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प्रश्न
Choose the correct alternative:
|byx + bxy| ≥ ______
विकल्प
|r|
2|r|
r
– r
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उत्तर
2|r|
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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.
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 = 1.9 and bXY = - 0.25
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
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 certain bivariate data the following information is available.
| X | Y | |
| Mean | 13 | 17 |
| S.D. | 3 | 2 |
Correlation coefficient between x and y is 0.6. estimate x when y = 15 and estimate y when x = 10.
For 50 students of a class, the regression equation of marks in statistics (X) on the marks in accountancy (Y) is 3y − 5x + 180 = 0. The variance of marks in statistics is `(9/16)^"th"` of the variance of marks in accountancy. Find the correlation coefficient between marks in two subjects.
For a bivariate data, `bar x = 53`, `bar y = 28`, byx = −1.5 and bxy = −0.2. Estimate y when x = 50.
Two lines of regression are 10x + 3y − 62 = 0 and 6x + 5y − 50 = 0. Identify the regression of x on y. Hence find `bar x, bar y` and r.
For certain X and Y series, which are correlated the two lines of regression are 10y = 3x + 170 and 5x + 70 = 6y. Find the correlation coefficient between them. Find the mean values of X and Y.
Find the line of regression of X on Y for the following data:
n = 8, `sum(x_i - bar x)^2 = 36, sum(y_i - bar y)^2 = 44, sum(x_i - bar x)(y_i - bar y) = 24`
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 the regression equation X on Y is 3x + 2y = 26, then bxy equal to
Choose the correct alternative:
If r = 0.5, σx = 3, σy2 = 16, then bxy = ______
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 bxy < 0 and byx < 0 then ‘r’ is > 0
Corr(x, x) = 1
State whether the following statement is True or False:
Regression coefficient of x on y is the slope of regression line of x on y
If n = 5, ∑xy = 76, ∑x2 = ∑y2 = 90, ∑x = 20 = ∑y, the covariance = ______
|bxy + byx| ≥ ______
Arithmetic mean of positive values of regression coefficients is greater than or equal to ______
Given the following information about the production and demand of a commodity.
Obtain the two regression lines:
| Production (X) |
Demand (Y) |
|
| Mean | 85 | 90 |
| Variance | 25 | 36 |
Coefficient of correlation between X and Y is 0.6. Also estimate the demand when the production is 100 units.
If n = 5, Σx = Σy = 20, Σx2 = Σy2 = 90 , Σxy = 76 Find Covariance (x,y)
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
| 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`
