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Calculating the Median of ACE Results: A Continuous Data Approach, Study notes of Statistics

How to calculate the median of ace results when students may want to respond with a score between two whole numbers. A formula and a worked example to help understand the process.

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2021/2022

Uploaded on 09/12/2022

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Explanation of Median for ACE results
The ACE evaluation response options are the whole numbers from 1 6.
However, it is feasible that a student would want to respond with an answer between any
2 of these whole numbers. For example, a student may not want to endorse 6, but doesn’t
want to go as low as 5. Perhaps the student’s satisfaction is really 5.4. If we had an
infinitely precise instrument, we could actually determine this. However, we do not. To
help compensate for this we must treat this data as continuous and not discrete.
The median for continuous data requires more arithmetic. The formula for this
calculation is given below:
i
f
fbN
LMedian
m
2
where: L = lower limit of interval containing median
fb
= sum of frequencies below interval containing the median
fm = frequency of the interval containing the median
N = Number of students
i = size of the interval containing the median
The steps to using this formula are as follows:
1. Divide the number of cases by 2 to find the median score.
2. From the lowest interval, sum the frequencies until the interval is found that
contains the median score.
3. Subtract the sum of the frequencies below the interval containing the median
score from N/2
4. Divide the difference by the frequency of scores in the interval that contains the
median score.
5. Multiply the quotient by the size of each interval.
6. Add the product to the lower limit of the interval containing the median score.
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Explanation of Median for ACE results

The ACE evaluation response options are the whole numbers from 1 – 6. However, it is feasible that a student would want to respond with an answer between any 2 of these whole numbers. For example, a student may not want to endorse 6, but doesn’t want to go as low as 5. Perhaps the student’s satisfaction is really 5.4. If we had an infinitely precise instrument, we could actually determine this. However, we do not. To help compensate for this we must treat this data as continuous and not discrete.

The median for continuous data requires more arithmetic. The formula for this calculation is given below:

i f

N fb Median L m

where: L = lower limit of interval containing median

 fb^ = sum of frequencies below interval containing the median

fm = frequency of the interval containing the median N = Number of students i = size of the interval containing the median

The steps to using this formula are as follows:

  1. Divide the number of cases by 2 to find the median score.
  2. From the lowest interval, sum the frequencies until the interval is found that contains the median score.
  3. Subtract the sum of the frequencies below the interval containing the median score from N/
  4. Divide the difference by the frequency of scores in the interval that contains the median score.
  5. Multiply the quotient by the size of each interval.
  6. Add the product to the lower limit of the interval containing the median score.

Worked example:

An instructor receives these scores on an ACE question: {1, 1, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6}

This table summarizes the key pieces of data that are required:

Score Real Limit Frequency Relative Frequency

Cumulative Frequency

Relative Cumulative Frequency 6 5.5-6.5 2 0.095 21 1. 5 4.5-5.5 10 0.476 19 0. 4 3.5-4.5 3 0.143 9 0. 3 2.5-3.5 3 0.143 6 0. 2 1.5-2.5 1 0.048 3 0. 1 0.5-1.5 2 0.095 2 0.

The steps to using this formula are as follows:

  1. Divide the number of cases by 2 to find the median score.  The number of cases we have is N = 21. So, we divide 21 by 2.
  2. From the lowest interval, sum the frequencies until the interval is found that contains the median score.  Here, the interval will be that contains the median will be 4.5 – 5. because the rcf is ≥ 0.50.
  3. Subtract the sum of the frequencies below the interval containing the median score from N/  The sum of the frequencies below the interval containing the median are: 3 + 3 + 1 + 2 = 9
  4. Divide the difference by the frequency of scores in the interval that contains the median score.
  5. Multiply the quotient by the size of each interval.  Here this is 1.
  6. Add the product to the lower limit of the interval containing the median score.  Here the lower limit of the interval containing the median score is 4.5.

Median