Correlation Test

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Correlation Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

It is moderate degree of relation if

A
$$-0.7\gt r\gt -0.99$$

B
$$0.7\lt r\lt 0.99$$

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

D
$$r=-1$$
Question 2
1 / -0

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 :

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

B
$$\dfrac{1}{4}$$

C
$$-\dfrac{1}{2}$$

D
$$\dfrac{1}{2}$$

Solution

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
$$\therefore r=-\sqrt{\cfrac{-3}{2}\times\cfrac{-1}{6}}$$
$$\therefore r=\cfrac{-1}{2}$$
Question 3
1 / -0

Coefficient of correlation is the measure of

A
Central tendency

B
Dispersion

C
Both central tendency and dispersion

D
Neither central tendency nor dispersion

Solution

Dispersion is the extent to which the distribution is stretched or squeezed.
The most common measures of dispersion are variance, standard deviation, mean deviation, correlation coefficient, range etc.
Question 4
1 / -0

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

A
0.48

B
-0.48

C
0.50

D
-0.50

Solution

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.
Question 5
1 / -0

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:

A
$$-0.5$$

B
$$+0.5$$

C
$$-0.6$$

D
$$+0.6$$

Solution

$$Corr(X,Y)=0.6$$
$$\therefore Corr(U,V)=\dfrac{Corr(2-3X,4Y+1)}{\sqrt{Var(U)Var(V)}}=\dfrac{Corr(-3X,4y)}{\sqrt{-3^2}\sqrt{4^2}}=\dfrac{-12Corr(X,Y)}{12}=-0.6$$
Question 6
1 / -0

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

A
Product Moment Correlation

B
Spearman's Rank Correlation

C
Concurrent Deviation Method

D
Karl Pearson's Co-efficient of Correlation

Solution

Karl pearson's coefficient of correlation determines how strongly the two variables are related to each other i.e. measure the linear correlation between the variables.
Question 7
1 / -0

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.

A
$$0.5515$$

B
$$0.4485$$

C
$$0.3995$$

D
$$0.2348$$

Solution

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.
Question 8
1 / -0

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

A
regression analysis

B
correlation analysis

C
dispersion

D
skewness

Solution

Correlation analysis refers to the connection between two or more things or interdependence between two variables, i.e., how strongly variables are connected.
Question 9
1 / -0

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?

A
$$\dfrac{\sigma_x \sigma_y}{\sigma_x^2 + \sigma_y^2}$$

B
$$\dfrac{\sigma_x + \sigma_y}{2\sigma_x \sigma_y}$$

C
$$\dfrac{\sigma_x^2 - \sigma_y^2}{\sigma_x^2 + \sigma_y^2}$$

D
$$\dfrac{\sigma_y - \sigma_x}{\sigma_x \sigma_y}$$

Solution

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

Question 10

1 / -0

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

$$0.3697$$

$$0.4673$$

$$0.6303$$

$$0.7834$$

Solution

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 $$0.6303$$

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Re-Attempt Weekly Quiz Competition

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

Correlation Test - 14

Result

Correlation Test - 14

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Question 1

$$-0.7\gt r\gt -0.99$$

$$0.7\lt r\lt 0.99$$

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

$$r=-1$$

Question 2

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

$$\dfrac{1}{4}$$

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

$$\dfrac{1}{2}$$

Solution

Question 3

Central tendency

Dispersion

Both central tendency and dispersion

Neither central tendency nor dispersion

Solution

Question 4

0.48

-0.48

0.50

-0.50

Solution

Question 5

$$-0.5$$

$$+0.5$$

$$-0.6$$

$$+0.6$$

Solution

Question 6

Product Moment Correlation

Spearman's Rank Correlation

Concurrent Deviation Method

Karl Pearson's Co-efficient of Correlation

Solution

Question 7

$$0.5515$$

$$0.4485$$

$$0.3995$$

$$0.2348$$

Solution

Question 8

regression analysis

correlation analysis

dispersion

skewness

Solution

Question 9

$$\dfrac{\sigma_x \sigma_y}{\sigma_x^2 + \sigma_y^2}$$

$$\dfrac{\sigma_x + \sigma_y}{2\sigma_x \sigma_y}$$

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

$$\dfrac{\sigma_y - \sigma_x}{\sigma_x \sigma_y}$$

Solution

Question 10

$$0.3697$$

$$0.4673$$

$$0.6303$$

$$0.7834$$

Solution

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