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Determination of the Optimum Number of Items to Retain in a Test Measuring a Single Ability

Published online by Cambridge University Press:  01 January 2025

B. J. Bedell
B. J. Bedell*
Affiliation:
Colonial Social Science Research Laboratory, Trinidad, B.W.I.
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Abstract

It should be possible to make a test a more accurate measuring instrument by discarding some of the items which have been found, by item analysis, to be inefficient. If this is so, fresh and more reliable scores from the already marked scripts would be obtained by neglecting the discarded items, and, in the future, the abbreviated test would be used. The present article offers two solutions, based on slightly different premises, to the problem of which items to discard when the test has been designed to measure a single ability. In a practical example these yield almost the same result, which is also the result obtained by a less rigorous grouping method described, which is by no means laborious to use.

Type
Original Paper
Information
Psychometrika , Volume 15 , Issue 4 , December 1950 , pp. 419 - 430
Copyright
Copyright © 1950 The Psychometric Society

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References

* This seems better than finding k for r1+2+...+k,1+2+...n greatest, as this would yield a test of k items correlating most highly with a criterion containing discarded items.

† From the well known formula for partial correlation, But when 3 is kept constant, on the assumption that 3 is solely responsible for the correlation between 1 and 2, r12.3 = 0.

* For is the sum of the elements in the N.E. quadrant and this with σj2 replacing the principal diagonal elements is the sum of the elements in the N.W. or S.E. quadrants

† For if there is no correlation,

. (This is truly so if k → ∞, so and only approximately so in samples.) Whence,

or,

, i.e.,

. Writing σj2, ej2, M σ2 for σj, ej, Mσ, the other relation is proved.

* If p is the probability of an item being answered correctly and q is the probability of its not being answered correctly, assuming a binomial distribution,

and

or,

. Also,

. If p be taken to have the values .2, .3, .4, .5, .6, .7, .8, M σ2 works out to be .2100 and M σ2 to be .2084. Or if p be taken to have an infinite number of values between .2 and .8,

and

* To obtain the value of x making y = f(x) a maximum, being given the coordinates (x1y1, (x2y2), (x3y3) of points near the maximum value of y, the value of x, (X) , making y a maximum for the parabola passing through the given points may be determined from

where

and

.

* The value of ej corresponding to the maximum rkk.

* The value of ej corresponding to the maximum rkk.