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Consider the two regression lines 3x+2y=26 and 6x+y=31, the regression coefficient of y on x is

Sagot :

yhixd

The regression lines 3x+2y=26 and 6x+y=31 are linear regressions

The mean values are 4 and 7 and the correlation coefficient between x and y is 0.25

The standard deviation of x is 2/13

The mean value and the correlation

We have the equations to be:

3x+2y=26 and 6x+y=31

Make y the subject in the second equation

y = 31 - 6x

Substitute y = 31 - 6x in the first equation

3x+2[31 - 6x] = 26

Expand

3x+ 62 - 12x = 26

Collect like terms

3x - 12x = 26 - 62

Evaluate

-9x = -36

Divide by - 9

x = 4

Substitute x = 4 in y = 31 - 6x

y = 31 - 6 * 4

y = 7

This means that the mean values are 4 and 7

To determine the correlation coefficient, we make y the subject in 3x+2y=26 and x the subject in 6x+y=31.

So, we have:

y = 13 - 3x/2 and x = 31/6 - 1/6y

The above means that:

Bxy = -1/6 and Byx = -3/2

The correlation coefficient is then calculated as:

r^2 = Bxy * Byx

r = -1/6 * -3/2

r = 0.25

Hence, the correlation coefficient between x and y is 0.25

The standard deviation of x

We have:

Var(y) = 4

In (a), we have:

y = 13 - 3x/2

To solve further, we make use of:

Var(y) = Var(ax + b) = a^2Var(x)

This gives

Var(y) = Var(13 - 3x/2) = 13^2 * Var(x)

So, we have:

Var(y) = 13^2 * Var(x)

Substitute 4 for Var(y)

4 = 13^2 * Var(x)

Divide both sides by 13^2

4/13^2 = Var(x)

Express 4 as 2^2

(2/13)^2 = Var(x)

So, we have:

Var(x) = (2/13)^2

Take the square root of both sides

SD(x) = 2/13

Hence, the standard deviation of x is 2/13

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