ORGANIZING THE DATA IN A FREQUENCY TABLE

Transcription

1 ORGANIZING THE DATA IN A FREQUENCY TABLE Suppose the scores obtained by 5 students on a standardized test are as follows: 68, 55, 61, 55, 43, 59, 55, 58, 77, 6, 56, 53, 58, 7, 57, 62, 5, 69, 44, 63, 48,79, 67, 56, 72, 47, 63, 59, 66, 57, 56, 73, 52, 61, 54, 63, 6, 74, 41, 53, 56, 69, 64, 57, 72, 51, 62, 59, 48, 57. You have to construct a frequency table for this dataset. This is a relatively smaller dataset since it has only 5 values. Therefore, we decide to have just five (i.e., c = 5 ) class intervals. The minimum of the values in the dataset is 41, and the maximum, 79. The range, R = max min = 79 41=38. Then, the class interval, H = R/ c = 38/5= 7.6. We can round up the value of H to 8 or 1. Since the value of 1 is more convenient to handle in calculations than 8, we let H = 1. So far, we have decided on the values of c and H, and now we need to pick a value for the lower boundary of the first class. Suppose we let the lower boundary of the first class be equal to 3. Then, the upper boundary of the first class would be 4 (note that U = L+ H for any class), and the minimum value does not fall in the first class. On the other hand, if we pick a value of 4 for the lower boundary of the first class, then, the maximum does not fall in the last class (verify this by listing all the five class intervals). So, a convenient value would be 35. Now, the minimum falls in the first class, and the maximum, in the last class. The next step is to scan each of the values in the given dataset, and place them in the appropriate class interval. To understand this classification process better, suppose that the 5 values in the dataset represent the lengths of 5 steel bars, and there are 5 storage bins labeled 35-45, 45-55, 55-65, 65-75, and You are asked to put each steel bar into one of the bins, depending on its length. Because of this analogy, some software systems refer to the class intervals as bins. For this reason, we will occasionally use the word bin to mean the same thing as the class interval. The first value in the dataset is 68, and it falls in the fourth class interval. We draw a vertical line under the tally marks column against the fourth class interval. Also, draw a line across the first value in the dataset, to indicate that it has been already scanned. The second value in the dataset is 55. Do we place it in the second bin or the third bin? This dilemma has arisen because we have chosen the system of overlapping class intervals, wherein the lower boundary of a class is equal to the upper boundary of the previous class. Of course, we can specify nonoverlapping intervals and avoid this dilemma. But, the variables we deal with are often continuous, and for theoretical reasons, it is better to retain the overlapping class intervals. Then, how do we resolve this dilemma? The first time you encounter a value of 55, place it in the second bin, the second time in the third bin, and the third time in the second bin and so on. In other words, we place the value of 55 alternately in the second and third bins. A similar rule applies for any other boundary value such as 65 or 75. We continue the scanning of the given values in this fashion. When the scanning of the first eleven values is completed, the dataset and the partially constructed frequency table should appear as follows: 1

2 68, 55, 61, 55, 43, 59, 55, 58, 77, 6, 56, 53, 58, 7, 57, 62, 5, 69, 44, 63, 48,79, 67, 56, 72, 47, 63, 59, 66, 57, 56, 73, 52, 61, 54, 63, 6, 74, 41, 53, 56, 69, 64, 57, 72, 51, 62, 59, 48, 57. Number Partial Frequency Table Interval Tally Marks L U When we scan all the 5 values, the completed frequency table appears as follows: Number Completed Frequency Table Interval Tally Marks Frequency L U We can now answer the fundamental question: how are the test scores distributed? From the table, we see that there are three students in the first class with scores between 35 and 45, eleven students in the second class with scores between 45 and 55, twenty four students in the third class with scores between 55 and 65, ten students in the fourth class with scores between 65 and 75, and two students in the fifth class with scores between 75 and 85. In other words, this table gives us the frequency distribution of the test scores (or lengths of the steel bars). In essence, it tells us how frequently an object (i.e., a steel bar or test score) falls into a bin or a class interval. 2

3 Instead of just 5 students, suppose that several thousands of students have taken the standardized test, and the pattern of their collective performance is similar to that of the 5 students observed here. We need to generalize the results obtained earlier so that we can draw useful conclusions about larger number of students. For this purpose, we define the relative frequency of a class as follows. Relative Frequency = Frequency/Total Frequency Total frequency is nothing but the total number of objects in the dataset (i,e., sum of all the class frequencies). In addition, we also calculate the cumulative relative frequency of each class. Cumulative Relative Frequency of a = Sum of the relative frequencies from the first class to the given class. Or Cumulative Relative Frequency of a = Cumulative Relative Frequency of the previous class + Relative Frequency of the given class. Further, Percentage = Relative Frequency * 1 Cumulative Percentage = Cumulative Relative Frequency * 1 The relative frequencies of the bins and related values are given in the following table. Number Relative Frequency Table Tally Marks Frequency Interval Relative Frequency L U (6%) (22%) (48%) (2%) (4%) Cumulative Relative Frequency.6 (6%).28 (28%).76 (76%).96 (96%) 1. (1%) The results obtained here are also illustrated in the following figures. It may be noted that we could have used the observed frequencies in place of the relative frequencies in drawing these graphs. (Can you calculate the values of the class cumulative frequencies?) 3

4 Histogram of Test Scores Relative Frequency Boundaries 3 Frequency Polygon 25 Frequency Midpoints 4

5 Cumulative Percentage Polygon 12 1 Cumulative Percentage Boundaries A histogram is nothing but a bar chart, where two consecutive bars share a common boundary (i.e., touch each other). The polygons are regular graphs, where the X-axis represents the test scores. In case of the frequency polygon, we plot the points at the midpoints of the class intervals. midpoint X = L+ U 2 In case of the cumulative percentage polygon, we plot the points at the class boundaries. Note that the sum of all relative frequencies must always be equal to 1.. Question: Suppose that 258, students have taken standardized test, and the distribution of their test scores is approximately the same as given in the Table above. Can you calculate the number of students in each of the five classes? 5

6 THREE THEORETICAL DISTRIBUTIONS 1. Introduction: A frequency table tells us how the objects are distributed among the classes. There are three well-known theoretical distributions often encountered in the analysis of continuous variables. 2. Uniform Distribution: Suppose that the monthly contributions of the members to the savings plan of a national union of workers follows uniform distribution. Further, assume that there are 3, members contributing to the plan every month. The class intervals and the class frequencies describing the distribution of the data are given in Table 1. The corresponding histogram is given in Figure 1. Table 1 Data on the Monthly Savings of Union Members L U f total Frequencies Intervals Figure 1. Histogram of the Monthly Contributions 6

7 3. Normal Distribution: Suppose that the diameter measurements of 4, silicon wafers follow normal distribution. The class intervals and the class frequencies describing the distribution of the diameter measurements are given in Table 2. The corresponding histogram is given in Figure 2. Table 2 Data on the diameters of silicon wafers L U f total Frequencies Intervals Figure 2. Histogram of the Diameter Measurements 7

8 4. Exponential Distribution: As a third example, suppose the failure times of an electronic component follow an exponential distribution. Further, assume that there are 1, points in the dataset. The class intervals and the class frequencies describing the distribution of the data are given in Table 3. The corresponding histogram is given in Figure 3. Table 3 Data on the life-times (failure-times) of an electronic component L U f Frequencies Intervals Figure 3. Histogram of Failure Times 8