Y - 154.2 = 3.22(X – 66.8) or

Y = 3.22X - 60.9 (1)

The variables X and Y above are not chosen without any a priori supposition about some linear correlation, "Height and Weight" have to follow each other to some degree but precisely which degree?

The figure shows the regression line (Y on X), the line for Yg and the deviation, one example of the explained and the correspondent unexplained deviation.

Figure:

We take (21) from the link "The curve fitting and The Method Least Square":

is the regression of X on Y (21 LS)

b = 616.32/191.68 = 3.22

b ’ = 616.32/2659.68 = 0.232

This implicates that the equation of the regression line also might be expressed as:

Y - 154.2 = 3.22(X - 66.8) or (1)

In the figure the example of explained variation refers to linear regression of Y on X, and unexplained variation is the rest of the total variation. X is called the regressor, Y the regressand.

Explained variation S (Yj –Yg)^2 or measured by the variance b ^2S xj^2/1

(Number of degrees of Freedom =1)

In the example with "height and weight" variance = 3.22^2 * 191.68 =1,987

Unexplained variation S (Yj -Y)^2 or measured by the variance s^2 = S (Yj -Y)^2/n-2

(number of degrees of freedom = n-2 12-2=10, explained in the link ls)

In the example: 677.4

Total variation S (Yj – Yg)^2 or S y^2

In the example: 2,659.68 - the divergence to the sum explained and unexplained is cause by roundings. (number of degrees of freedom = n-1=12-1=11)

(Explained variance)/(unexplained variance), F ratio to be tested = b ^2S xj^2/s^2

= [b ^2S xj^2]/[ S (Yj -Y)^2/n-2]

In the example: 1,987/677.4 = 2.93. (number of degrees of freedom is: 1)

Total number of degrees of freedom 1+6=7

The second method, the t-ratio is preferable if a confidence interval also is desired. The two tests are equivalent because the t statistic is related. There are three equivalent ways of testing the null hypothesis that the regressor has no effect on Y: F test and the t test of b =0, and also the test of r =0 in a bivariate normal population for various sample sizes n. Forget the last mentioned for the moment (but I have be correct).

The estimat b can be tested by t-distribution as mentioned:

= 3.22 +- 2.45 * 8.2/Ö 191.68

= 3.22+-1.45

Since b =0 is not included in the confidence interval, the null hypothesis must be rejected at the 5% level.

r ^2 = [S (Y – Yg)^2]/ [S (Yj – Yg)^2] = (explained variation of Y)/(total variation of Y)

r = Ö r ^2 is called the coefficient of correlation.

The maximum value of the right side is 1 and r has the limits +-1. If r^2 =0 the regression line explains nothing. When r =0, then b =0. These are just two equivalent ways of formally stating in this case that there is no observed linear relation between X and Y.

Would you propose to calculate the regression line and measure the correlation with Y as the regressor on X as the regressand?

Here I would perfer to use pure logic in the starting point: The reason why the "weight" is high or low might be determined to some degree by the "height", but not the other way round. Notice, the regression technic does certainly not provide such knowledge. And notice in a lot of cases the logic not so smooth as it is in our example.

Compare r with standard deviation (s) (the rules are not proved here)

Similiar for the regresion on Y:

Not seldom like in the example above we find that one of the regressions is so dominating of practical interest that it is equivocal clear the regression must express some relevant knowledge. In other hand the method of just measuring the values might affect the equipment of measurement so that you in reality is the testing the equipment, and not the environment with or without knowing it or with or without admitting it.