# Module3-trends in evaluation Unit-6;Interpreting Test Results

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 Module3-trends in evaluation Unit-6;Interpreting Test Results DEFINITION OF STATISTICS:- It is a scientific study of dealing with the numerical data that has been gathered by the administration of some kind of test in a classified or tabulated manner involving the method of collection of data, their classification, analysis, and interpretation in the field of enquiry. USES OF STATAISTICS TO A TEACHER:- Presents facts (data) in a definite form. General pattern of the set of scores is arrived by calculating statistical measures. Presents in graphical form. Find nature of the group by comparing the performances. Give meaning to interpretation of the data. Performance of the group is described in terms of deviation from the normal distribution. For correction. To verify hypotheses. To draw conclusion and predict. For research. IMPORTANT TERMS POPULATION:- All the people/ object from which sample is selected is called population. SAMPLE:- The group under actual study or part of the population which we select for study is called as sample. SCORE:- The marks assigned after assessment of a test is called score. The mathematical meaning of a score is an interval which extends 0.5 units below to 0.5 units above. it is also called face value of the score e.g. Score of 50 has a face value 49.5---50.5 SERIES:- When raw score are arranged in a define order, a series is formed. ORGANIZATION OF RAW SCORES GIVE MEANING TO THE DATA (SCORES). FREQUENCY DUSTRIBUTION:- To arrange numerical data(raw scores) according to magnitude/size. RAW SCORES OBTAINTED FROM A TEST 13, 7, 44, 29, 11, 22, 33, 41, 17, 23, 31, 10, 9, 24, 34, 39, 2, 15, 25, 27, 28, 26, 32, 47, 42, 41, 23, 24, 22, 21, 28, 27, 17, 16, 12, 28, 26, 18. STEPS IN FREQUENCY DISTRIBUTION DETERMINE THE RANGE: Rang is the difference between highest and lowest score. Range = 47-7 = 40 DECIDING SZE / LENGTH OF THE CLASS- INTERVAL(C-I) : Length can be 2,3, 4, 5, 10. No. of C.I should be between 5 to 10. Length of C.I = = Range / total no. of C.I, = 40/8 = = 5 TALLY THE FREQUENCY: Write the C.f in ascending / descending order converting highest and lowest score. Put tally for each score in front of the C.I where it falls till all the scores are covered. The no. of tallies in a C.I represents the frequency of that C.i Frequency is symbolized as (f). LIMITS OF THE C.I. : EXCLUSIVE LIMITS/ EXPRESSED LIMITS-50---59 EXACT LIMITS/ACTUAL LIMITS----49.5------59.5 MID POINT OF THE C.I.: MID-POINT(X) = LOWEST LIMIT+HIGHEST LIMIT 2 Mid point of the C.I( 45---49) = 45 + 49 = 47 2 FREQUENCY DISTRIBUTION TABLE .C I TALLIES FREQUENCY(F) MID P (X) 45 – 49 I 1 47 40 – 44 IIIII 5 42 35 – 39 II 2 37 30 – 34 IIII 4 32 25 – 29 IIIIIIIIII 10 27 20 – 24 IIIIIII 7 22 15 – 19 IIIII 5 17 10 – 14 IIII 4 12 5 – 9 II 2 7 _________ N=40 ADVANTAGES OF FREQUENCY DISTRIBUTION Makes data easier to interpret. Provides clear picture of the group. Forms basis of graphical representation. Helps in comparing the groups. Essential for calculating different statistically measures. LIMITATIONS Not possible to located the score of a particular student. For calculation, mid-point is taken as representative, which may not be true. Some information is lost. After the tabular representation of data, the next task is make a numerical representation of by statistical descriptive measures. DESCRIPTIVE MEASURES A-MEASURES OF CENTERAL TENDENCY (C. T) It indicates the central position of the performance of the group or the average performance of the group. It indicates how the performance of the group on an average is, when arranged in order of size. Since the representative / average value is usually found near the center of the data, they are commonly called measures of central tendency. This is used for comparison of data. IMPORTANCE OF C.T: T is an average representative score. It enables us to compare groups. For calculation of other statistical measures. MEAN / ARITHMETIC MEAN When data are ungrouped, M = ∑ X, N It is the sum of the separate scores divided by total numbers. X= score, N=total no. of scores. e.g. 5, 7, 8,9,11, M=5+7+8+9+11 40=8 5 5 When data is grouped –: Assumed mean method .C I (F) M ID P ( x ‘ fx’ 45 – 49 1 4 4 40 – 44 5 3 15 35 – 39 2 2 4 30 – 34 4 1 4 25 – 29 10 27 0 0 20 – 24 7 -1 -7 15 – 19 5 -2 -10 10 – 14 4 -3 -12 5 – 9 2 -4 -8 A.M.= Assumed mean =Midpoint of the C.I. with highest no. of frequencies = 27 X’ = Deviation of assumed means from the mid- point of the C.I. Fx’ = Deviation multiplied by the frequencies of that C.I. Fx’ = Sum of fx’ = -10 C = Correction = fx’ = -10= -0.25 --- ------ N 40 T.M = true means = A.m + cxi T.M = 27 –(0.25x5) 27-1.25 25.75 STEPS Choose an Assumed Mean, usually mid- point of the C.I having highest frequencies is chosen.i. e. A. M. Multiply deviation (x’) with the frequencies of that C.I from the assumed mean i. e x’ = X- A.M. Multiply deviation (x’) with the frequencies of that C.I i.e f x x’ = fx’ Find the sum of fx’ i.e fx’ Use the formula for correction ( c). C = fx’/N Calculate True Mean (T.M), T.M = A.M + C INTERPRETATION : MEAN IS THE AVERAGE SCORE OF THE VDISTRIBUTION . IN THE GIVEN DISTRIBUTION MEANS IS 25.75 THAT MEANS MOST OF THE SCORES LIE AROUND 25.75 i.e most of the students have scored 25.75 in the given test. It is the central value of the distribution. USES When greatest stability is needed. Other statistical measures like S.D are to be computed. When the distribution is symmetrical about the center. MEDIAN(Mdn)/ POSITIONAL AVERAGE/ COUNTING AVERAGE It is the mid point of the series. OR Median is the point below which 50% of score lie. WHEN DATA IS UNGROUND = (N+1)th SCORE 2 WHEN N IS ODD : 6,6,8,9,10, 11, 12. N=7 , 7+1 = 4TH SCORE 2 Median is the 4th score in the series i. e score 9 WHEN IS EVEN : 7, 8, 9, 10, 11, 12. N= 6, 6+1 = 3.5TH SCORE 2 Median is the 3.5th score in the series i. e average of the 3rd and 4th score i.e 9+10 = 9.5 2 WHEN DATA ARE GROUPED .CI f F(cf) 45---49 1 40 40---44 5 39 35---39 2 34 30---34- 4 32 25---29 10 28 --- N/2=20,falls in this C.I 20---24 7 18 15---19 5 11 10---14 4 6 5----9 2 2 ----------- N=40 F(cf)-- Cumulative frequency N/2=40/2=20 MEDIAN=L+(N/2-F)x i fm L= Lower limit of the C.I in which N/2 falls.=24.5 N/2=Half of the total no. of scores. =20 F=Cumulative Frequency of the C>I below the N/2 (Median) C.I =18 fm= Frequency of the C.I in which N/2 (Median) falls,=10 i= Length of the C.I = 5 MEDIAN (Mdn)=24.5+ (20-18)x 5 10 = 24.5+2x5= 24.5 +1=25.5 10 INTERPRETATION: Median is the positional average of the distribution. it divides the distribution into two equal halves. The median=25.5 that means 50% of the scores lie above 25.5 and 50% of scores lie below 25.5.it is the 50th percentile of the distribution OR N/2 students have scored above 25.5 and 20 students have scored below 25.5. USES: WHEN EXACT MID POINT (50%) OF THE DISTRIBUTION IS NEEDED. TO avoid influence of the extreme scores. When distribution is interpreted in terms of percentile. B-MEASURES OF VARIABILITY VARIABILITY REFER TO THE SPREAD OF THE separate scores around the measures of central tendency. B-INTERPRETATION W.r.t .N.P.C M + 1 S.D = 68.26 % cases. M + 2 S.D = 95.00 % cases. M + 3 S.D = 99.00 % cases. M = 25.75, S.D = 10.17 M + 1S.D = 25.75 + 10.17 = 35.92 & 25.75 – 10.17 = 15.58 Range is 15-36, The number of cases with this range are 5,7,10,4,2. Total number of the cases = 5+7+10+4+2 = 28 N= 40 Out of 40 ________28 cases ... out of 100 _________ 28 x 100 = 70 % This means between M + 1 S.D, 70 % cases lie i.e distribution is more or less normal. Otherwise if the value of M + 1 S.D = 68.26 % cases is lower or higher than the distribution is not normal or it is skewed. C- PERCENTILES & PERCENTILE RANKS SUBJECT STUDENT A STUDENT B ENGLISH 60 40 MATHS 85 95 HISTORY 70 35 _______________ ___________ 215 170 72% 57% I CLASS II CLASS Are this I class & II class logical? The marks in different subjects are neither comparable nor combinable. If the scores of the individual are to be compared, the unit of measurement must be the same. ... if the scores are to be combined, they should be converted in to Derived scores having common units of measurement They are PERCENTILES & PERCENTILE RANKS PERCENTILES The points that divide a distribution into 100 equal divisions are called percentiles. It is a point on the score below which a given percentage of scores lie. It is also known as centiles. It is symbolically represented as Pp. P50, it is called 50th percentile. It means 50% of the cases lie below this point. If P50 = 72, than, it is said 50% of the cases lie below the score 72. or 50 % of the students have scored below the marks 72 and 50 % of the students have scored the marks above 72. PERCENTILE RANKS A percentile rank indicates a pupil’s relative position in a group in terms of the % of pupils scoring below him. It is represented as PR. e.g N=50 A student X gets a score of 60 in a test. P40 = 60, find the PR P40 means that 40% of the students have scored below 60. Out of 50______ 40 students. ... out of 100 ________ 40 x 100 = 80% that means 80th percentile = 60. 50 The X who has scored 60 marks his percentile rank is 80. Example :- P60 = 27 it is the 60th percentile. 60% of the case lie below the score 27 or 40% of the cases lie above the score 27, and the individual scoring 27 has a percentile rank 60. It is also written as PR60 = 27. USES: * For comparing the norms of the group. * For preparing the profiles of characteristics of the individuals. * For calculation of other statistics. Problem: In a test following results were found:- RAM’S PR = 65 & MOHAN’S PR = 79, compare and interpret their performances. Interpretation: 65 % of the students have scored below RAM. & 79% of the students have scored 65% SAMPLE QUESTION Observe the given distribution table and answer the questions given below:- Class interval frequencies 80-84 2 75-79 3 70-74 5 65-69 12 60-64 8 55-59 6 50-54 4 ------------ N= 40 1) Calculate the mean/median of the given distribution and interpret the result. (5 ) ii) If SD= 8.2, interpret with respect to Normal Probability Curve. (2) iii) If P43 =74, Interpret. (1) iv) In a Geography test, Raja’s PR is 36 and Rani’s PR is 63, interpret and compare their performance. (2) Calculation of mean by assumed mean method T.M = A.M + Cxi 67+(-15x5) --------- 40 67-1.87 65.13 Interpretation of mean : Mean is the average score of the distribution. In the given distribution Mean=65.13 that means most of the scores lie around . It is the centvalue of the distribution. Calculation of median MEDIAN=L+(N/2-F)x i fm =64.5+(20-18)x5 ------------- 12 64.5+(2)x5 ------ 12 64.5+10 ---- 12 64.5+0.83 65.33 Median is the positional average of the distribution. It divides the distribution into two equal halves. The Median =65.33 , that means 50% of scores lie below 65.33 and 50% scores lie above 65.33 . Here, N/2=20. ie.20 students have scored below 65.33 and 20 students have scored above 65.33 According to Normal Probability Curve Between M + 1S.D. and M – 1 S.D. lie 68.26 % cases. Here, M = 65.33 and S. D. = 8.2 M + S. D = 65.33 + 8.2 = 73.53 M – S.D. = 65.33 - 8.2 = 57.13 Therefore the range is 57.13 to 73.53 Total frequency between this range = 6+8+12+5 = 31 , N = 40 40 ------- 31 100 ------ ? = 77.5 % Conclusion: As number of cases between M +1 S. D. and M – 1 S. D. is 77.5 %, given distribution is not normal . Percentile and Percentile Ranks Q.- If P65 = 36, Interpret. It is the 65th percentile. thus, 65% students in the class have scored below 36 and 35% have scored above it. Q. In a Math test Sudha’s PR=75 and Sheela’s PR=85. Compare and interpret their performance. Interpretation- 75% students have scored less than Sudha and 85% students have scored less than Sheela in a math test. Comparison- Less number of students have scored above Sheela than Sudha. Conclusion- In a math test Sheela’s performance is better than Sudha’s performance. --------------------------------------------------------------------------------------------------------------- | Page

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