MCQ Questions for Class 11 Commerce Economics Correlation Quiz 3

Total expenditure on all these items in

$1998$ was approximately what percent of the total expenditure in

$2002$ ?

0%
$62$ %
0%
$66$ %
0%
$69$ %
0%
$71$ %

Explanation

Required percentage

$= \left [\dfrac {(288 + 98 + 3.00 + 23.4 + 83)}{(420 + 142 + 3.96 + 49.4 + 98)} \times 100\right ]$ %

$= \left [\dfrac {485.4}{713.36}\times 100\right ]$ %

$\approx 69.45$ %.

If increase in value of one variable is accompnied by the proportional decrease in second variable, then corrrelation between two variables is

0%
perfect positive
0%
perfect negative
0%
imperfect positive
0%
imperfect negative

The correlation between number of hours and efficiency of work is

0%
perfect positive
0%
perfect negative
0%
imperfect positive
0%
imperfect negative

Explanation

relationship between their output and their working hours

$\therefore$ The correlation between number of hours and efficiency of work is imperfect negative

Option D is correct

Find Karl pearson's correlation coefficient.

Marks in english	10	25	13	25	22	11	12	25	21	20
Marks in hindi	12	22	16	15	18	18	17	23	24	17

0%
$0.99$
0%
$0.6$
0%
$.235$
0%
$0.462$

Explanation

$\bar { x } =\cfrac { \sum { x } }{ n }=\cfrac{184}{10}=18.4$

$\bar { y } =\cfrac { \sum {y } }{ n }=\cfrac{182}{10}=18.2$

$r=\cfrac { \sum { (x-\bar { x } )(y-\bar { y } ) } }{ \sqrt { \sum { { (x-\bar { x } ) }^{ 2 } } \sum { { (y-\bar { y } ) }^{ 2 } } } } =\cfrac { 97.44 }{ \sqrt { 348.4\times 127.50 } } =0.462$

1020361_576706_ans_3b10f1106bd7498b9331d2a56dd50a70.png

Correlation measures____________.

0%
the degree of variability between the two variables
0%
the nature of the relationship between the two variables
0%
functional relationship between two variables
0%
all of the above

	Height	Long Jump	High Jump
A	158	324	175
B	165	365	185
C	162	380	180
D	170	400	184
E	175	350	199
F	163	350	172
G	178	425	189

Find rank correlation between height & long jump and height & high jump.

0%
$0.9 \ and \ 0.25$
0%
$0.86 \ and\ 0.52$
0%
$0.52 \ and \ 0.86$
0%
$0.89 and \ 0.63$

Explanation

$Height$	$Rank(Height)$	$Long\,Jump$	$Rank(Long\,jump)$	$d$	$d^2$
$158$	$7$	$324$	$6$	$1$	$1$
$165$	$4$	$365$	$4$	$0$	$0$
$162$	$6$	$380$	$3$	$3$	$9$
$170$	$3$	$400$	$2$	$1$	$1$
$175$	$2$	$350$	$6$	$4$	$16$
$163$	$5$	$350$	$5$	$0$	$0$
$178$	$1$	$425$	$1$	$0$	$0$
					$\sum d^2=26$

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

$=1-\dfrac{6\times 26}{7((7)^2-1)}$

$=0.52$

$Height$	$Rank(Height)$	$High\,jump$	$Rank(High\,jump)$	$d$	$d^2$
$158$	$7$	$175$	$6$	$1$	$1$
$165$	$4$	$185$	$3$	$1$	$1$
$162$	$6$	$180$	$5$	$1$	$1$
$170$	$3$	$184$	$4$	$1$	$1$
$175$	$2$	$199$	$1$	$1$	$1$
$163$	$5$	$172$	$7$	$2$	$4$
$178$	$1$	$189$	$2$	$1$	$1$
					$\sum d^2=10$

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

$=1-\dfrac{6\times 10}{7((7)^2-1)}$

$=0.86$

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.

0%
True
0%
False

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.

Chemistry	28	6	9	43	10	17	47	23	35
Biology	8	49	12	4	31	30	33	45	23

The marks obtained in Chemistry and Biology are as follows;

0%
$0.1$
0%
$-0.35$
0%
$-0.86$
0%
$0.78$

Explanation

$\rho=1-\cfrac{6\sum{{d}^{2}}}{n({n}^{2}-1)}=\cfrac{6\times{162}^{2}}{9\times 80}=-0.35$
1158220_576793_ans_cd82f645818c4da6b35e39a80ec02c2d.png

When two regression coefficients bear same algebraic signs, then correlation coefficient is:

0%
Positive
0%
Negative
0%
According to two signs
0%
Zero

Explanation

The '+' or '-' sign is given to the correlation coefficient based on the signs of the two regression coefficients.

$r=\pm \sqrt{b_{yx}*b_{xy}}$

where

$r$ is the correlation coefficient

$b_{yx},b_{xy}$ are the regression coefficients.

Calculate Spearman's rank coefficient for the data:

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

0%
$0.56$
0%
$0.45$
0%
$-0.76$
0%
$0.76$

Explanation

$\rho=1-\cfrac{6\sum{{d}^{2}}}{n({n}^{2}-1)}=1-\cfrac{6\times 40}{10(100-1)}=1-\cfrac{240}{990}=0.758\approx 0.76$
1174817_576779_ans_077138b183224620a616464a3716ed46.png

It is moderate degree of relation if

0%
$-0.7\gt r\gt -0.99$
0%
$0.7\lt r\lt 0.99$
0%
$0.5\lt r \lt0.699$
0%
$r=-1$

It is high degree of relation if

0%
$-0.7\gt r\gt -0.99$
0%
$0.7\lt r\lt 0.99$
0%
$0.5\lt r \lt0.699$
0%
$r=1$

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 :

0%
$-\dfrac{1}{4}$
0%
$\dfrac{1}{4}$
0%
$-\dfrac{1}{2}$
0%
$\dfrac{1}{2}$

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

$\therefore r=-\sqrt{\cfrac{-3}{2}\times\cfrac{-1}{6}}$

$\therefore r=\cfrac{-1}{2}$

Coefficient of correlation is the measure of

0%
Central tendency
0%
Dispersion
0%
Both central tendency and dispersion
0%
Neither central tendency nor dispersion

The regression coefficients of a bivariate distribution are -0.64 and -0.Then the correlation coefficient of the distribution is

0%
0.48
0%
-0.48
0%
0.50
0%
-0.50

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.

The correlation coefficient between two variables

$X$ and

$Y$ is found to be

$0.6$ . All the observations on

$X$ and

$Y$ are transformed using the transformations

$U = 2 - 3X$ and

$V = 4Y + 1$ . The correlation coefficient between the transformed variables

$U$ and

$V$ will be:

0%
$-0.5$
0%
$+0.5$
0%
$-0.6$
0%
$+0.6$

Explanation

$Corr(X,Y)=0.6$

$\therefore Corr(U,V)=\dfrac{Corr(2-3X,4Y+1)}{\sqrt{Var(U)Var(V)}}=\dfrac{Corr(-3X,4y)}{\sqrt{-3^2}\sqrt{4^2}}=\dfrac{-12Corr(X,Y)}{12}=-0.6$

Which of the following methods of measuring correlation is the most popular method?

0%
Product Moment Correlation
0%
Spearman's Rank Correlation
0%
Concurrent Deviation Method
0%
Karl Pearson's Co-efficient of Correlation

In a dance competition, the marks given by two judges to

$10$ participants are given below.

Participant	A	B	C	D	E	F	G	H	I	J
1st Judge	1	5	4	8	9	6	10	7	3	2
2nd Judge	4	8	7	6	5	9	10	3	2	1

Find the rank correlation coefficient.

0%
$0.5515$
0%
$0.4485$
0%
$0.3995$
0%
$0.2348$

Explanation

Rank Correlation Coefficient,

$r_s=1-\cfrac{6\sum{\rho_i^2}}{n(n^2-1)}$

where,

$\rho_i=$ Difference between two ranks of each observation.

Here,

$n=10$

$r_s=1-\cfrac{6\sum{\rho_i^2}}{n(n^2-1)}$

$=1-\cfrac{6(9+9+9+4+16+9+0+16+1+1)}{10(100-1)}$

$=1-\cfrac{6\times 74}{10\times 99}$

$=1-0.4484848$

$=0.5515$

Hence, A is correct option.

The degree of relationship between the two variables is measured by _________.

0%
regression analysis
0%
correlation analysis
0%
dispersion
0%
skewness

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?

0%
$\dfrac{\sigma_x \sigma_y}{\sigma_x^2 + \sigma_y^2}$
0%
$\dfrac{\sigma_x + \sigma_y}{2\sigma_x \sigma_y}$
0%
$\dfrac{\sigma_x^2 - \sigma_y^2}{\sigma_x^2 + \sigma_y^2}$
0%
$\dfrac{\sigma_y - \sigma_x}{\sigma_x \sigma_y}$

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,

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

Following are the marks of

$10$ students obtained in Physics and Chemistry in an examination. Find the rank-correlation coefficient.

x	43	96	74	38	35	43	22	56	35	80
y	30	94	84	13	30	18	30	41	48	95

0%
$0.3697$
0%
$0.4673$
0%
$0.6303$
0%
$0.7834$

Explanation

Let the ranks of students obtained in Physics be

$x$ and the ranks of students obtained in Chemistry be

$y$ .

$X$	$Y$	Rank $X$ $(x)$	Rank $Y$ $(y)$	$d=x-y$	$d^2$
$43$	$30$	$5.5$	$7$	$-1.5$	$2.25$
$96$	$94$	$1$	$2$	$-1$	$1$
$74$	$84$	$3$	$3$	$0$	$0$
$38$	$13$	$7$	$10$	$-3$	$9$
$35$	$30$	$8.5$	$7$	$1.5$	$2.25$
$43$	$18$	$5.5$	$9$	$-3.5$	$12.25$
$22$	$30$	$10$	$7$	$3$	$9$
$56$	$41$	$4$	$5$	$-1$	$1$
$35$	$48$	$8.5$	$4$	$4.5$	$20.25$
$80$	$95$	$2$	$1$	$1$	$1$
			$\sum$	$0$	$58$

In the $X$ series $43$ has repeated twice and given ranks $5.5$ instead of $5$ and $6$ . For this the correction factor is $\dfrac{2(4-1)}{12}=\dfrac{1}{2}$ .

Also $35$ has repeated twice and given ranks $8.5$ instead of $8$ and $9$ . For this the correction factor is $\dfrac{2(4-1)}{12}=\dfrac{1}{2}$ .

In the $Y$ series $30$ has repeated thrice and given ranks $7$ instead of $6,7,8$ . For this the correction factor is $\dfrac{3(9-1)}{12}=2$ .

So the total correction factors $C.F=\dfrac{1}{2}+\dfrac{1}{2}+2=3$

The rank correlation coefficient is given by

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

$=1-\dfrac{6(58+3)}{10(100-1)}$

$=1-\dfrac{276}{10 \times 99}$

$=1-\dfrac{366}{990}$

$=1-0.3696$

$=0.6303$

Therefore the rank correlation coefficient is

$0.6303$

What is the formula to calculate Spearman's rank correlation coefficient when ranks are not repeated.

0%
$R=\dfrac{6\sum d_{i}^{2}}{n(n^{2}-1)}$
0%
$R=1-\dfrac{6\sum d_{i}^{2}}{n(n^{2}-1)}$
0%
$R=-\dfrac{6\sum d_{i}^{2}}{n(n^{2}-1)}$
0%
$R=1+\dfrac{6\sum d_{i}^{2}}{n(n^{2}-1)}$

Explanation

When ranks are not repeated Speraman's rank coefficient

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

$'n'$ is the number of observations

$'{ d }_{ i }'$ is the difference in rank for

${i}^{th}$ observation.

So option

$B$ is corrrect.

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

0%
$0.7048$
0%
$0.7500$
0%
$0.7638$
0%
None of the above

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.

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?

0%
$0.4$
0%
$0.5$
0%
$0.6$
0%
$0.7$

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

$:-$

$\Rightarrow r=\dfrac{30}{\sqrt{25\times 144}}$

$\Rightarrow r=0.5$

Correlation coefficient is denoted by the letter________.

0%
'C'
0%
'L'
0%
'Co'
0%
'r'

Which of the following is advisable to use while interpreting the value of co-efficient correlation?

0%
Square of the co-efficient of correlation
0%
Probable error
0%
Scatter diagram
0%
Correlation graph

The formula of Karl Pearson's co-efficient of correlation is ________.

0%
$r = 1 - \dfrac{6\Sigma{{D}_2}}{{{N}_{3}} - N}$
0%
$r = \dfrac{{\Sigma}{x}{y}}{{N}{\sigma}{x}{\sigma}{y}}$
0%
$r = \dfrac {{\sigma}{x}}{{\sigma}{y}}$
0%
$r = \dfrac {{\sigma}{y}}{{\sigma}{x}}$

Defects of Karl Pearson's method is/are________.

0%
it consumes more time
0%
the value of correlation is affected by extreme values
0%
if the data are not homogeneous, the co-efficient of correlation will give a wrong picture
0%
all of the above

The formula of product moment correlation is ______________.

0%
$r=\dfrac{S_{xy}}{\sqrt{S_xS_y}}$ ; S= Standard error
0%
$r = \dfrac {{\sigma}_{xy}}{{N}{\sigma}_{x}{\sigma}_{y}}$ ; N=number of variables
0%
$r = \dfrac {{-r}^2}{\sqrt{N}}$
0%
$r = \dfrac {{6}{\sigma}^2_{{d}}}{n({n}^2 - {1})}$

What are the assumptions of correlation co-efficient?
I. There is non-linear relationship between two variables.
II. The two variables have cause and effect relationship.
III. Each variable is affected by a large number of independent causes.
IV. If the error is reduced to minimum, the co-efficient of correlation is more reliable.

0%
I, II and III
0%
II, Ill and IV
0%
I, II and IV
0%
All of the above

CBSE Questions for Class 11 Commerce Economics Correlation Quiz 3 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Economics Correlation Quiz 3 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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