Correlation and Regression MCQ Quiz in தமிழ் - Objective Question with Answer for Correlation and Regression - இலவச PDF ஐப் பதிவிறக்கவும்

The geometric mean between the two regression coefficients is equal to the correlation coefficient.

R = √(byx*bxy) = √(-0.8 * -0.2) = √0.16 = - 0.40

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

Therefore, If the two regression coefficients are -0.8 and -0.2, then the value of the coefficient of correlation is - 0.40.

Concept:

\({\rm{Coefficient\;of\;correlation}} = {\rm{\;}}\sqrt {{b_{yx}} \times {b_{xy}}} \)

Here, b_yx and b_xy are regression coefficients.

Or the slopes of the equation y on x and x on y are denoted as b_yx and b_xy

Calculation:

Given two lines of regression,

x + 3y = 11

⇒ y = 11/3 − x/3

So, b_yx = − 1/3

Again,

2x + y = 7

⇒ x = 7/2 - y/2

So, b_xy = −1/2

\({\rm{Coefficient\;of\;correlation}} = {\rm{\;}}\sqrt {{b_{yx}} \times {b_{xy}}} = \;\sqrt {\frac{{ - 1}}{3} \times \frac{{ - 1}}{2}} = \;\sqrt {\frac{1}{6}} = \;0.408\)

Since b_xy and b_yx are negative.

r = −0.408

Concept:

1) Coefficient of correlation = \(r=√{{b_{xy}\times b_{yx}}}\) 

Where byx and bxy are regression coefficients or the slope of the equation 'y on x' and 'x on y' are denoted as b_yx and b_xy

2) Standard deviation = \(√{Varience}\)

3) \(\frac{σ_y}{σ_x}=\frac{b_{yx}}{r},\) σ_y and σ_x are the standard deviations of y and x series respectively. 

Calculation:

Here, The two lines of regression are 8x - 10y = 66 and 40x - 18y = 214

For line 10y = 8x - 66

⇒ b_yx = 8/10 = 4/5

And, for line 40x - 18y = 214

⇒ 40x = 18y + 214

⇒ b_xy = 18/40 = 9/20 

Now, Coefficient of correlation = \(r=√{b_{xy}\times b_{yx}}\)

r = ± \(√({\frac{4}{5}\times\frac{9}{20})}=\pm{√{(\frac{9}{25}})}=\pm\frac{3}{5}\)

Since, b_xy > 0 and b_yx > 0.

So, r = 3/5

Here, variance of x series is 9

Standard deviation of x series is σ_x = √9 = 3

We know, \(\frac{σ_y}{σ_x}=\frac{b_{yx}}{r}\)

So, \(\sigma_y =\frac{b_{yx}}{r}\times σ_x\)

⇒ \(\sigma_y =\frac{\frac{4}{5}}{\frac{3}{5}}\times 3 =4\)

∴ The standard deviation of the y series = 4.

CONCEPT:

\({b_{yx}} = r.\frac{{{\sigma _y}}}{{{\sigma _x}}}\)

Where \({\sigma _x}\) = standard deviation of x; \({\sigma _y}\) = standard deviation of y

Regression Line y on x is given as \(x - \bar x = {b_{xy}}\left( {y - \bar y} \right)\)

CALCULATION:

\({b_{yx}} = r.\frac{{{\sigma _y}}}{{{\sigma _x}}} = 0.5 \times \frac{5}{{10}} = 0.25\)

Regression Line y on x is given as \(x - \bar x = {b_{xy}}\left( {y - \bar y} \right)\)

y – 10 = 0.25(x - 20)

y – 10 = 0.25x - 5

y = 0.25x – 5 + 10

y = 0.25x + 5

CONCEPT:

If there are two variables x and y we can have the following two types of regression lines,

1. Regression equation of y on x (y dependent, x independent)
2.  Regression equation of x on y (x dependent, y independent)

• Regression equation of y on x will be \({\bf{Y}} - {\bf{\bar Y}} = {{\bf{b}}_{{\bf{yx}}}}\left( {{\bf{X}} - {\bf{\bar X}}} \right)\)

Where b_yx stands for regression coefficient of y on x .Here y depends on x and y is a dependent and x is an independent variable. This equation will be of the form y = ax + b and it is used to estimate the value of y given the value of x. The slope of this equation is b_yx.

• Regression equation of x on y will be \({\bf{X}} - {\bf{\bar X}} = {{\bf{b}}_{{\bf{xy}}}}\left( {{\bf{Y}} - {\bf{\bar Y}}} \right)\)

Where b_xy stands for regression coefficient of ‘x’ on ‘y’. Here x depends on y and x is a dependent and y is an independent variable

This equation will be of the form x = by + a and it is used to estimate the value of x given the value of y

The slope of this equation is b_xy.

• Correlation coefficient is the geometric mean between regression coefficients i.e., \({\rm{r}} = \pm \sqrt {{{\rm{b}}_{{\rm{yx}}}}{{\rm{b}}_{{\rm{xy}}}}} \)

CALCULATION

Given equations are X + Y = 5, X + 2Y = 3.

On subtracting we get            ⇒ - y = 2

Putting y = -2, in equation (1)

we have x ± 2 = 5

⇒ x = 5 + 2          ⇒ x = 7

∴ x = 7and y = -2

∴ \({\rm{\bar x}} = 7{\rm{\;and\;\bar y}} = - 2\)

Suppose x + y – 5 = 0 is regression equation of y on x

⇒ 1x + 1y – 5 = 0

⇒ 1y = -1x + 5

\( \Rightarrow {\rm{y}} = \frac{{ - 1}}{1}{\rm{x}} + \frac{5}{1}\)

⇒ y = -1x + 5

∴ b_yx = -1

Suppose X + 2Y -3 = 0 is regression equation of x on y

⇒ 1x + 2y – 3 = 0

⇒ 1x = -2y + 3

\( \Rightarrow {\rm{x}} = \frac{{ - 2}}{1}{\rm{y}} + \frac{3}{1}\)

⇒ x = -2y + 3

∴ b_xy = -2

b_yx⋅b_xy = 2 which is > 1

So, our guessing is wrong so we interchange our guessing

Now, Suppose X + 2Y - 3 = 0 is regression equation of y on x

⇒ 1x + 2y-3 = 0.

⇒ 2y = -1x + 3

\( \Rightarrow {\rm{y}} = \frac{{ - 1}}{2}{\rm{x}} + \frac{3}{2}\)

⇒ y = -0.5x + 1.5

∴ b_yx=-0.5

And suppose X + Y – 5 = 0 is regression equation of x on y

⇒ 1x + 1y – 5 = 0

⇒ 1x = -1y + 5  \( \Rightarrow {\rm{x}} = \frac{{ - 1}}{1}{\rm{y}} + \frac{5}{1}\)

⇒ x = -1y + 5

∴ b_xy = -1

\({\rm{r}} = {\rm{\;}}\sqrt {{{\rm{b}}_{{\rm{xy}}}}{{\rm{b}}_{{\rm{yx}}}}} = \sqrt { - 0.5 \times - 1} = \sqrt {0.5} = {\rm{\;}} - 0.707\)

Concept:

Coefficient of Correlation: The geometric mean between two regression coefficient is equal to coefficient of correlation i.e.,

\({\rm{r}} = \sqrt {{{\rm{b}}_{{\rm{yx}}}}{{\rm{b}}_{{\rm{xy}}}}} \)

Some other properties of regression coefficient:

• The regression coefficient of y on x is represented by b_yx and x on y as b_xy.
• Both regression coefficient must have same sign.
• If one regression coefficient is greater than unity than others will be lesser than unity.
• Arithmetic mean of both regression coefficient:
• \(\frac{{{{\rm{b}}_{{\rm{yx}}}} + {{\rm{b}}_{{\rm{xy}}}}}}{2} \ge {\rm{r}}\)

Calculation:

Given: b_yx = 0.2 & b_xy = 1.8

To find: coefficient of correlation (r)

So, coefficient of correlation = r

\(= \sqrt {0.2 \times 1.8} \)

= √(0.36)

= 0.6

Concept:

\({\rm{Coefficient\;of\;correlation\;}} = {\rm{\;r\;}} = {\rm{\;}}\sqrt {{{\rm{b}}_{{\rm{yx}}}} \times {{\rm{b}}_{{\rm{xy}}}}} \) ,

Where b_yx and b_xy are regression coefficients or the slopes of the equation y on x and x on y are denoted as b_yx and b_xy

Note: If b_xy and b_yx are negative then the coefficient of correlation would be negative and if If b_xy and b_yx are positive then the coefficient of correlation would be positive.

Calculation:

Given:

b_xy = -0.64 and b_yx = -0.36

We know that,

\(\begin{array}{l} {\rm{Coefficient\;of\;correlation\;}} = {\rm{\;r\;}} = {\rm{\;}}\sqrt {{{\rm{b}}_{{\rm{yx}}}} \times {{\rm{b}}_{{\rm{xy}}}}} \\ {\rm{r\;}} = {\rm{\;}}\sqrt { - 0.64 \times - 0.36} = \; \pm 0.48 \end{array}\)

Here b_xy and b_yx are negative then the coefficient of correlation would be negative

∴ r = -0.48

Concept:

Correlation coefficients are used to measure how strong a relationship is between two variables. 

The coefficient of correlation does not change with a change in origin or change in scale. 

Calculation:

Given, r(X, Y) = 0.7

Z = X - 5 and W = 3Y

So, r(Z, W) = r(X - 5, 3Y)

= r(X, Y)

As the correlation coefficient does not change so the coefficient will remain the same for Z and W 

So, the correlation coefficient between Z and W = 0.7

Concept:

Equation of regression of Y on X:

The equation of regression of Y on X with given values of \(\rm \bar X, \bar Y,\sigma_{x}\mbox{ and }\sigma_{y}\) is given as follows:

\(\rm \rm \dfrac{y-\bar Y}{\sigma_y} =r_{xy} \dfrac{x-\bar X}{\sigma_x} \)

Calculation:

It is given that X̅ = 65, Y̅ = 67, σX = 2.5, σY = 3.5 and r(X, Y) = 0.8 .

Using the equation of regression of Y on X we write:

\(\rm \rm \rm \frac{y-\bar Y}{\sigma_y} =r_{xy} \frac{x-\bar X}{\sigma_x} \\ \left(\frac{y-67}{3.5}\right) =0.8 \times \left(\frac{x-65}{2.5}\right)\\ y - 67 = 1.12(x-65)\\y-67=1.12x-72.8\\y=1.12x-5.8\)

Therefore, the required equation is y = 1.12x - 5.8

Concept:

• Coefficient of correlation \({\rm{\;}} = {\rm{\;r\;}} = {\rm{\;}}\sqrt {{{\rm{b}}_{{\rm{yx}}}} \times {{\rm{b}}_{{\rm{xy}}}}} \)
  Where, b_yx and b_xy are regression coefficients or the slopes of the equation y on x and x on y are denoted as b_yx and b_xy 

Calculation:

Given:

Regression coefficient of Y on X = b_yx = -6

And correlation coefficient between X and Y = r = -1/2

We have to find the value of regression coefficient of X on Y = b_xy

We know that Coefficient of correlation\({\rm{\;}} = {\rm{\;r\;}} = {\rm{\;}}\sqrt {{{\rm{b}}_{{\rm{yx}}}} \times {{\rm{b}}_{{\rm{xy}}}}} \) 

Squaring both sides, we get

⇒ r² = b_yx × b_xy

\( \Rightarrow {\rm{\;}}{\left( {\frac{{ - 1}}{2}} \right)^2} = {\rm{\;}} - 6{\rm{\;}} \times {\rm{\;}}{{\rm{b}}_{{\rm{xy}}}}\)