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Assignment 7 Name:Grace Lin Class:碩專二甲 Number:Na2c0023 P.137 APPLICATION EXERCISES QA. Answer A1.--- According to the figure above, an IQ score of 85 would represent 16 percentile score. A2.--- About 81.85 persentage of students scored between 70 and 115. 34.13+34.13+13.59= 81.85 A3-1--- About 5 standard deviations would Iliana be above the mean. 177-100=77 77⌯15=5.13 A3-2--- It depends on what kind of test did she take. If the test is Wechsler Intelligence Scale Fourth Edition, it is a formal norm referenced test, it means that Iliana has got the figure of mental giftedness. She is really intelligent when she had a score of 177. Otherwise, we don’t know what kind of test it is or what is the detail of the test, it is not enough to judge whether Iliana is Intelligent or not. The only thing we can say is that Iliana had got a high score. Her score is almost the highest. A4-1--- Iliana’s z score wuld be 5 (177-100) ⌯15=5.13 A4-2--- Iliana’s T score wuld be 101 (10x5.13)+50=101.3 A4-3--- Iliana’s CEEB score wuld be 1013 (100x5.13)+500=1013 QB. Answer Student Raw score z score T score CEEB score A 64 c 70 h B 50 d e i C a -1.0 F j D b -1.5 g 350 (10c)+50=70 c=2 e=(10d)+50 =(100)+50 =50 e=50 64-50=14 --- 2 (standard deviation) c-2=d 2-2=0 d=0 f=(10-1)+50 =40 f=40 50-1(standard deviation) =50-7 =43 g=(10-1.5)+50 =35 g=35 a=43 h=(100c)+500 50-1.5(standard deviation) =50-10.5 =39.5 b=39.5 =(1002)+500 =700 h=700 i=(100d)+500 =(1000)+500 =500 i=500 j=(100-1)+500 =-100+500 =400 j=400 Student Raw score z score T score CEEB score A 64 +2 70 700 B 50 0 50 500 C 43 -1.0 40 400 D 39.5 -1.5 35 350 QC. Answer Test k Raw scores M S Standardized scores M S A 110 60 25 500 100 B 75 60 15 50 10 C 50 11 4 0 1 k = number of items on the test C1 a.---z scores = Test C (-3 ≤ z ≤ 3) b.---T scores = Test B (20 ≤ T ≤ 80) c.---CEEB scores = Test A (200 ≤ CEEB ≤ 800) C2 a.--- Test A has the largest standard deviation: 25. b.--- Test C has the lowest mean: 11. c.--- Test B has a negatively skewed distribution Test A has a normal skewed distribution. Test C has a positively skewed distribution. C3 a.--- a raw score of 11 equals a z of 0 The raw score of 11 = mean = a z of 0 b.--- a raw score of 7 equals a T score of 40 (7-11) ⌯4=-1 T=(10-1)+50 =40 c.--- a raw score of 19 equals a CEEB score of 700 (19-11) ⌯4=2 CEEB=(1002)+500 =700 QD. Answer Student Raw Score z T CEEB Shenan 77 2.07 71 707 Robert 75 1.55 65 655 Randy 72 0.77 58 577 Mitsuko 72 0.77 58 577 Millie 70 0.26 53 526 Kimi 70 0.26 53 526 Kazumoto 69 0.00 50 500 Kako 69 0.00 50 500 Joji 69 0.00 50 500 Jeanne 69 0.00 50 500 Issaku 68 -0.26 47 474 Iliana 68 -0.26 47 474 Dean 67 -0.52 45 448 Corky 64 -1.29 37 371 Bill 64 -1.29 37 371 Archie 61 -2.07 29 293 Mean 69 0 50 500 3.87 1.00 10.00 100.00 SD QE. Answer Student Mean SD Mode 1 2 3 4 5 6 7 8 9 Raw Score 92 93 86 83 40 70 90 24 99 10 11 12 z T 0.43 0.48 0.11 -0.05 -2.30 -0.73 0.32 -3.14 0.79 54 55 51 50 27 43 53 19 58 CEEB 543 548 511 495 270 427 532 186 579 83 99 88 -0.05 0.79 0.22 50 58 52 495 579 522 13 14 15 16 17 18 19 80 100 90 97 97 96 90 -0.20 0.85 0.32 0.69 0.69 0.64 0.32 48 58 53 57 57 56 53 480 585 532 569 569 564 532 20 21 22 23 24 85 94 87 100 50 0.06 0.53 0.16 0.85 -1.78 51 55 52 58 32 506 553 516 585 322 83.88 19.08 90 62 -0.09 1.08 49.08 10.81 490.83 108.12 midpoint 90 median -2.03 Skewness 3.72 Kurtosis The skewness = -2.03 Midpoint< Mean< Median= Mode Kurtosis=3.72 The data of the test in my class is not normally distributed. It is a negatively skewed distribution. The Kurtosis is Leptokurtic.