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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
- Obtain the two regression equations.
- What is the likely sales when the advertising budget is ₹ 15 lakh?
- What should be the advertising budget if the company wants to attain sales target of ₹ 120 lakh?
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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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Mean of x = 53
Mean of y = 28
Regression coefficient of y on x = – 1.2
Regression coefficient of x on y = – 0.3
a. r = `square`
b. When x = 50,
`y - square = square (50 - square)`
∴ y = `square`
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`x - square = square (25 - square)`
∴ x = `square`
Mean of x = 25
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`sigma_x` = 4
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r = 0.5
byx = `square`
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∴ y = `square`
If byx > 1 then bxy is _______.
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| x | y | |
| Mean | 53 | 142 |
| Variance | 130 | 165 |
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