As there are six tests in each age group from III to X, each test in this part of the scale counts 2 months toward mental age. There are eight tests in group XII, which, because of the omission of the 11-year group, have a combined value of 24 months, or 3 months each. Similarly, each of the six tests in XIV has a value of 4 months (24 ÷ 6 = 4). The tests of the “average adult” group are given a value of 5 months each, and those of the “superior adult” group a value of 6 months each. These values are in a sense arbitrary, but they are justified in the fact that they are such as to cause ordinary adults to test at the “average adult” level.
The calculation of mental age is therefore simplicity itself. The rule is: (1) Credit the subject with all the tests below the point where the examination begins (remembering that the examination goes back until a year group has been found in which all the tests are passed); and (2) add to this basal credit 2 months for each test passed successfully up to and including year X, 3 months for each test passed in XII, 4 months for each test passed in XIV, 5 months for each success in “average adult,” and 6 months for each success in “superior adult.”
For example, let us suppose that a child passes all the tests in VI, five of the six tests in VII, three in VIII, two in IX, and one in X. The total credit earned is as follows:—
Taking a subject who tests higher, let us suppose the following tests are passed: All in X, six of the eight in XII, two of the six in XIV, and one of the six in “average adult.” The total credit is as follows:—
| Years | Months | |
|---|---|---|
| Credit presupposed, years I to IX | 9 | |
| Credit earned in X, 6 tests passed, 2 months each | 1 | |
| Credit earned in XII, 6 tests passed, 3 months each | 1 | 6 |
| Credit earned in XIV, 2 tests passed, 4 months each | 0 | 8 |
| Credit earned in “average adult,” 1 success, 5 months | 5 | |
| Total credit | 12 | 7 |
One other point: If one or more tests of a year group have been omitted, as sometimes happens either from oversight or lack of time, the question arises how the tests which were given in such a year group should be evaluated. Suppose, for example, a subject has been given only four of the six tests in a given year, and that he passes two, or half of those given. In such a case the probability would be that had all six tests been given, three would have been passed; that is, one half of all. It is evident, therefore, that when a test has been omitted, a proportionately larger value should be assigned to each of those given.
If all six tests are given in any year group below XII, each has a value of 2 months. If only four are given, each has a value of 3 months (12 ÷ 4 = 3). If five tests only are given, each has a value of 2.4 months (12 ÷ 5 = 2.4). If in year group XII only six of the eight tests are given, each has a value of 4 months (24 ÷ 6 = 4). If in the “average adult” group only five of the six tests are given, each has a value of 6 months instead of the usual 5 months. In this connection it will need to be remembered that the six “average adult” tests have a combined value of 30 months (6 tests, 5 months each); also that the combined value of the six “superior adult” tests is 36 months (6 × 6 = 36). Accordingly, if only five of the six “superior adult” tests are given, the value of each is 36 ÷ 5 = 7.2 months.
For example, let us suppose that a subject has been tested as follows: All the six tests in X were given and all were passed; only six of the eight in XII were given and five were passed; five of the six in XIV were given and three were passed; five of the six in “average adult” were given and one was passed; five were given in “superior adult” and no credit earned. The result would be as follows:—
| Years | Months | |
|---|---|---|
| Credit presupposed, years I to IX | 9 | |
| Credit earned in X, 6 given, 6 successes | 1 | |
| Credit earned in XII, 6 given, 5 passed. Unit value of each test given is 24 ÷ 6 = 4. Total value of the 5 tests passed is 5 × 4 or | 1 | 8 |
| Credit earned in XIV, 5 tests given, 3 passed. Unit value of each of the 5 given is 24 ÷ 5 = 4.8. Value of the 3 passed is 3 × 4.8, or | 0 | 14+ |
| Credit earned in “average adult,” 5 tests given, 1 passed. Unit value of the 5 tests given is 30 ÷ 5 = 6. Value of the 1 success | 0 | 6 |
| Credit earned in “superior adult” | 0 | 0 |
| Total credit | 13 | 4+ |