The document discusses various methods for organizing and analyzing test score data, including: 1) Organizing scores in ascending or descending order. Ranking scores from highest to lowest. 2) Creating a stem-and-leaf plot to separate scores into "stems" and "leaves". 3) Calculating measures of central tendency (mean, median, mode) and using frequency distributions to analyze grouped score data.

1. ORGANIZING TEST SCORES FOR STATISTICAL ANALYSIS Organizing Test Scores By Ordering Ordering refers to the numerical arrangement of numerical observations or measurements. 2 Ways by which test scores can be arranged: Ascending order – from lowest to highest Descending order – from highest to lowest.

2. Example The following are scores obtained by 10 students in their quizzes in Math for the first grading period: 110, 130, 90, 140, 85, 115, 125, 95, 135, 100

3. Organizing Test Scores By Ranking Ranking is the process of determining the relative position of scores, measures or values based on magnitude, worth, quality, or importance.

4. Steps to Rank Test Scores Arrange the scores from highest to lowest. Assign serial numbers for each score. The last serial number has to correspond to the total number of scores arranged in descending order. 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

5. R = (SN 1 + SN 2 + SN k ... ) NTS Where: R = rank SN 1 = serial number of the first score SN 2 = serial number of the second score SN k = other serial number NTS = number of tied scores

6. Example Rank the following scores obtained by 20 first year high school students in spelling: 15 14 10 9 8 8 7 6 2 4 8 7 8 10 9 14 12 4 4 6

7. Organizing Test Scores through a Stem-and-Leaf Plot In this method of organizing scores, a numerical score is separated into two parts, the stem which is the first leading digit of the scores, while the trailing digit is the leaf .

8. Procedures Split each numerical score or value into two sets of digits. 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 labeled by the appropriate stem number.

9. Example Construct a stem-and-leaf plot of the following periodic test results in Biology 30 74 80 57 32 40 55 59 31 77 82 59 90 51 54 62 33 46 65 49 92 69 66 41 42 50 68 48 57 71 68 59 60 53 49 63 53 44 81 68

10. STEM LEAF 3 4 5 6 7 8 9 0 1 2 3 0 1 2 4 6 8 9 9 0 1 3 3 4 5 7 7 9 9 9 0 2 3 5 6 8 8 8 9 1 4 7 0 1 2 0 2

11. Organizing Data by Means of a Frequency Distribution Frequency Distribution is table showing the number of times a score occurs. 2 Types of Frequency Distribution Single Value Grouped Frequency Distribution

12. Preparing Single Value Frequency Distribution Steps Arrange the scores in descending order. List them in the x column of the table. Tally each score in the tally column Add the tally marks at the end of each row. Write the sum in the frequency column. Sum up all the row total tally mark (n = ___)

13. Example Prepare a single values frequency distribution for the spelling test of First High School students given below. 14 8 8 6 14 6 14 6 6 4 2 6 6 4 9 2 4 3 10 10 6 9 8 2 4 4 5 6 3 8

14. Preparing Grouped Frequency Distribution Find the lowest and highest score. Compute the range. Determine the class size and class interval. Desired number of classes is from 10 to 15. Determine the score at which the lowest interval should begin. It should be multiple of the class size. Record the limits of all class intervals, with the highest score value at the top. Tally raw score in the appropriate class intervals. Convert each tally to frequency.

15. Prepare a grouped frequency distribution for the following scores obtained by 50 students in a periodical test in Mathematics 84 80 68 87 86 70 79 90 67 80 82 62 85 86 86 61 86 87 91 78 72 96 89 84 78 88 78 78 82 76 70 86 85 88 70 79 75 89 73 86 72 68 82 89 81 69 77 81 77 83

16. Setting Class Boundaries and Class Marks Class boundary is the integral limit of a class. The apparent limits of a class are comprised of an upper and lower limit. The real or exact limits of scores in a class extend from one-half of the smallest unit of the measurement below the value of the score to one-half unit above.

17. Class mark is the midpoint of a class in grouped frequency distribution. It is used when the potential score is to be represented by one value if other measures are to be calculated. CM = (LL + UL) 2

18. Derived Frequencies from Grouped frequency Distribution 3 Types of Frequency Distributions that may derived from a frequency distribution table. Relative frequency distribution indicates what percent of scores fall within each of the classes. RF = ( F / N ) 100

19. Cumulative frequency distribution indicates the number of scores that lie above or below a class boundary. 2 Types Less than cumulative frequencies are obtained by adding the successive frequencies from the bottom to the top of the distribution. Greater than cumulative frequencies are calculated by adding the successive frequencies from the top to the bottom of the distribution.

20. Exercises 1. Rank the following test scores. Indicate on the space opposite each score its appropriate rank. 83 87 91 80 64 83 75 90 98 86 72 71 79 80 65 84 83 93 90 68 2. Construct a stem-and-leaf plot for the test scores above.

21. A group of 24 high students who took the entrance examination test obtained the following scores on numerical ability test: 26 21 29 32 24 17 23 29 17 20 26 23 21 7 28 25 14 23 18 16 18 31 32 27 Prepare a single value frequency distribution of these entrance examination scores

22. Prepare a grouped frequency distribution for the scores obtained by 40 first year high school students in their unit test in Math . 68 65 71 64 69 66 66 60 71 67 67 63 66 65 69 74 66 64 62 62 65 63 67 66 72 68 62 70 63 60 67 62 72 64 68 65 61 61 64 75

23. Measures of Central Tendency - A measure of central tendency is a typical value or a representative value of a set of data. Example: Average grade The median score The most popular

24. Mean Referred to as the average For a given set of data, the mean M is the sum of all n values divided by the total frequency . M = Sx / n where, M = mean Sx = sum of test scores n = total number of test scores

25. 1 . Solve for the mean of 10 , 9, 8, 12, 15 2. Find the mean of the following test scores in Science: 15, 21, 16, 17, 20, 21, 17 and 19. 3. Calculate the mean of the following test scores: 45, 35, 20, 23, 43, 42, 40, 36

26. Exercise The teacher gave five tests in Math. Bea got the following scores in the 1 st four tests: 82, 76, 79, and 81. What must be her score in the 5 th test so that her average is 80?

27. The test scores of eleven students are shown below. Find their average score. 45, 50, 55, 38, 39, 2, 48, 53, 50, 40, 52

28. Disadvantage The mean is easily influenced by extreme values (very high or very low) What to do? - Set aside or disregard the extreme value & solve for the mean. - Do not use the mean. Use other measures

29. The Weighted Mean Incorporates into the formula the weight of each term. Example: Find the average grade of each of two students whose grades in five subjects are as follows: Subject Grade Units (1) (2) A 85 79 2 B 90 88 3 C 95 95 5 D 83 94 3 E 82 89 1

30. Weight as Percentage Example: The grading scheme in Statistics is 30% Quizzes, 10% R, 10% A/SW, 50% M.E. Find the grades of students A & B. Q R A/SW M.E. A 93 95 96 98 B 90 85 80 95

31. Weight as Fraction Example: A teacher gives one quiz and one long exam and would like to determine the weighted average of a student. The grades and the corresponding fractional weights are: Grade (x) Fractional Weight (W) Quiz 90 1 / 3 Exam 98 2 / 3

32. The Median The median (Md) is the middle value in a set of observations arranged from highest to lowest or vice versa. Example: Test scores of 15 junior students arranged from lowest to highest 14 14 15 15 15 16 16 17 17 17 18 18 19 20 20

33. Find the median of 8, 12, 5, 6, 13, and 15 Find the median of the set: 25, 28, 22, 20, 18, 23, 30, 24 Find the median score of 11 students in English test 6, 4, 9, 7, 3, 11, 12, 5, 10, 6, 8

34. The Mode The mode (Mo) is the observation which occurs most often in a set of values. The value with the highest frequency. Example: Find the mode of this set: 24 27 32 29 31 35 27 32 24 25 30 24

35. Find the mode of the following set of data: 6, 4, 9, 7, 3, 11, 12, 5, 10, 6, 8 Find the mode of the test scores of 15 students 15 15 15 14 15 14 19 19 17 16 14 18 20 20 19

36. Choosing the Measure of Central Tendency The mean is appropriate for interval and ratio variables. The mode is preferred for groups of data which do not tend to group around a central point. The median best measures the central tendency of groups which contain extreme values.

37. Mean For Grouped Test Scores Frequency-Class Mark Method Calculate the class mark of each class interval Multiply each class mark by its corresponding frequency Sum up the cross products of the class mark and frequency of each class Count the number of cases or total number of scores Plug into computational formula the values obtained in step 3 and 4.

38. Using the Deviation Method M = Sfcm / n where: M = mean f = frequency of a class cm = class mark n = total number of scores or cases Sfcm = sum of the cross products of the frequency and class mark

39. Finding the Mean for Grouped Test Scores Score Class Mark Frequency 75-79 70-74 65-69 60-64 55-59 50-54 45-49 40-44 35-39 30-34 77 72 67 62 57 52 47 42 37 32 1 4 2 5 7 9 11 8 4 3

40. Deviation Method of Solving the Mean M = AM + ( Sfd / n )i where M = mean AM = assumed mean f = class frequency d = class deviation score Sfd = sum of the cross products of class frequency and deviation score i = class size n = total number of scores

41. Finding the Median from Grouped Data Cumulate the class frequencies from the lowest to the highest interval Compute the locator of the class containing the median by dividing the total number of scores by 2 Locate the cumulative frequency (CF) that approximate n / 2 Locate the median class based on the CF in step 3. The median class is just above it

42. Determine the frequency of the median class (f) Determine the class boundary of the median class Determine the class size (i) Substitute the obtained values into the formula

43. Md = L + ( n / 2 – CF) I f where: L = lower class boundary of the median class n / 2 = locator of the median class n = total number of scores CF = cumulative frequency before the median class f = frequency of the median class i = class size

44. Compute the mean and median of the test scores of second year high school students in Filipino. Score Frequency 90 – 92 87 – 89 84 – 86 81 – 83 78 – 80 75 – 77 72 – 74 69 – 71 66 – 68 63 – 65 60 – 62 2 7 10 6 8 4 3 4 3 1 2

45. Mode for Grouped Data Mo = 3Md – 2M where: Mo = mode Md = median M = mean

46. The table below gives the score distribution of 90 students in a Statistics test Score Frequency 55 – 59 50 – 54 45 – 49 40 – 44 35 – 39 30 – 34 25 – 29 20 – 24 15 – 19 10 – 14 2 5 10 12 15 16 13 10 4 4