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It refers to the process of examining the students response to each item in the test.
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Purposes and Elements of Item Analysis1. To select the best available items for the final form of the test2. To identify the structural defects in the items3. To detect learning difficulties of the class as a whole; and4. To identify the areas of weaknesses of the students in need of remediation
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Characteristics of an item Desirable characteristics can be retain for subsequent use Undesirable characteristics is either to revised or rejected
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Three main Elements in an Item- Analysis1.Difficulty level of the items2. Discrimination power of each item3. Examination of the effectiveness of distracters
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Difficulty index - refers to the proportion of the number of students in the upper and lower groups who answered an item correctly. Therefore it can be obtain by adding the proportion in the upper and lower groups who got the item right and divide it by 2.
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Index of Discrimination- it is the percentage of high-scoring individuals responding correctly vs. the number of low-scoring individuals responding correctly to an item.
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• Maximum Positive Discriminating Power of an item – it is indicated by an index of 1.00 and is obtain when all the groups answered correctly and no one in the lower group did. • Zero Discriminating power – is obtain when an equal number of students in both groups got the item right • Negative Discriminating Power of an item – it is obtain when more students in the lower group got the item right than in the upper group.
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Measures of attractiveness. To measure the attractiveness of the incorrect option in a multiple choice test, we count the number of the students who selected the incorrect option in both the upper and lower groups. The incorrect options should attract less of the upper group than the lower group.
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PREPARING DATA FOR ITEM ANALYSIS1. Arrange test scores from highest to lowest.2. Get one-third of the papers from highest scores and the other one-third from the lowest scores.3. Record separately the number of times each alternative was chosen by the students in both groups.4. Add the number of correct answer to each item made by the combined upper and lower groups.
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5. Compute the index of difficulty for each item,index of difficulty = No. of students responding correctly to an item x 100 Total no. of students in the upper and lower groups6. Compute the index of discrimination index of discrimination=Upperncr – Lowerncr No. of students per group
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Difficulty of a test item can be interpreted with the use of... Range Difficulty Level 20 & below very difficult 21-40 difficult 41-60 average 61-80 easy 81-above very easy
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Discrimination Index Range Verbal Description0.40 and above very good item0.30-0.39 good item0.20-0.29 fair item0.09-0.19 poor item
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CORRELATING TEST SCORESCORRELATION- the relationship between two or more paired-factors or two or more sets of tests scoresCORRELATION COEFFICIENT- a numerical measure of the linear relationship between two factors on sets of scores
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Obtained Correlation coefficientcan be interpreted with the use of….Correlation Coefficient Degree of Relationship 0.00-0.20 negligible 0.21-0.40 low 0.41-0.60 moderate 0.61-0.80 substantial 0.81-1.00 high to very high
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Pearson’s Product-Moment Correlation1. Compute the sum of each set of scores (SX.SY).2. Square each score and sum the squares (SX2 ,SY2 ).3. Count the number of scores in each group (N).4. Multiply each X score by its corresponding Y score.5. Sum the cross product of X and Y (SXY).6. Calculate the correlation, following the formula:
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Spearman Rho1. Rank the scores in distribution X, giving the highest score a rank of 1.2. Repeat the process for the scores in distribution Y.3. Obtain the difference between the two sets of ranks (D).4. Square each of these differences and sum up squared differences (SD2 )
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5. Solve for Rho following the formula: Rho=1-{6 SD2 } {N3 –N}Where: rho= rank- order correlation coefficient D= difference between paired ranks SD2 = sum of squared differences between paired ranks N= No. of paired ranks
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Organizing Test Scores for Statistical Analysis1.Organizing test scores by ordering2.Organizing test scores by ranking3.Organizing test scores through a stem- and leaf plot4.Organizing data by means of a frequency distribution
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Preparing Single Value Frequency Distribution1. Arrange the scores in descending order. List them in the X column of the table.2. Tally each score in the tally column.3. Add the tally marks at the end of each row. Write the sum in the frequency column.4. Sum up all the row total tally marks (N=___).
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Shapes of the frequency Polygons1. Normal- bell- shaped curve.2. Positive skewed- most scores are below the mean and there are extremely high scores. (mean is greater than the mode)3. Negatively skewed- most scores are above the mean and there are extremely low scores. (mean is lower than the mode).
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4. Leptokurtic- highly peaked and the tails are more elevated above the baseline.5. Mesokurtic- moderately peaked6. Platykurtic- flattened peak7. Bimodal curve- curve with two peaks or mode.
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8. Polymodal curve- curve with three or more modes9. Rectangular Distribution- there is no mode
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Skewness- degree of symmetry of the scores kurtosis – degree of peakness or flatness of the distribution curve
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Sk= 3( M –Md) SDK= Q (P90 – P10)• Normal distribution – 0.263• Platykurtic - > 0.263• Leptokurtic - < 0.263
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Organizing Test Scoresfor Statistical Analysis
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Organizing Test Scores By Ordering Ordering refers to the numerical arrangement of numerical observations or measurements.There are two ways of ordering:1. Ascending Order2. Descending Order
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the following are the scores obtained by 10 students in their quizzes in English for the first grading students.A B C D E F G H I J110 130 90 140 85 87 115 125 95 135
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Organizing test scores by ranking Ranking is another way by which test scores can be organized. It is process of determining the relative position of scores, measures of values based on magnitude, worth, quality, or importance,
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Steps in ranking test scores: Arrange the test scores from highest to lowest Assign serial number for each score. Assign the rank of 1 to the highest score and the lowest rank to the lowest score. In case there are ties, get the average of the serial numbers of the tied scores. R= ( SN1 + SN2 + SN3 .... SN N)
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Example: Rank the following scores obtained by 20 ist year high school students in spelling. 15 the rank of 12, 8,7, and Find 14 10 9 8 8 7 6 2 4 4 8 7 8 10 9 14 12 4 6
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Organizing Test Scores Through A stem and Leaf Plot It is a method of graphically sorting and arranging data to reveal its distribution. It is a method of organizing a scores, a numerical score is separated into two parts, usually the first one or two digits and the other digits. The stem is the first leading digit of the scores while the trailing digit is the leaf
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Procedures: Split each numerical score or value into two sets of digit. The first or leading set of digits is the stem, and the second or trailing set of digits is the leaf List all possible stem digits from lowest to highest. For each score in the mass of data, write down the leaf numbers on the line labelled by the appropriate stem number
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Illustrate the stem and leaf plot on the following periodical test results in biology.30 74 80 57 3231 77 82 59 9033 46 65 49 9242 50 68 48 57
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Organizing Data by means of frequency distribution Preparing Single value Frequency Distribution1. Arrange the scores in descending order. List them in the x column of the table.2. Tally each score in the tally column.3. Add the tally marks at the end of each row. Write down the sum in the frequency column.4. Sum up all the row total tally marks
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Prepare a single value frequency distribution for the spelling test scores of grade 3 pupils14 2 6 8 8 6 6 9 8 64 2 14 9 4 6 2 4 14 45 6 3 6 6 10 10 4 3 8
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Preparing Group Frequency Distribution Steps Find the lowest and the highest score. Compute the range. Determine the class interval Determine the score at which the lowest interval should begin. Record the limits of all class interval Tally the raw scores in the appropriate class interval Convert each tally to frequency.
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Setting the class boundaries and class limits Class boundary is the integral limit of a class. These integral limit should be apparent or real. • The apparent limits of a class are comprised of an upper and lower limit Class mark is the midpoint of a class in a grouped frequency distribution. • It is used when the potential score is to be represented by one value if other measures are to be calculated
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Derived Frequencies From Grouped Frequency Distribution Relative frequency distribution indicates what percent of scores falls within each of the classes. RF = ( F/N) 100 .
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Computation of relative frequency Class frequenc Relativeinterval y Frequen cy 75-77 1 2.5 72-74 3 7.5 69-71 5 27.5 66-68 12 30 63-65 11 25.5 60-62 8 20 40 100
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Cumulative Frequency distribution indicates the number of scores that lie above or below a class boundaryTypes:1. <cf- are obtained by adding the successive frequencies from the bottom to the top of the distribution2. >cf- are obtained by adding the frequencies from top to bottom
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1. MEAN It is often called arithmetic average.
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2. Median It is the score that occurs at a point on the scale below which 50 % of the scores fall and above which the other 50 % of the scores occur.
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3. Mode It is the most recurring score in a set of test scores
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Measure of Dispersion To determine the size of the distribution of the test scores or the portion of it.
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Range It is the simplest and the easiest measure of dispersion. It simply measure how far the highest score from the lowest score It is considered as the least satisfactory measure of dispersion For ungrouped data we have: R= Hs - Ls
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Example Determine the range of the test score of nine students in a community development course test.Sol: R = 43-19 = 24
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Compute the range of the following frequency distribution of the test scores in Math Class interval Frequency 60-64 1 55-59 5 50-54 4 45-49 5 40-44 7 35-39 8 30-34 4 25-29 3 20-24 2 15-19 1
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Compute the inter quartile range ofthe following frequency distribution of the test scores in Math Class interval Frequency 60-64 1 55-59 5 50-54 4 45-49 5 40-44 7 35-39 8 30-34 4 25-29 3 20-24 2 15-19 1
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The quartile Deviation It devides the difference of the 3rd and 1st quartile into two. It is the average distance from the median to the two quartiles QD = Q3- Q2 2
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Example Determine the quartile deviation of the test score of nine students in a community development course test.Sol: 15 / 2 = 7.5
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Compute the quartile deviation ofthe following frequency distribution of the test scores in Math Class interval Frequency 60-64 1 55-59 5 50-54 4 45-49 5 40-44 7 35-39 8 30-34 4 25-29 3 20-24 2 15-19 1
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