change in the value of one variable predicts a change in the opposite direction in the second variable. 

It is moderate degree of relation if 

  1. $-0.7\gt r\gt -0.99$

  2. $0.7\lt r\lt 0.99$

  3. $0.5\lt r \lt0.699$

  4. $r=-1$


Correct Option: C

It is high degree of relation if 

  1. $-0.7\gt r\gt -0.99$

  2. $0.7\lt r\lt 0.99$

  3. $0.5\lt r \lt0.699$

  4. $r=1$


Correct Option: B

The formula for correlation coeficient $r$ of two variables $x$ and $y$ is 

  1. $u _i=\dfrac{x _i-a}{h}$,

  2. $\dfrac{1}{n}\sum(x-\overline x)(y-\overline y)$

  3. $\dfrac{d _xd _y}{\sqrt{{\sum d _x}^2.{\sum d _y}^2}}$

  4. $\dfrac{\sum xy}{\sigma _x \sigma _y}$


Correct Option: C

Write True/False in the following statement:
The value of correlation coefficient lies between $-2$ to $+2$

  1. True

  2. False


Correct Option: B
Explanation:

The value of correlation coefficient lies between $-1$ to $+1$.

Hence the given statement is false.

State the following statement is true or false

The value of coefficient of correlation is greater than $1$.

  1. True

  2. False


Correct Option: B
Explanation:
Value of coeff. of correlation lies between $-1$ and $+1$.
Depending on the strength & direction of linear relationship between two variable. 
Coeff. of correlation $\ngtr   1$
$\therefore $ Given statement is false.

State the following statement is True or False
The value of correlation coefficient is always $2$.

  1. True

  2. False


Correct Option: B
Explanation:

Correlation coefficient is the measure of the degree of linear relationship between two variables, usually labelled X and Y. While in regression the emphasis is on predicting one variable from the other, in correlation the emphasis is on the degree to which a linear model may describe the relationship between two variables. In regression the interest is directional, one variable is predicted and the other is the predictor; in correlation the interest is non-directional, the relationship is the critical aspect.

$-1\le r\le 1$

Hence statement is false

Consider the following statements :
1. Two independent variables are always uncorrelated.
2. 
The coefficient of correlation between two variables X and Y is positive. When X decreases then Y decreases.
Which of the above statements is/are correct ?

  1. 1 only

  2. 2 only

  3. Both 1 and 2

  4. Neither 1 nor 2


Correct Option: C
Explanation:

Two independent variables have coefficient of correlation $0$, So they are always uncorrelated.

Positive coefficient means they both move in same direction, so if $X$ decreases then $Y$ also decreases.
So, both the statements are correct.
Hence, C is correct.

State the following statement is true or false

The value of coefficient of correlation is always $2$.

  1. True

  2. False


Correct Option: B
Explanation:
Value of coefficient of Correlation is always between $-1 $ and $+1$, depending on the strength and direction of a linear relationship between the variables.
Value of correlation coefficient lies between $-1$ and $+1$.
Therefore, the given statement is FALSE. 

Calculate the correlation coefficient between the corresponding values of X and Y in the following table:

X 2 4 5 6 8 11
Y 18 12 10 8 7 5
  1. $-0.65$

  2. $-0.82$

  3. $-0.92$

  4. $-0.48$


Correct Option: C
Explanation:
$X\\ 2\\ 4\\ 5\\ 6\\ 8\\ 11\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { x } =36$             $Y\\ 18\\ 12\\ 10\\ 08\\ 07\\ 05\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { y } =60$           $x=x-\overline { x } \\ -4\\ -2\\ -1\\ \quad 0\\ \quad 2\\ \quad 5\\ \ _ \ _ \ _ \ _ \ _ \ _ \\ \quad 0$              $Y=y-\overline { y } \\ \quad 8\\ \quad 2\\ \quad 0\\ -2\\ -3\\ -5\\ \ _ \ _ \ _ \ _ \ _ \\ \quad 0$               $XY\\ -32\\ -4\\ \quad 0\\ \quad 0\\ -6\\ -25\\ \ _ \ _ \ _ \ _ \ _ \ _ \\ -67$          ${ X }^{ 2 }\\ 16\\ 4\\ 1\\ 0\\ 4\\ 25\\ \ _ \ _ \ _ \ _ \ _ \\ \quad 50$         ${ Y }^{ 2 }\\ 64\\ 4\\ 0\\ 4\\ 9\\ 25\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { { Y }^{ 2 }=106 } $

Therefore, $\overline { a } =\cfrac { 36 }{ 6 } \\ \quad =6$
$\overline { y } =\cfrac { 60 }{ 6 } \\ \quad =10$

Therefore, $=\quad \cfrac { \sum { XY }  }{ \sqrt { \sum { { X }^{ 2 } } \sum { { Y }^{ 2 } }  }  } \\ =\cfrac { -67 }{ \sqrt { 50*106 }  } \\ =-0.92$

State the following statement is True or False
If two regression coefficients are $0.8$ and $0.2$ respectively, then correlation coefficient is  equal to zero.

  1. True

  2. False


Correct Option: B
Explanation:
Let $r _x=0.8, r _{y}=0.2$
Coeff. of correlation = Geometric Mean of Regression Coeff. 
$=\sqrt { { r } _{ x }.{ r } _{ y } } $

$=\sqrt { 0.8\times 0.2 } $ 

$=\sqrt { 0.16 } $

$\therefore $   CorrelationCoeff. $= 0.4\neq 0$

$\therefore $ The given statement is false.

Two variates, x and y, are uncorrelated and have standard deviations $\sigma _x$ and $\sigma _y$ respectively. What is the correlation coefficient between x + y and x - y?

  1. $\dfrac{\sigma _x \sigma _y}{\sigma _x^2 + \sigma _y^2}$

  2. $\dfrac{\sigma _x + \sigma _y}{2\sigma _x \sigma _y}$

  3. $\dfrac{\sigma _x^2 - \sigma _y^2}{\sigma _x^2 + \sigma _y^2}$

  4. $\dfrac{\sigma _y - \sigma _x}{\sigma _x \sigma _y}$


Correct Option: C
Explanation:
Let $u = (x + y); v = (x - y)$
$\overline{u} = (\overline{x} + \overline{y}); \overline{v} = (\overline{x} - \overline{y})$

$cov(u,v) = E{(u - \overline{u})(v - \overline{u})}=E\{(x - \overline{x}) + ( y - \overline{y})\}\times \{(x - \overline{x}) - ( y - \overline{y})\}$

= $E\{(x - \overline{x} )^2 - ( y - \overline{y})^2\} = \sigma^2 _x - \sigma^2 _y$
Also, $var(u) = E\{( u - \overline{u}\} = E( x - \overline{x}) + (y - \overline{y})\}^2 = \sigma^2 _x + \sigma _y^2$

Also, $var(u) = E\{( v - \overline{v}\} = E( x - \overline{x}) + (y - \overline{y})\}^2 = \sigma^2 _x + \sigma _y^2$

Thus, $\rho=\dfrac{cov(u,v)}{\sigma _u\times\sigma _v}=\dfrac{\sigma^2 _x - \sigma^2 _y}{\sigma^2 _x + \sigma^2 _y}$

The regression coefficients of a bivariate distribution are -0.64 and -0.36. Then the correlation coefficient of the distribution is

  1. 0.48

  2. -0.48

  3. 0.50

  4. -0.50


Correct Option: B
Explanation:

Solution:

We have,
$b _{xy}=-0.64$ and $b _{yx}=-0.36$
$\therefore$ Correlation coefficient $=\sqrt{b _{xy}\times b _{yx}}$
$=\pm\sqrt{(-0.64)(-0.36)}=\pm0.48$
$\Longrightarrow \sigma=-0.48$
[$\because b _{xy}$ and $b _{yx}$ both are negative.]
Hence, B is the correct option.

For two variables x and y. the two regression coefficients are $b _{x}= -\dfrac{3}{2}$ and $b _{y}=-\dfrac{1}{6}$.
The correlation coefficient between x and y is : 

  1. $-\dfrac{1}{4}$

  2. $\dfrac{1}{4}$

  3. $-\dfrac{1}{2}$

  4. $\dfrac{1}{2}$


Correct Option: C
Explanation:

We know that, for two variables $x$ and $y$

$r=\pm \sqrt { { b } _{ x }{ b } _{ y } } $
Where ${b} _{x} $ and ${b} _{y}$ are the regression coefficients 
and $r$ is the correlation coefficient between $x$ and $y$
Also, the sign of $r$ is same as the sign of regression coefficients
$\therefore r=-\sqrt{\cfrac{-3}{2}\times\cfrac{-1}{6}}$
$\therefore r=\cfrac{-1}{2}$

If the covariance between x and y is $30$, variance of x is $25$ and variance of y is $144$, then what is the correlation coefficient?

  1. $0.4$

  2. $0.5$

  3. $0.6$

  4. $0.7$


Correct Option: B
Explanation:
Given:-
$Cov(x,y)=30$
$V(x)=25$
$V(y)=144$
As we know formula of Corelation coefficient is$:-$
Let $r$ be Corelation coefficient of $x,y$ 
Then,
$r=\dfrac{Covariance(x,y)}{\sqrt{V(x)\times V(y)}}$
on solving$:-$
$\Rightarrow r=\dfrac{30}{\sqrt{25\times 144}}$
$\Rightarrow r=0.5$

For two variables $x$ and $y$ regression equations are given as $7x-3y-18=0$ and $4x-y-11=0$ then the correlation coefficient between $x$ and $y$ is 

  1. $0.7048$

  2. $0.7500$

  3. $0.7638$

  4. None of the above


Correct Option: C
Explanation:

Let $7x-3y-18=0$ represents the regression line of $y$ on $x$
$\Rightarrow y=-6+\dfrac { 7 }{ 3 } x\ \Rightarrow { b } _{ yx }=\dfrac { 7 }{ 3 } $
Then $4x-y-11=0$ is the regression line of $x$ on $y$.
$\Rightarrow x=\frac { 11 }{ 4 } +\dfrac { y }{ 4 } \ \Rightarrow { b } _{ xy }=\dfrac { 1 }{ 4 } $
Both ${ b } _{ yx }$ and ${ b } _{ yx }$ are positive. 
$\Rightarrow r=\sqrt { { b } _{ yx }\times { b } _{ xy } } \ \Rightarrow r=\sqrt { \dfrac { 7 }{ 3 } \times \dfrac { 1 }{ 4 }  } =0.7638$
So, option C is correct.

Calculate the coefficient of correlation between $x$ and $y$ for the data

x 1 2 3 4 5 6 7 8 9 10
y 3 10 5 1 2 9 4 8 7 6
  1. $0.12$

  2. $0.19$

  3. $0.22$

  4. $0.62$


Correct Option: B
Explanation:
$x\\ 1\\ 2\\ 3\\ 4\\ 5\\ 6\\ 7\\ 8\\ 9\\ 10\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { x=55 } $         $y\\ 3\\ 10\\ 5\\ 1\\ 2\\ 9\\ 4\\ 8\\ 7\\ 6\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \sum { y=55 } $                 $X=x-\overline { x } \\ -4.5\\ -3.5\\ -2.5\\ -1.5\\ -0.5\\ \quad 0.5\\ \quad 1.5\\ \quad 2.5\\ \quad 3.5\\ \quad 4.5\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \quad 0$                          $XY\\ \quad 11.25\\ -15.75\\ \quad 1.25\\ \quad 6.75\\ \quad 1.75\\ \quad 1.75\\ -1.25\\ \quad 6.25\\ \quad 1.75\\ \quad 2.25\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \quad 16\\ $                   ${ \quad X }^{ 2 }\\ 20.25\\ 12.25\\ 06.25\\ 02.25\\ 00.25\\ 00.25\\ 02.25\\ 06.25\\ 12.25\\ 20.25\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ 82.50$

${ \quad Y }^{ 2 }\\ 06.25\\ 20.25\\ 00.25\\ 12.25\\ 12.25\\ 02.25\\ 06.25\\ 02.25\\ 00.25\\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \\ \quad 82.50$

Therefore, $\overline { x } =\cfrac { 55 }{ 10 } \\ \quad =5.5$
$\cfrac { \sum { y }  }{ 10 } =5.5$

Therefore, $r=\cfrac { \sum { XY }  }{ \sqrt { \sum { { X }^{ 2 }\sum { { Y }^{ 2 } }  }  }  } \\ =\cfrac { 16 }{ 82.5 } \\ =0.19$
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