<p> Procedure on developing local norms</p><p>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. </p><p>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)</p><p>285 256 329 298 312 290 212 301 321 279</p><p>Step 2</p><p>Determine two descriptive statistics: average or mean (M) and standard deviation SD</p><p>ΣX/N, where:</p><p>Σ = Sum of</p><p>X = raw scores</p><p>N = number of students in the sample</p><p>From our example</p><p>Σ = Sum of = 285 + 256 + . . . +279 = 2883 </p><p>M = ΣX/N = 2883/10 = 288.3</p><p>Standard deviation SD = SD = √ Σ(x – M)2 / N</p><p>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</p><p>SD = √ Σ(X – M)2 / N</p><p>Σ = sum of</p><p>X = each raw score M = the average</p><p>N= number of children tested </p><p>From the above example</p><p>Σ(X – M)2 = 10508.1</p><p>Σ (X – M)2 / N = 10508.1/10 = 1050.81</p><p>SD = √ Σ (X – M)2 / N; √1050.81 = 32.42</p><p>SD = 32.42 </p><p>Next Step</p><p>Ask the computer to convert the raw scores in “z’ scores.</p><p>A z score is ….</p><p>Z = (X-M)/SD</p><p>Z score for each raw score is obtained by </p><p>Z= (285-288.3)/32.42 = -.10 . . .</p><p>Z= (279-288.3)/32.42 = -.29</p><p>Because it is very comfortable to read whole numbers, we transform the Z scores in T scores. </p><p>A T score is defined with the following formula: </p><p>T = Z(SD) + M</p><p>T = -.10(10) + 50 = 49.0</p><p>.A T score is the result of a mean of 50 and a standard deviation of 10. . . T = -.29(10) + 50 = 46.1</p><p>Finally, we rounded each T score to whole numbers and we have</p><p>For the first</p><p>Z = -.10</p><p>Z = -.10(10) + 50 = 49 For the last Z score</p><p>Z = -.29 Z = -.29(10) + 50 = 46.1 = 46</p><p>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.</p><p>T = z(10) + 50 </p><p>The above table illustrates the standardization for the 10 raw scores.</p><p>We recommend to make selection based in those students who reached at least 1SD.</p><p>In this set of 10 scores, only two scores were identified having a equal or better 1SD</p><p>Those scores were 321 and 329. </p><p>Students identified with those scores strongly recommended to further School District testing.</p><p>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?</p>