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प्रश्न
If the regression equations are 8x – 10y + 66 = 0 and 40x – 18y = 214, the mean value of y is ______
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उत्तर
17
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
Information on v:ehicles [in thousands) passing through seven different highways during a day (X) and number of accidents reported (Y) is given as follows :
`Sigmax_i` = 105, `Sigmay_i` = 409, n = 7, `Sigmax_i^2` = 1681, `Sigmay_i^2` = 39350 `Sigmax_iy_i` = 8075
Obtain the linear regression of Y on X.
For the given lines of regression, 3x – 2y = 5 and x – 4y = 7, find:
(a) regression coefficients byx and bxy
(b) coefficient of correlation r (x, y)
For the following bivariate data obtain the equations of two regression lines:
| X | 1 | 2 | 3 | 4 | 5 |
| Y | 5 | 7 | 9 | 11 | 13 |
From the data of 20 pairs of observations on X and Y, following results are obtained.
`barx` = 199, `bary` = 94,
`sum(x_i - barx)^2` = 1200, `sum(y_i - bary)^2` = 300,
`sum(x_i - bar x)(y_i - bar y)` = –250
Find:
- The line of regression of Y on X.
- The line of regression of X on Y.
- Correlation coefficient between X and Y.
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 data obtained on X, the length of time in weeks that a promotional project has been in progress at a small business, and Y, the percentage increase in weekly sales over the period just prior to the beginning of the campaign.
| X | 1 | 2 | 3 | 4 | 1 | 3 | 1 | 2 | 3 | 4 | 2 | 4 |
| Y | 10 | 10 | 18 | 20 | 11 | 15 | 12 | 15 | 17 | 19 | 13 | 16 |
Find the equation of the regression line to predict the percentage increase in sales if the campaign has been in progress for 1.5 weeks.
If for bivariate data `bar x = 10, bar y = 12,` v(x) = 9, σy = 4 and r = 0.6 estimate y, when x = 5.
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`
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`
Regression equation of X on Y is ______
Choose the correct alternative:
The slope of the line of regression of y on x is called the ______
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
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`
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`
