Advertisements
Advertisements
प्रश्न
For the following bivariate data obtain the equations of two regression lines:
| X | 1 | 2 | 3 | 4 | 5 |
| Y | 5 | 7 | 9 | 11 | 13 |
उत्तर
| X = xi | Y = yi | `"x"_"i"^2` | `"y"_"i"^2` | xi yi |
| 1 | 5 | 1 | 25 | 5 |
| 2 | 7 | 4 | 49 | 14 |
| 3 | 9 | 9 | 81 | 27 |
| 4 | 11 | 16 | 121 | 44 |
| 5 | 13 | 25 | 169 | 65 |
| 15 | 45 | 55 | 445 | 155 |
From the table, we have
n = 6, ∑ xi = 15, ∑ yi = 45, `sum "x"_"i"^2 = 55`, `sum "y"_"i"^2 = 445`, ∑ xi yi = 155
`bar x = (sum x_i)/"n" = 15/5 = 3`
`bar y = (sum y_i)/"n" = 45/5 = 9`
Now, for regression equation of Y on X,
`"b"_"YX" = (sum"x"_"i" "y"_"i" - "n" bar "x" bar "y")/(sum "x"_"i"^2 - "n" bar"x"^2)`
`= (155 - 5 xx 3 xx 9)/(55 - 5(3)^2) = (155 - 135)/(55 - 45) = 20/10 = 2`
Also, `"a" = bar y - "b"_"XY" bar x` = 9 - 2(3) = 9 - 6 = 3
The regression analysis of Y on X is
Y = a + bYX X
∴ Y = 3 + 2X
Now, for regression equation of X on Y,
`"b"_"XY" = (sum"x"_"i" "y"_"i" - "n" bar "x" bar "y")/(sum "y"_"i"^2 - "n" bar"y"^2)`
`= (155 - 5xx3xx9)/(445 - 5(9)^2) = (155 - 135)/(445 - 405) = 20/40 = 0.5`
Also, `"a"' = bar x - "b"_"XY" bar y`
= 3 - (0.5)(9) = 3 - 4.5 = - 1.5
The regression equation of X on Y is
X = a' + bXY Y
∴ X = - 1.5 + 0.5Y
∴ X = 0.5Y - 1.5
APPEARS IN
संबंधित प्रश्न
Given that the observations are: (9, -4), (10, -3), (11, -1), (12, 0), (13, 1), (14, 3), (15, 5), (16, 8). Find the two lines of regression and estimate the value of y when x = 13·5.
Find the feasible solution for the following system of linear inequations:
0 ≤ x ≤ 3, 0 ≤ y ≤ 3, x + y ≤ 5, 2x + y ≥ 4
Compute the product moment coefficient of correlation for the following data:
n = 100, `bar x` = 62, `bary` = 53, `sigma_x` = 10, `sigma_y` = 12
`Sigma (x_i - bar x) (y_i - bary) = 8000`
Calculate the Spearman’s rank correlation coefficient for the following data and interpret the result:
| X | 35 | 54 | 80 | 95 | 73 | 73 | 35 | 91 | 83 | 81 |
| Y | 40 | 60 | 75 | 90 | 70 | 75 | 38 | 95 | 75 | 70 |
Given the following data, obtain a linear regression estimate of X for Y = 10, `bar x = 7.6, bar y = 14.8, sigma_x = 3.2, sigma_y = 16` and r = 0.7
bYX is ______.
The equation of the line of regression of y on x is y = `2/9` x and x on y is x = `"y"/2 + 7/6`.
Find (i) r, (ii) `sigma_"y"^2 if sigma_"x"^2 = 4`
If for a bivariate data byx = – 1.2 and bxy = – 0.3 then find r.
From the two regression equations y = 4x – 5 and 3x = 2y + 5, find `bar x and bar y`.
The equations of the two lines of regression are 3x + 2y − 26 = 0 and 6x + y − 31 = 0 Find
- Means of X and Y
- Correlation coefficient between X and Y
- Estimate of Y for X = 2
- var (X) if var (Y) = 36
Find the equation of the line of regression of Y on X for the following data:
n = 8, `sum(x_i - barx).(y_i - bary) = 120, barx = 20, bary = 36, sigma_x = 2, sigma_y = 3`
In the regression equation of Y on X, byx represents slope of the line.
Choose the correct alternative:
If the lines of regression of Y on X is y = `x/4` and X on Y is x = `y/9 + 1` then the value of r is
Choose the correct alternative:
u = `(x - 20)/5` and v = `(y - 30)/4`, then bxy =
Choose the correct alternative:
y = 5 – 2.8x and x = 3 – 0.5 y be the regression lines, then the value of byx is
State whether the following statement is True or False:
The equations of two regression lines are 10x – 4y = 80 and 10y – 9x = 40. Then bxy = 0.9
State whether the following statement is True or False:
y = 5 + 2.8x and x = 3 + 0.5y be the regression lines of y on x and x on y respectively, then byx = – 0.5
State whether the following statement is True or False:
If equation of regression lines are 3x + 2y – 26 = 0 and 6x + y – 31= 0, then mean of X is 7
Among the given regression lines 6x + y – 31 = 0 and 3x + 2y – 26 = 0, the regression line of x on y is ______
The age in years of 7 young couples is given below. Calculate husband’s age when wife’s age is 38 years.
| Husband (x) | 21 | 25 | 26 | 24 | 22 | 30 | 20 |
| Wife (y) | 19 | 20 | 24 | 20 | 22 | 24 | 18 |
The equations of the two lines of regression are 6x + y − 31 = 0 and 3x + 2y – 26 = 0. Calculate the mean values of x and y
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 means are 136 and 148 respectively. The sum of product of deviations from respective means is 122. Obtain the regression equation of x on y
If n = 5, Σx = Σy = 20, Σx2 = Σy2 = 90, Σxy = 76 Find the regression equation of x on y
The regression equation of x on y is 40x – 18y = 214 ......(i)
The regression equation of y on x is 8x – 10y + 66 = 0 ......(ii)
Solving equations (i) and (ii),
`barx = square`
`bary = square`
∴ byx = `square/square`
∴ bxy = `square/square`
∴ r = `square`
Given variance of x = 9
∴ byx = `square/square`
∴ `sigma_y = square`
The management of a large furniture store would like to determine sales (in thousands of ₹) (X) on a given day on the basis of number of people (Y) that visited the store on that day. The necessary records were kept, and a random sample of ten days was selected for the study. The summary results were as follows:
`sumx_i = 370 , sumy_i = 580, sumx_i^2 = 17200 , sumy_i^2 = 41640, sumx_iy_i = 11500, n = 10`
Complete the following activity to find, the equation of line of regression of Y on X and X on Y for the following data:
Given:`n=8,sum(x_i-barx)^2=36,sum(y_i-bary)^2=40,sum(x_i-barx)(y_i-bary)=24`
Solution:
Given:`n=8,sum(x_i-barx)=36,sum(y_i-bary)^2=40,sum(x_i-barx)(y_i-bary)=24`
∴ `b_(yx)=(sum(x_i-barx)(y_i-bary))/(sum(x_i-barx)^2)=square`
∴ `b_(xy)=(sum(x_i-barx)(y_i-bary))/(sum(y_i-bary)^2)=square`
∴ regression equation of Y on :
`y-bary=b_(yx)(x-barx)` `y-bary=square(x-barx)`
`x-barx=b_(xy)(y-bary)` `x-barx=square(y-bary)`
Out of the two regression lines x + 2y – 5 = 0 and 2x + 3y = 8, find the line of regression of y on x.
For a bivariate data `barx = 10`, `bary = 12`, V(X) = 9, σy = 4 and r = 0.6
Estimate y when x = 5
Solution: Line of regression of Y on X is
`"Y" - bary = square ("X" - barx)`
∴ Y − 12 = `r.(σ_y)/(σ_x)("X" - 10)`
∴ Y − 12 = `0.6 xx 4/square ("X" - 10)`
∴ When x = 5
Y − 12 = `square(5 - 10)`
∴ Y − 12 = −4
∴ Y = `square`
