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

Explanation:

Variance is independent of change of origin as the change in origin is uniformly added to all the values and hence the mean also and hence, when (x − \(\rm \overline{x}\))² is calculated, it remains unaffected. But, change of scale alters all values unevenly and hence, variation changes. 

The correlation coefficient is dependent on the choice of both origin and scale of observation. 
Hence, both (I) and (II) is correct. 

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

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