Procedure on developing local norms

Step 1. Collect all raw scores of the 77 items per test by classroom or by grade level. (Local norms can be developed by classroom or by grade level. A school campus might wish to use only classroom norms, or it might wish to normalize the scores by grade level. Additionally, a school district might wish to norm its data by grade level across the district.

The following example, is a real sample of 10 students attending a bilingual kindergarten class who were screened by a teacher using the Hispanic Bilingual Gifted Screening Instrument (HBGSI) (Irby, Lara- Alecio, 2002)

285 256 329 298 312 290 212 301 321 279

Step 2

Determine two descriptive statistics: average or mean (M) and standard deviation SD

ΣX/N, where:

Σ = Sum of

X = raw scores

N = number of students in the sample

From our example

Σ = Sum of = 285 + 256 + . . . +279 = 2883

M = ΣX/N = 2883/10 = 288.3

Standard deviation SD = SD = √ Σ(x – M)2 / N

Σ X (X-M) Σ (X-M)2 (X-M)2/SD = Z T= Z(10)+50 Rounded 285 -3.3 10.89 -.10 49.0 49 256 -32.3 1043.29 -1.00 40.0 40 329 40.7 1656.49 1.26* 62.6 63 298 9.7 94.09 .30 53.0 53 312 23.7 561.69 .73 57.3 57 290 1.7 2.89 .05 50.5 51 212 -76.3 5821.69 -2.35 26.50 27 301 12.7 161.29 .39 53.90 54 321 32.7 1069.29 1.01* 60.10 60 279 -9.3 86.49 -.29 46.10 46 2883 1050.81 SD=32.42

SD = √ Σ(X – M)2 / N

Σ = sum of

X = each raw score M = the average

N= number of children tested

From the above example

Σ(X – M)2 = 10508.1

Σ (X – M)2 / N = 10508.1/10 = 1050.81

SD = √ Σ (X – M)2 / N; √1050.81 = 32.42

SD = 32.42

Next Step

Ask the computer to convert the raw scores in “z’ scores.

A z score is ….

Z = (X-M)/SD

Z score for each raw score is obtained by

Z= (285-288.3)/32.42 = -.10 . . .

Z= (279-288.3)/32.42 = -.29

Because it is very comfortable to read whole numbers, we transform the Z scores in T scores.

A T score is defined with the following formula:

T = Z(SD) + M

T = -.10(10) + 50 = 49.0

.A T score is the result of a mean of 50 and a standard deviation of 10. . . T = -.29(10) + 50 = 46.1

Finally, we rounded each T score to whole numbers and we have

For the first

Z = -.10

Z = -.10(10) + 50 = 49 For the last Z score

Z = -.29 Z = -.29(10) + 50 = 46.1 = 46

Ask the computer to convert the z scores in “T” scores. One of the main objectives for doing this is to convert any number in positive.

T = z(10) + 50

The above table illustrates the standardization for the 10 raw scores.

We recommend to make selection based in those students who reached at least 1SD.

In this set of 10 scores, only two scores were identified having a equal or better 1SD

Those scores were 321 and 329.

Students identified with those scores strongly recommended to further School District testing.

What we really need is not all this explanation—that is for you.. But we just need the computer to calculate the T score for each child’s HBGSI and then to tell the teacher whether the score is one standard deviation above the mean. One standard deviation above the mean would be a score of 60 or above. Anyone who makes that would get a yes to go ahead for further testing. What do you think?