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You are given the following information about advertising expenditure and sales. Correlation coefficient between X and Y is 0.8 - Mathematics and Statistics

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प्रश्न

You are given the following information about advertising expenditure and sales.

  Advertisement expenditure
(₹ in lakh) (X)
Sales (₹ in lakh) (Y)
Arithmetic Mean 10 90
Standard Mean 3 12

Correlation coefficient between X and Y is 0.8

  1. Obtain the two regression equations.
  2. What is the likely sales when the advertising budget is ₹ 15 lakh?
  3. What should be the advertising budget if the company wants to attain sales target of ₹ 120 lakh?
योग
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उत्तर

Given: `bar x = 10, bar y = 90, sigma_x = 3, sigma_y = 12`, r = 0.8

`"b"_"YX" = "r" sigma_y/sigma_x = 0.8 xx 12/3 = 0.8 xx 4` = 3.2

`"b"_"XY" = "r" sigma_x/sigma_y = 0.8 xx 3/12 = 0.8 xx 0.25` = 0.2

(i) The regression equation of Y on X is

`("Y" - bar y) = "b"_"YX" ("X" - bar x)`

∴ (Y - 90) = 3.2 (X - 10)

∴ Y - 90 = 3.2 X - 32

∴ Y = 3.2 X - 32 + 90

∴ Y = 3.2 X + 58         .....(i)

The regression equation of X on Y is

`("X" - bar x) = "b"_"XY" ("Y" - bar y)`

∴ (X - 10) = 0.2 (Y - 90)

∴ X - 10 = 0.2 Y - 18

∴ X = 0.2 Y - 18 + 10

∴ X = 0.2 Y - 8         .....(ii)

(ii) For X = 15, from equation (i) we get

Y = 3.2 (15) + 58 = 48 + 58 = 106

∴ Likely sales is ₹ 106 lakh when advertising budget is ₹ 15 lakh.

(iii) For Y = 120, from equation (ii) we get

X = 0.2 (120) - 8 = 24 - 8 = 16

∴ To attain sales target of ₹ 120 lakh, advertising budget must be ₹ 16 lakh.

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Properties of Regression Coefficients
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Linear Regression - Exercise 3.2 [पृष्ठ ४७]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 3 Linear Regression
Exercise 3.2 | Q 4 | पृष्ठ ४७

संबंधित प्रश्न

For bivariate data. `bar x = 53`, `bar y = 28`, byx = −1.2, bxy = −0.3. Find the correlation coefficient between x and y.


For bivariate data. `bar x = 53, bar y = 28, "b"_"YX" = - 1.2, "b"_"XY" = - 0.3` Find estimate of Y for X = 50.


Bring out the inconsistency in the following:

bYX = bXY = 1.50 and r = - 0.9 


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∑ x = 8500, ∑ y = 9600, σX = 60, σY = 20, r = 0.6

Estimate the expenditure on food and entertainment when expenditure on accommodation is Rs 200.


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Mean 13 17
S.D. 3 2

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For bivariate data, the regression coefficient of Y on X is 0.4 and the regression coefficient of X on Y is 0.9. Find the value of the variance of Y if the variance of X is 9.


For a bivariate data, `bar x = 53`, `bar y = 28`, byx = −1.5 and bxy = −0.2. Estimate y when x = 50.


In a partially destroyed record, the following data are available: variance of X = 25, Regression equation of Y on X is 5y − x = 22 and regression equation of X on Y is 64x − 45y = 22 Find

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4y − 15x + 500 = 0
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The following results were obtained from records of age (X) and systolic blood pressure (Y) of a group of 10 men.

  X Y
Mean 50 140
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and `sum (x_i - bar x)(y_i - bar y) = 1120`. Find the prediction of blood pressure of a man of age 40 years.


Choose the correct alternative:

If r = 0.5, σx = 3, `σ_"y"^2` = 16, then byx = ______


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If byx = 1.5 and bxy = `1/3` then r = `1/2`, the given data is consistent


State whether the following statement is True or False: 

If bxy < 0 and byx < 0 then ‘r’ is > 0


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State whether the following statement is True or False:

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Given the following information about the production and demand of a commodity.
Obtain the two regression lines:

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(₹ in lakhs)
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(₹ in lakhs)
Mean 10 90
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If n = 5, Σx = Σy = 20, Σx2 = Σy2 = 90 , Σxy = 76 Find Covariance (x,y) 


x y `x - barx` `y - bary` `(x - barx)(y - bary)` `(x - barx)^2` `(y - bary)^2`
1 5 – 2 – 4 8 4 16
2 7 – 1 – 2 `square` 1 4
3 9 0 0 0 0 0
4 11 1 2 2 4 4
5 13 2 4 8 1 16
Total = 15 Total = 45 Total = 0 Total = 0 Total = `square` Total = 10 Total = 40

Mean of x = `barx = square`

Mean of y = `bary = square`

bxy = `square/square`

byx = `square/square`

Regression equation of x on y is `(x - barx) = "b"_(xy)  (y - bary)`

∴ Regression equation x on y is `square`

Regression equation of y on x is `(y - bary) = "b"_(yx)  (x - barx)`

∴ Regression equation of y on x is `square`


Mean of x = 53

Mean of y = 28

Regression coefficient of y on x = – 1.2

Regression coefficient of x on y = – 0.3

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`y - square = square (50 - square)`

∴ y = `square`

c. When y = 25,

`x - square = square (25 - square)`

∴ x = `square`


Mean of x = 25

Mean of y = 20

`sigma_x` = 4

`sigma_y` = 3

r = 0.5

byx = `square`

bxy = `square`

when x = 10,

`y - square = square (10 - square)`

∴ y = `square`


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`sum(x - overlinex)^2` = 1200, `sum(y - overliney)^2` = 300, `sum(x - overlinex)(y - overliney)` = – 250

Find: 

  1. byx
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  3. Correlation coefficient between x and y.

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