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
<h3>The mean value and the correlation</h3>
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
<h3>The standard deviation of x</h3>
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
Read more about regression lines at:
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