# Interpolation

Annotation category:
Chapter 6

 Note:

Following the same procedure, the difference between 0.0678 and 0.0194 = 0.0484.

Now, the computed corelation coefficient of 0.467 is 67% (0.67) of the way between 0.40 and 0.50, Thus, 0.67 x 0.0484 = 0.0324 which, if we subtract it from 0.0678, will give us the answer: 0.0678 - 0.324 = 0.0354 or about 3.5% probability that the result would occur by chance if there is actually no correlation between the two quantities.

Note that if we had chosen to extrapolate between the columns first and the rows second, we would have gotten exactly the same answer. The starting matrix:

 ro= 0.4 0.5 N 20: 0.081 0.025 25: 0.048 0.011

would be collapsed by interpolating between the columns for both row 20 and row 25:

 ro= 0.467 N 20: 0.0435 25: 0.0232

and then interpolating between the rows: 0.0435 - 2/5 x (0.0435 - 0.0232) = 0.0354.

It should be emphasized that this interpolated value is not strictly correct because the function which produces the table is not a linear one; i.e., the difference between 0.3 and 0.4 is not the same as the difference between 0.4 and 0.5, as you can readily verify for yourself by looking at the table. Over small intervals, however, a linear interpolation is usually sufficiently accurate.

 Find this term in: par # ---- 20