Advertisements
Advertisements
Question
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.
Advertisements
Solution
Given, n = 50, X = marks in Statistics,
Y = marks in Accountancy,
Regression equation of X on Y is
3y – 5x + 180 = 0,
`bar y = 44, sigma_X^2 = 9/16 sigma_Y^2`
Now, 3y – 5x + 180 = 0 is the regression equation of X on Y.
∴ The equation becomes 5X = 3Y + 180
i.e., X = `3/5` Y + `180/5`
Comparing it with X = bXY Y + a', we get
`b_(XY) = 3/5, a = 180/5` = 36
a = `barx - b_(XY) bary`
∴ 36 = `bar x - 3/5 xx 44`
∴ 36 = `bar x` – 26.4
∴ `bar x` = 36 + 26.4 = 62.4
Also, `sigma_X^2 = 9/16 sigma_Y^2`
∴ `sigma_X^2/sigma_Y^2 = 9/16`
∴ `sigma_X/sigma_Y = 3/4`
`b_(XY) = r xx sigma_X/sigma_Y`
∴ `3/5 = r xx 3/4`
∴ `3/5 xx 4/3` = r
∴ r = `4/5` = 0.8
∴ Mean marks in statistics `(barx)` are 62.4 and correlation coefficient (r) between marks in the two subjects is 0.8.
APPEARS IN
RELATED QUESTIONS
For bivariate data. `bar x = 53, bar y = 28, "b"_"YX" = - 1.2, "b"_"XY" = - 0.3` Find estimate of X for Y = 25.
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.
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?
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
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 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.
From the two regression equations, find r, `bar x and bar y`. 4y = 9x + 15 and 25x = 4y + 17
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.
The two regression equations are 5x − 6y + 90 = 0 and 15x − 8y − 130 = 0. Find `bar x, bar y`, 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.
The two regression lines between height (X) in inches and weight (Y) in kgs of girls are,
4y − 15x + 500 = 0
and 20x − 3y − 900 = 0
Find the mean height and weight of the group. Also, estimate the weight of a girl whose height is 70 inches.
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.
If bYX = − 0.6 and bXY = − 0.216, then find correlation coefficient between X and Y. Comment on it.
Choose the correct alternative:
If byx < 0 and bxy < 0, then r is ______
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 bxy < 0 and byx < 0 then ‘r’ is > 0
State whether the following statement is True or False:
Corr(x, x) = 0
State whether the following statement is True or False:
Cov(x, x) = Variance of x
The value of product moment correlation coefficient between x and x is ______
Arithmetic mean of positive values of regression coefficients is greater than or equal to ______
| 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`
The regression equation of y on x is 2x – 5y + 60 = 0
Mean of x = 18
`2 square - 5 bary + 60` = 0
∴ `bary = square`
`sigma_x : sigma_y` = 3 : 2
∴ byx = `square/square`
∴ byx = `square/square`
∴ r = `square`
For a bivariate data:
`sum(x - overlinex)^2` = 1200, `sum(y - overliney)^2` = 300, `sum(x - overlinex)(y - overliney)` = – 250
Find:
- byx
- bxy
- Correlation coefficient between x and y.
