Students Ability to Apply Different Mathematical Concepts - Term Paper Example

? Diana C. Davila Analysis of Test Results Critical Assignment 2 Richard Celestino, M.A. # EDF 3430-81043-3 Thursday 4:00PM-6:50PMEducational Measurement and Evaluation Following the assessment of the Numerical and Place value unit, students should be able to learn and become familiar with numeration and place value of numbers. To determine the learning level of the students, an assessment was devised that measured the students’ ability to apply different mathematical concepts to real world applications. The objectives after the completion of the course are to learn how to apply number analysis, knowledge, application and comprehension of mathematical concepts. This is to be done by comparing and ordering numbers, use of vocabulary, matching of standard forms of words, recognizing greater or lesser numbers, rounding off numbers, and drawing graphs. This assessment was conducted by having students complete a test that measured the qualities mentioned above. From an item analysis of the questions and answers, 10 questions were selected that tested all the items. The numbers of correct answers per student and per question are presented and an analysis of the answers is presented. The validity of the results is determined by calculating the Range, Rank, Mean, Median, Mode, Variance, and Standard Deviation of the test. This analysis should be used to determine the level of knowledge of each student and the level of difficulty of each question. After an analysis of the test results, a conclusion can then be made whether each test item should be retained, improved or removed from the unit syllabus. This analysis is presented at the end of this assessment. The Questions for Analysis Directions: Read each instruction carefully and give the correct answer. Numeration and Place Value Chapter Test Multiple Choice 1) Use the graph below. Circle the letter to show the best answer. How many dolphins and whales are in the aquarium? (A) 1 (C) 5 (B) 4(D) 6 Round to the nearest ten. Circle the letter to show the best answer. 2) 37 3) 75 4) 23 (A) 13 (A) 14 (A) 13 (B) 30 (B) 70 (B) 20 (C) 35 (C) 74 (C) 25 (D) 40(D) 80 (D) 30 Round to the nearest hundred. Circle the letter to show the best answer. 5) 440 6) 214 7) 304 (A) 400(A) 200 (A) 290 (B) 404 (B) 210 (B) 300 (C) 445 (C) 215 (C) 305 (D) 500 (D) 300 (D) 310 8) 504 (A) 500 (B) 505 (C) 510 (D) 600 9) Use the graph below. Circle the letter to show the best answer. How many seals and penguins are in the aquarium? (A) 1 (B) 3 (C) 8 (D) 9 10) Choose which of the following is correct. . (A) 85____>____106 (B) 64________46 Data Calculation and Analysis The test shown above was presented to 20 students, and the answers for each question per student are presented in the following table. Each correct answer by astudent is indicated by "1" and each incorrect answer is indicated by "0". The total number ofcorrect responses by each student is listed in the Total row. The numbers of correct responses per question are in the (+)column and the numbers of incorrect responses per question are in the (-)column. Table 1:Data from Test Assignment S1 S2 S3 S4 S5 S6 S7 S8 S9 S10 S11 S12 S13 S14 S15 S16 S17 S18 S19 S20 + - Q1 1 1 1 0 1 1 1 1 0 0 0 1 1 0 0 1 0 1 0 1 12 8 Q2 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 4 16 Q3 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 1 0 0 8 12 Q4 1 0 0 0 0 0 1 0 1 0 1 0 0 1 1 1 1 1 0 1 10 10 Q5 0 1 1 1 0 1 0 1 0 1 1 0 1 0 0 0 1 0 1 0 10 10 Q6 0 0 1 0 1 0 0 0 1 0 0 1 0 0 0 1 0 1 0 0 6 14 Q7 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 5 15 Q8 1 1 1 0 0 1 0 1 0 1 0 0 0 1 0 0 1 0 0 1 9 11 Q9 1 1 1 0 1 0 0 1 1 0 1 0 0 0 1 0 0 1 1 0 10 10 Q10 0 1 0 1 1 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 8 12 Total 5 6 5 4 5 3 5 5 3 3 4 3 4 3 3 4 4 6 3 4 82 118 Data analysis can be done using a number of descriptive measures, and the most common of these measures are the mean, mode, median, quartiles, range, and standard deviation. The calculation of the standard deviation will help in the determination of the level of difficulty of each question, and the basic statistics will help in the determination of the level of understanding of each student. The basic statistics for each student are shown in the table below, and they include the mean, mode, median, range, and rank. Table 2: Basic Statistics Student Score Rank Mean Median Mode Stdev Range S2 6 1.5 4.1 4 3 1.02 3 S18 6 1.5 4.1 4 3 1.02 3 S1 5 5 4.1 4 3 1.02 3 S3 5 5 4.1 4 3 1.02 3 S5 5 5 4.1 4 3 1.02 3 S7 5 5 4.1 4 3 1.02 3 S8 5 5 4.1 4 3 1.02 3 S4 4 10.5 4.1 4 3 1.02 3 S11 4 10.5 4.1 4 3 1.02 3 S13 4 10.5 4.1 4 3 1.02 3 S16 4 10.5 4.1 4 3 1.02 3 S17 4 10.5 4.1 4 3 1.02 3 S20 4 10.5 4.1 4 3 1.02 3 S6 3 17 4.1 4 3 1.02 3 S9 3 17 4.1 4 3 1.02 3 S10 3 17 4.1 4 3 1.02 3 S12 3 17 4.1 4 3 1.02 3 S14 3 17 4.1 4 3 1.02 3 S15 3 17 4.1 4 3 1.02 3 S19 3 17 4.1 4 3 1.02 3 The data provided in the above table is arranged from the highest-ranking student to the lowest ranked, where the rank indicates the total number of correct responses that a student had. For example, the highest ranked students, S2 and S18, had a score of 6 out of a possible 10 in the exam. The other students are distributed across the table following this characteristic. The distribution shows that there are two students with ranks of 1.5, five students with a rank of 5, six students with a rank of 10.5, and seven students with ranks of 17. After the scores are listed from the highest to the lowest, it is then possible to determine the range of the scores; the difference between the highest and lowest score. From the above table, the highest score is identified as 6 and the lowest as 3, therefore, the range is 3, as indicate above. The measures of central tendency calculated, the mean, mode, and range are used to determine the cluster of students around a central figure. The mean indicates the average number of correct responses that a student provides, meaning that it indicates the tendency of a student to get a correct response if a test is administered. This is calculated by dividing the total possible number of correct responses by the correct responses given by the student, and from the table, the highest mean is from the highest ranked student. For example, the mean is calculated as the total scores for the students divided by the number of students, Total scores = sum of all the scores for the students = 82 Number of students = 20 Therefore, mean = 82/20 = 4.1. The total mean for the distribution is found by dividing the total number of correct responses by the total number of students taking the test, which is 4.1. The other measure of central tendency is the median, which is identified by finding the middle score for the 20 students, which in this case is 4. The median is found by identifying the middle scores in the distribution table, which are the scores for students 10 and 11. Their scores are then added together and the answer divided by two. Scores = 4 and 4 Median = (4+4)/2 = 4. The mode is also found by identifying the commonest score for the distribution, and from the table, this figure is identified as 3. The mode is identified by counting out the number that appears the most times in the distribution table, and as identified above, the number is 3. The mode is found by identifying the umber that occurs most in the distribution, which in this case is 3. To determine the validity of the assessment, the three measures of central tendency; the mean, mode, and median should be clustered together round a common figure, and from the table, it is evident that this fact holds. Therefore, this test can be determined to be hard enough for the students. Variance The variance of the test scores is calculated using the test scores calculated above, which is indicated in the following table. Table 3: Variance Student Score Mean Difference Squared S2 6 4.1 1.9 3.61 S18 6 4.1 1.9 3.61 S1 5 4.1 0.9 0.81 S3 5 4.1 0.9 0.81 S5 5 4.1 0.9 0.81 S7 5 4.1 0.9 0.81 S8 5 4.1 0.9 0.81 S4 4 4.1 -0.1 0.01 S11 4 4.1 -0.1 0.01 S13 4 4.1 -0.1 0.01 S16 4 4.1 -0.1 0.01 S17 4 4.1 -0.1 0.01 S20 4 4.1 -0.1 0.01 S6 3 4.1 -1.1 1.21 S9 3 4.1 -1.1 1.21 S10 3 4.1 -1.1 1.21 S12 3 4.1 -1.1 1.21 S14 3 4.1 -1.1 1.21 S15 3 4.1 -1.1 1.21 S19 3 4.1 -1.1 1.21       Total 19.8       Variance= Total Squared/Number of scores = 19.8/20 0.99       Standard Deviation= Square root of variance = v0.99 0.99499 The variance for each student refers to the difference between the individual test scores and the class mean. This result is then squared to eliminate the negatives, and the summation of the squares divided by the number of scores gives the variance. The standard deviation is found by finding the square root of the variance, which indicates the average deviation of each student from the class mean. From the test scores identified, the variance is 0.99, therefore, the standard deviation is found to be 0.99. When the standard deviation is added to the mean, it gives a value of 5.09, and when one standard deviation is subtracted from the mean, the value is 3.11. An analysis of the scores by the students indicates that there are seven scores above the mean, and no scores between the mean and one standard deviation less. The total of the scores within one standard deviation of the mean is 7, which is 35% of the total number of students taking the test. The interpretation of these findings indicates that there are two scores higher than one standard deviation from the mean. The other indication is that the lowest score is 3, and since the difference between the mean and one standard deviation is 3.11, there are seven scores below one standard deviation from the mean. This indicates that there are more scores to the left of the distribution than the right, or that the mean is bigger than the median. This indicates that the distribution is right skewed, which indicates that the data is not symmetric about the mean. The significance of this is that the students do not have a similar learning ability; more students are expected to have lower scores than the ones who exceed the mean. A discrimination index can also be calculated from the basic statistics identified above, an index that determines whether the test achieves its objectivity in the paper. The discrimination index helps an examiner determine whether the test distinguishes between students who have learned the required unit objectives and students who have not acquired the objectives. The discrimination index can be calculated by comparing the scores from the low and high-scoring students. The index is calculated by identifying the top and lower quartiles for the data and calculating the number of correct responses per group. Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Top Scores S6 1 0 0 0 1 0 0 1 0 0 S9 0 0 0 1 0 1 0 0 1 0 S10 0 0 0 0 1 0 0 1 0 1 S12 1 0 1 0 0 1 0 0 0 0 S14 0 0 1 1 0 0 0 1 0 0 Total 2 0 2 2 2 2 0 3 1 1 Bottom Scores S18 1 0 1 1 0 1 1 0 1 0 S2 1 0 1 0 1 0 0 1 1 1 S8 1 1 0 0 1 0 0 1 1 0 S7 1 0 1 1 0 0 1 0 0 1 S5 1 0 1 0 0 1 0 0 1 1 Total 5 1 4 2 2 2 2 2 4 3 To calculate the discrimination index, the number of student responses is divided by the number of students in the group being considered, and the difference for each conversion, the discrimination index can be calculated. The ratings for the discrimination index are: Discrimination Index Rating Scale Less than 0.0 Major Flaw 0.0-0.19 Poor 0.2-0.39 Fair 0.4-0.59 Good Greater than 0.6 Excellent Table 5: Discrimination Index Question Top Five Conversion Bottom Five Conversion Discrimination Rating Q1 = 2/5= 0.4 = 5/5= 1 -0.6 Major Flaw Q2 = 0/5= 0 = 1/5 = 0.2 -0.2 Major Flaw Q3 = 2/5= 0.4 = 4/5= 0.8 -0.4 Major Flaw Q4 = 2/5= 0.4 = 2/5= 0.4 0 Major Flaw Q5 = 2/5= 0.4 = 2/5= 0.4 0 Major Flaw Q6 = 2/5= 0.4 = 2/5= 0.4 0 Major Flaw Q7 = 0/5= 0 = 2/5= 0.4 -0.4 Major Flaw Q8 = 3/5= 0.6 = 2/5= 0.4 0.2 Fair Q9 = 1/5 = 0.2 = 4/5= 0.8 -0.6 Major Flaw Q10 = 1/5 = 0.2 = 3/5= 0.6 -0.4 Major Flaw The discrimination index calculated above indicates that there are only one question is fair for the students, all the other questions are majorly flawed, therefore, they need t o be revised. An item difficulty index can also be calculated for each question, which is done by considering the correct number of answers per question. The number of correct responses is divided by the number of students to achieve the item difficulty score. The rating for this is shown as: Item Difficulty Index Less than 0.29 Too Difficult 0.3-0.39 Acceptable 0.4-0.49 Optimal More Than 0.5 Too Easy Table 6: Difficulty Rating Scale Question Conversion Rating Q1 = 12/20 = 0.6 Too Easy Q2 = 4/20 = 0.2 Too Difficult Q3 = 8/20 = 0.4 Optimal Q4 = 10/20 = 0.5 Too Easy Q5 = 10/20 = 0.5 Too Easy Q6 = 6/20 = 0.3 Acceptable Q7 = 5/20 = 0.25 Too Difficult Q8 = 9/20 = 0.45 Optimal Q9 = 10/20 = 0.5 Too Easy Q10 = 8/20 = 0.4 Optimal The difficulty-rating index indicates that 2 of the questions are too difficult, 1 question is acceptable, 3 questions are optimal, and 4 questions are too easy. Therefore, the too easy and too difficult questions have to be revised. From the item difficulty rating calculated, it is evident that 6 questions need to be revised, since they are either too easy or too difficult for the students. The too difficult questions need to be revised since they present a challenge to the students and do not give the students a chance to express what they have learnt. Conversely, the too easy questions do not reflect the learning that the students have undergone. This means that the students do not get to apply the learning objectives taken in the course. The easy questions are revised by adding some element of difficulty into them, while the difficult questions are simplified to some extent. Question Revisions Numeration and Place Value Chapter Test Multiple Choice 1) Use the graph below. Circle the letter to show the best answer. How many dolphins and whales are in the aquarium? (A) 1 (C) 5 (B) 4(D) 6 The question might be too easy for the students, since the numbers are already given, and there are only six numbers for the students to consider. This makes the student get a higher chance of getting the correct answer. Round to the nearest ten. Circle the letter to show the best answer. 2) 37 3) 75 4) 23 (A) 13 (A) 14 (A) 13 (B) 30 (B) 70 (B) 20 (C) 35 (C) 74 (C) 25 (D) 40(D) 80 (D) 30 The options presented might be confusing for the students, so the questions need to be revised. Round to the nearest hundred. Circle the letter to show the best answer. 5) 440 6) 214 7) 304 (A) 400(A) 200 (A) 290 (B) 404 (B) 210 (B) 300 (C) 445 (C) 215 (C) 305 (D) 500 (D) 300 (D) 310 Round to the nearest hundred. Circle the letter to show the best answer. 8) 504 (A) 500 (B) 505 (C) 510 (D) 600 9) Use the graph below. Circle the letter to show the best answer. How many seals and penguins are in the aquarium? (A) 1 (C) 8 (B) 3 (D) 9 10) Choose which of the following is correct. . (A) 85____>____106 (B) 64________46 Explain: The symbol always points to the smaller number. Discussion The assessment above covered four major aspects of a student’s learning, data analysis, comprehension, knowledge, and application. The analysis of the test scores for each student and each question indicated that some questions have to be revised to match applicable criteria, a factor that is identified by the item discrimination and difficulty indices calculated above. The analysis of the test results is done on a statistical basis, where the responses from each student are analyzed to determine the difficulty of each question. From an analysis of the test results, it is evident that the course is sufficient to teach the students, though some questions in the test have to be revised to meet required standards. As explained above, some questions have to be revised either to be challenging enough for the students or to be a little simpler. The main basis for the revision is got from the item analysis and the basic statistics calculated. The basic statistics indicate that it is more likely that a student would fail the exam or get a low score. This is indicated by the right-skewed nature of the distribution of the scores, which means that there are more scores to the left of the distribution. Basically, this means that more students would score a low figure in the test, which is an indication that some of the questions in the test have to be changed. The implication of the test scores identified above indicates that some questions have to be changed by the teachers and that the parents of the pupils need to teach their children when they are away from school. The other factor is that some of the questions were identified as too easy and yet some students failed the questions. Work Cited Black, K. Business Statistics for Contemporary Decision-Making, 2003. New York: Wiley. Print. Nitko, A. J., & Brookhart, S. M. (2012). Educational assessment of students. (6 ed.). Pearson. Triola, M. Elementary Statistics, 2003. Boston: Addison Wesley. Print. Read More