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11th Business Maths Weekly Test - 3 ( Correlation and Regression Analysis )

MABS Institution

11th Business Maths Weekly Test - 3 ( Correlation and Regression Analysis )-Aug 2020

Time : 1.00 Hours

Part - A

Multiple Choice Question - Answer All Question

When one regression coefficient is negative, the other would be

Negative
Positive
Zero
None of them

If the values of two variables move in same direction then the correlation is said to be

Negative
positive
Perfect positive
No correlation

The coefficient of correlation describes

the magnitude and direction
only magnitude
only direction
no magnitude and no direction

If the values of two variables move in opposite direction then the correlation is said to be

Negative
Positive
Perfect positive
No correlation

The variable which influences the values or is used for prediction is called

Dependent variable
Independent variable
Explained variable
Regressed

The correlation coefficient is

r(X,Y)=\(\frac { { \sigma }_{ x }{ \sigma }_{ y } }{ cov(x,y) } \)
r(X,Y)=\(\frac { cov(x,y) }{ { \sigma }_{ x }{ \sigma }_{ y } } \)
r(X,Y)=\(\frac { cov(x,y) }{ { \sigma }_{ y } } \)
r(X,Y)=\(\frac { cov(x,y) }{ { \sigma }_{ x } } \)

The regression coefficient of Y on X

b_xy=\(\frac { N\Sigma dxdy-(\Sigma dx)(\Sigma dy) }{ N\Sigma dy^{ 2 }-(\Sigma dy)^{ 2 } } \)
b_yx=\(\frac { N\Sigma dxdy-(\Sigma dx)(\Sigma dy) }{ N\Sigma dy^{ 2 }-(\Sigma dy)^{ 2 } } \)
b_yx=\(\frac { N\Sigma dxdy-(\Sigma dx)(\Sigma dy) }{ N\Sigma dx^{ 2 }-(\Sigma dx)^{ 2 } } \)
b_xy=\(\frac { N\Sigma xy-(\Sigma x)(\Sigma y) }{ \sqrt { N\Sigma { x }^{ 2 }-(\Sigma x)^{ 2 }\times \sqrt { N\Sigma { y }^{ 2 }-(\Sigma y)^{ 2 } } } } \)

The correlation coefficient from the following data N=25, ΣX=125, ΣY=100, ΣX²=650, ΣY²=436, ΣXY=520

0.667
-0.006
-0.667
0.70

The regression coefficient of X on Y

b_xy=\(\frac { N\Sigma dxdy-(\Sigma dx)(\Sigma dy) }{ N\Sigma dy^{ 2 }-(\Sigma dy)^{ 2 } } \)
b_yx=\(\frac { N\Sigma dxdy-(\Sigma dx)(\Sigma dy) }{ N\Sigma dy^{ 2 }-(\Sigma dy)^{ 2 } } \)
b_xy=\(\frac { N\Sigma dxdy-(\Sigma dx)(\Sigma dy) }{ N\Sigma dx^{ 2 }-(\Sigma dx)^{ 2 } } \)
b_xy=\(\frac { N\Sigma xy-(\Sigma x)(\Sigma y) }{ \sqrt { N\Sigma { x }^{ 2 }-(\Sigma x)^{ 2 }\times \sqrt { N\Sigma y^{ 2 }-(\Sigma y)^{ 2 } } } } \)

If regression co-efficient of Y on X is 2, then the regression co-efficient of X on Y is

≤\(\frac{1}{2}\)
2
>\(\frac{1}{2}\)
1

Correlation co-efficient lies between

0 to ∞
-1 to +1
-1 to 0
-1 to ∞

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

-0.12
0.001
-1
-0.97

If X and Y are two variates, there can be atmost

One regression line
two regression lines
three regression lines
more regression lines

Scatter diagram of the variate values (X,Y) give the idea about

functional relationship
regression model
distribution of errors
no relation

If r(X,Y) = 0 the variables X and Y are said to be

Positive correlation
Negative correlation
No correlation
Perfect positive correlation

The person suggested a mathematical method for measuring the magnitude of linear relationship between two variables say X and Y is

Karl Pearson
Spearman
Croxton and Cowden
Ya Lun Chou

The variable whose value is influenced or is to be predicted is called

dependent variable
independent variable
regressor
explanatory variable

The correlation coefficient

r=±\(\sqrt { { b }_{ xy }\times { b }_{ yx } } \)
r=\(\frac { 1 }{ { b }_{ xy }\times { b }_{ yx } } \)
r=b_xy x b_yx
r=±\(\sqrt { \frac { 1 }{ { b }_{ xy }\times { b }_{ yx } } } \)

Example for positive correlation is

Income and expenditure
Price and demand
Repayment period and EMI
Weight and Income

The term regression was introduced by

R.A Fisher
Sir Francis Galton
Karl Pearson
Croxton and Cowden

Part - B

Generic 3 - Answer All Question

Calculate the correlation coefficient from the data given below:

X 1 2 3 4 5 6 7 8 9

Y 9 8 10 12 11 13 14 16 15

Calculate the correlation co-efficient for the following data.

X 5 10 5 11 12 4 3 2 7 1

Y 1 6 2 8 5 1 4 6 5 2

Calculate the regression equation of X on Y from the following data:

X 10 12 13 12 16 15

Y 40 38 43 45 37 43

Calculate the correlation co-efficient from the below data:

X 1 2 3 4 5 6 7 8 9

Y 9 8 10 12 11 13 14 16 5

Compute the co-efficient of correlation batween the variates x and y from the given data:
No. of pairs of x and y series=8.x-series A.M.=74.5, x-series assumed mean =69, x-series S.D=13.07, y-series S.D=15.85, sum of products of corresponding deviations of x and y series =2176.

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