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प्रश्न
Bring out the inconsistency in the following:
bYX = bXY = 1.50 and r = - 0.9
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उत्तर
Given, bYX = bXY = 1.50 and r = - 0.9
Here, the coefficient of regressions is positive and the coefficient of correlation is negative.
But, for consistent data they must have the same signs.
∴ The given data is inconsistent.
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संबंधित प्रश्न
From the data of 7 pairs of observations on X and Y, following results are obtained.
∑(xi - 70) = - 35, ∑(yi - 60) = - 7,
∑(xi - 70)2 = 2989, ∑(yi - 60)2 = 476,
∑(xi - 70)(yi - 60) = 1064
[Given: `sqrt0.7884` = 0.8879]
Obtain
- The line of regression of Y on X.
- The line regression of X on Y.
- The correlation coefficient between X and Y.
Bring out the inconsistency in the following:
bYX = 2.6 and bXY = `1/2.6`
For a certain bivariate data
| X | Y | |
| Mean | 25 | 20 |
| S.D. | 4 | 3 |
And r = 0.5. Estimate y when x = 10 and estimate x when y = 16
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| X | Y | |
| Mean | 85 | 90 |
| S.D. | 5 | 6 |
The coefficient of correlation between X and Y is 0.6. Also estimate the production when demand is 100.
Two samples from bivariate populations have 15 observations each. The sample means of X and Y are 25 and 18 respectively. The corresponding sum of squares of deviations from respective means is 136 and 150. The sum of the product of deviations from respective means is 123. Obtain the equation of the line of regression of X on Y.
An inquiry of 50 families to study the relationship between expenditure on accommodation (₹ x) and expenditure on food and entertainment (₹ y) gave the following results:
∑ 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.
The following data about the sales and advertisement expenditure of a firms is given below (in ₹ Crores)
| Sales | Adv. Exp. | |
| Mean | 40 | 6 |
| S.D. | 10 | 1.5 |
Coefficient of correlation between sales and advertisement expenditure is 0.9.
Estimate the likely sales for a proposed advertisement expenditure of ₹ 10 crores.
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
- Mean values of X and Y
- Standard deviation of Y
- Coefficient of correlation between X and Y.
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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 |
| Variance | 150 | 165 |
and `sum (x_i - bar x)(y_i - bar y) = 1120`. Find the prediction of blood pressure of a man of age 40 years.
The equations of two regression lines are 10x − 4y = 80 and 10y − 9x = − 40 Find:
- `bar x and bar y`
- bYX and bXY
- If var (Y) = 36, obtain var (X)
- r
Choose the correct alternative:
If byx < 0 and bxy < 0, then r is ______
Choose the correct alternative:
|byx + bxy| ≥ ______
Choose the correct alternative:
If r = 0.5, σx = 3, `σ_"y"^2` = 16, then byx = ______
Choose the correct alternative:
Both the regression coefficients cannot exceed 1
State whether the following statement is True or False:
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 u = x – a and v = y – b then bxy = buv
State whether the following statement is True or False:
Cov(x, x) = Variance of x
If n = 5, ∑xy = 76, ∑x2 = ∑y2 = 90, ∑x = 20 = ∑y, the covariance = ______
|bxy + byx| ≥ ______
If u = `(x - "a")/"c"` and v = `(y - "b")/"d"`, then bxy = ______
The equations of the two lines of regression are 2x + 3y − 6 = 0 and 5x + 7y − 12 = 0. Find the value of the correlation coefficient `("Given" sqrt(0.933) = 0.9667)`
For a certain bivariate data of a group of 10 students, the following information gives the internal marks obtained in English (X) and Hindi (Y):
| X | Y | |
| Mean | 13 | 17 |
| Standard Deviation | 3 | 2 |
If r = 0.6, Estimate x when y = 16 and y when x = 10
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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Mean of x = 18
`2 square - 5 bary + 60` = 0
∴ `bary = square`
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∴ byx = `square/square`
∴ byx = `square/square`
∴ r = `square`
The following results were obtained from records of age (x) and systolic blood pressure (y) of a group of 10 women.
| x | y | |
| Mean | 53 | 142 |
| Variance | 130 | 165 |
`sum(x_i - barx)(y_i - bary)` = 1170
