Standardized Tests (Continued)

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More on Percentile rank

Mathematically, the percentile rank is defined as:

p = [cfi + .5(fi)]/N * 100%

where cfi is the cumulative frequency, the counts accumulated by the current count and all previous ones, for all scores lower than the score of interest, fi is the frequency of scores in the interval of interest, and N is the number in the sample.

Steps to compute a percentile rank

  1. Construct a frequency (counts) distribution for the raw scores.

  2. For a given raw score, determine the cumulative frequency for all scores lower than the score of interest.

  3. Add half the frequency for the score of interest to the cumulative frequency value determined in step 2.

  4. Divide the total by N, the number of examinees and multiply by 100%.

Table 1 Raw Scores, Frequency Distribution,
& Percentile Rank
Raw Score frequency cumulative freq.Percentile rank
11 2 2 01
12 1 3 02
13 6 9 04
14 5 14 08
15 12 26 13
16 17 43 23
17 2164 36
18 28 92 52
19 19 111 67
20 15 126 79
21 10 136 87
22 5 141 92
23 3 144 95
24 4 148 97
25 2 150 99

For example, based on the data in Table 1, the percentile rank of a raw score of 17 is:

P = 43 + (.5)(21)/150 * 100% = 36

  • Percentile rank scale is a non-linear transformation of the raw score scale. This means that at different regions on the raw score scale, a gain of 1 point may correspond to gains of different magnitudes on the percentile rank scale.

  • Percentile rank scores are less stable (reliable) for scores in the central part of the distribution than for those in the extremes (the values in the two tails of the distribution).

Transformations of z-scores

Z scores have negative values, which can be difficult to interpret to test users. How can you explain an examinee that his z score is -1.5? For this reason it is often convenient to perform a linear transformation on z-scores to convert them to values that are easier to record or explain. The general form of such a transformation is:

y = m + k(z)

where Y is the derived score, and m and k are constant values arbitrarily chosen to suit the convenience of the test developer. The value chosen for the m will be the mean of the new distribution after transformation, and the value chosen for the k will be its standard deviation.

Teacher scores (T scores)

T = 50 + 10(z)

College Entrance Examination Board (CEEB)

y = 500 + 100(z)

Intelligence Quotient (IQ)

IQ = 100 + 15(z)

Stanines

Stanines are ranges or bands within fixed percentages of score fall.

  • They divide the normal curve into nine portions

  • Each portion is 1.5 standard deviation wide.

  • Mean = 5

  • Standard deviation = 2

Table 2 Stanines
Stanine Percentage of cases
1 4% (lowest)
2 7%
3 12%
4 17%
5 20%
6 17%
7 12%
8 7%
9 4% (highest)

Summary Figure

Grade and Age Equivalents

When children are tested on aptitude or achievement measures, the test user often wants a normative score which will indicate how a given child's performance compares with that of others at a particular age or grade level.

Misinterpretations of grade equivalent scores

  • "Alex obtained a math computation grade-equivalent (GE) score of 7.6 (seventh grade, sixth month) on the CAT. That means that even though she's only in fourth grade, she can do seventh-grade-level math." Wrong. Alex's obtained score is the score that the test developer estimates would be obtained by the average seventh-grader during the sixth month of the school. It just means that Alex's math is above average.

  • Comparisons among GE scores across areas for the same student also may be misleading. e.g. a GE of 3.9 in reading and 3.9 in science would appear to imply equal proficiency in the two subjects, but this is not necessarily true. Although the two scores are equal, the percentile ranks in the two subjects could be, for example, 68 in reading and 82 in science.

Questions

 1. If the z score is 1.5, what are the following scores?

  • T score
  • CEEB

2. Given the following data, compute the percentile rank.

  • cfi = 40
  • fi = 20
  • N = 100

3. My percentile rank in a test is 52 and John's rank in the same test is 30. "Ha! I am doing much better than John." Am I right? Why or why not?

4. Dr. Who developed a test and collected data on its reliability. These are what he got:

  • Internal consistency: 1.10
  • Test-retest reliability: .99
  • Equivalency : .99

Because the coefficients exceed the standards, he concluded that his test is highly reliable. Is he right? Why or why not?

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