A test norm is a set of scalar data describing the performance of a large number of people on that test. Test norms can be represented by two important statistics: Means and Standard Deviations. The most important measure in psychometrics is the arithmetical average or the mean. This is intuitive for many people. For example the average of 2,4 and 6 is 4. Or the average of 1,2,3 is 2. In psychometrics this average is important because it indicates he middle position in a distribution.
Formula for the Mean (or Arithmetical Average)
X = the mean
Σ X = the sum of the scores
N = the number of people
Use the data below to practice finding the mean from a set of raw scores on a Numerical Critical Reasoning Assessment. The mean for the Verbal assessment has already been calculated for you.
Name Verbal Numerical
Bilton 42 45
Jones 58 55
Clayton 54 50
Pearson 56 60
Pickering 45 40
|Calculating the mean for the
Verbal Test above
|Data for Practice calculation of the mean for the Numerical Test above
Interpreting test norms requires an understanding of Standard Deviations as well as Means. Once we understand both of these we have established the basis to describe performance with a high level of fluency and accuracy.
The Standard Deviation (SD) tells us how spread out the distribution is around a central point.
The standard deviation is, in effect, an average of the departure of people’s scores from the group mean. This is a measure of how spread out the group’s scores typically are. For example, a group whose scores clustered very tightly around the mean would have a low Standard Deviation. A group whose scores were very spread out would have a high Standard Deviation, as, on average, each score would differ more from the group mean.
The standard deviation can be calculated by the formula:
Where SD = standard deviation (also represented by sd or )
X = each individual raw score taken from the mean
N = number in group
The standard deviation gives an indication of the degree of dispersion of the scores from the mean, and an estimate of the variability of results in the total sample. It is usefully employed in comparing the variability of different groups. It is used to develop Test Norms because, unlike percentiles, measures based on how much an individual deviates from the mean can be mathematically manipulated and compared.
Let us suppose that we have two distributions, Group A and Group B, with the same mean but the performance of people Group A is more spread out. Group A will have a larger standard deviation than Group B, which has smaller individual differences. In Group B each score is closer to the mean and this means there is a smaller standard deviation.
Describing Performance using Z Score
The Z Score, based on the mean and standard deviation, indicates how many standard deviations above or below the mean a score is. A Z score is merely a raw score, which has been changed to standard deviation units.
The standard Z score is calculated by the formula:
Where Z = standard Z score
X = each individual raw score
X (bar) = mean score
SD= standard deviation
Usually when standard scores are used they are interpreted in relation to the normal distribution curve. Standard Z scores in standard deviation units are marked out either side of the mean. Those above the mean are positive, and those below the mean negative in sign. Therefore, by the calculation of the standard Z score an individual’s score can be viewed in relation to the rest of the distribution.
The standard score is very important when comparing scores from different tests. Before these scores can be properly compared they must be converted to a common scale such as the standard Z score. The Z-score can then be used to express an individual’s score on different tests in terms of norms. One important advantage in using the normal distribution curve as a basis for test norms is that the standard deviation has a precise relationship with the area described by the curve. One standard deviation above and below the mean will include approximately 68% of the sample, as illustrated here.
Z scores have a mean of 0 and a standard deviation of 1. Because of this, Z scores can be rather difficult and cumbersome to handle because most of them are decimals and approximately half of them can be expected to be negative. To remedy these drawbacks various transformed standard score systems have been derived. These simply entail multiplying the obtained Z score by a new standard deviation and adding it to a new mean. Both of these steps are devised to eradicate decimals and negative numbers.
Describing Performance Using T Scores
The t in t score stands for transformed. A T score is a transformed z score. T scores and stens (see below) take the Z Score and transform it into a more effective and user-friendly format. However critically, both still describe the level of deviation a particular score has from the mean.
The T score (transformed score) is the most common standardised norm system for ability tests. The T score is derived from the Z score, transformed to a new scale, so that the:
T score mean is set at 50 and the standard deviation at 10.
Calculate the T score using the following formula:
Z score X 10+50
If Z score = 2
T score = 2 X 10 + 50
= 20 + 50
Although the T score can easily be calculated, it is most often used by reference to a norm table. In order to identify the relevant T score, you find the raw score in the body of the table and read across to the right hand side. The advantage of T scores over percentiles is that they are a linear scale with equal units of measurement. Thus they can be mathematically manipulated and give results which can be directly compared both within and between individuals. Their disadvantage is that they are not so easily explained or meaningful to those not knowledgeable about testing.
Describing Performance Using Stens
Stens or Standard Tenths are the most commonly used scales for comparing individual differences. The normal distribution is divided into ten stens. Unlike percentiles, stens are equal units of measurement, and unlike percentiles they are not influenced by clustering around the midpoint.
A Sten is based on a mean of 5.5 and a standard deviation of 2.
Calculate the sten using the following formula:
Z score x 2 + 5.5
If Z score = 2
Sten = 2 x 2 + 5.5
Normally, the sten is taken to the nearest whole number, with a minimum value of 1 and a maximum of 10.
Stens are commonly used as the norm system for personality questionnaires. However, they are also a useful system for ability tests as they encourage us to think in bands of scores but still provide sufficient discrimination between applicants.