Learning Statistics

Grouped Frequency Distribution of Interval Data

To organize the interval- level scores into a grouped frequency distributions, we condense the separate scores into a fewer number of categories or groups. Each group contains more than one score value, called the class interval. The class interval contains the number of score values.

Examples: Given the following scores is a psychology test, make a frequency table. Raw Scores

Step 1. Compute the range. In the given data, the highest score is 82 and the lowest score is 28. The range (H – L) is 82 – 28 = 54 Step 2. Divide the range by 10 to 15 to determine the acceptable size of the interval. Let the lowest interval begin with a number which is a multiple of the interval size. In applying this step, we should decide how large each of the interval is going to be. Generally, the most acceptable number of intervals is as few as 5 to as many as 20. In estimating the number of intervals, a trial and error fashion is suggested. If we try 2 as the size of our class interval, we reject this because it results in too many categories of intervals.

If we try 12 as the size of our class interval, we also reject this because there are only 4 categories of intervals. The number 4 is not within the acceptable range of the number of class intervals or categories of class intervals. Since we need between 5 and 20 categories of intervals in our frequency distribution, we can take 10 to 15 and divide this into the range. In this case we may divide 54 by 10,11,12,13,14, and 15. If we select 10, it results in a value of 5.4 or 5. This value could be used as our interval size.

As we may have observed, most frequency distributions have odd numbers as the size of the interval. The advantage of using old numbers is that, the midpoints of the intervals will be whole numbers.

Step 3. Organize the class interval. See to it that the lowest interval begins with a number that is a multiple of the interval size. Since the lowest score is 28 and the size of your interval is 5, the lowest interval would begin with 25 and end at 29. These are the interval limits. You take note that the upper and lower limits (the exact or real limits)

Step 4. Tally each score to the category of class interval it belongs to

Step 5. Count the tally column and summarize it under column f. Then add your frequency which is the total number of cases (N)

Step 6. Compute the midpoint (M) for each class interval and put it under column M. You can obtain the midpoint by the formula below.Step 6. Compute the midpoint (M) for each class interval and put it under column M. You can obtain the midpoint by the formula below.

Step 6. Compute the midpoint (M) for each class interval and put it under column M. Where: M = the midpoint LS = the lowest score in the class interval HS = the highest score in the class interval

Step 7. Compute cumulative distributions for “less than” and “greater than”. Then put them under column “less than” cumulative and “more than” cumulative distribution. Cumulative frequencies can be obtained by adding the frequency for any class interval or category to the total frequency for all categories above and below it.

From the given less than (CF<) and greater than (CF>) cumulative frequency distribution, we can interpret that there is only 1 student whose score is less than 30, just as there are 4 students who scored less than 40. Those who scored less than a given score is evident in the CF less than column. Also, you can interpret this as that only 1 student has a score which is greater than 79, 2 students obtained a score that is greater than 74, and so forth.

Step 8. Compute the relative frequency distribution. Where: RF = the relative frequency CF = the class frequency TF = the total frequency

(Back to the Top)** Exercise:**

Below are scores in Statistics examination. Convert the following distribution of scores into a grouped frequency table and (a) determine the size of class intervals, (b) indicate the upper and lower limits of each class interval, (c) identify the midpoints of each class interval, (d) find the cumulative frequency “less than” and “greater than” for each class interval and (e) find the relative frequency for each class interval

(Back to the Top)

**Presentation and Analysis**

The table reveals the respondents’ length of service. From among the school heads, 5 or 20.83% had served the system from 31-40 years; likewise, 11 or 45.83 % served for 21-30 years, and another 5 0r 20.83% served from 11-20 years while 3 or 12.5% served from 1-10 years.

From the teacher-participants, Nine (9) or 13.24 % had served for 21-30 years while twenty-seven (27) or 39.71% had served from 11-20 years and the remaining thirty-two (32) or 47.05 had served from 1-10 years.

In general, school heads had a mean of 23 years of service while teachers had a mean of 12 years.

**Interpretation**

It could be gleaned that if compared with the length of service of the school head-participants, others could have made it in the school heads’ positions. But it should be understood that length of service or seniority does not necessarily mean an automatic qualification for headship position.

It could be said that the trend is to place qualified teachers as school heads regardless of the length of service. It could be said also that other factors other than length of service or eligibility or educational attainment are necessary to weigh the credentials of those who would like to assume higher positions than a classroom teacher.

(Back to the Top)