Research on Accuracy of Mean Value in Frequency Table

How to calculate Mean Value in Frequency Table

Example Data:

10, 20, 30, 45, 50, 55, 60, 65, 70, 75, 85, 90, 95

Frequency Table of this data
#	Class	Class Value	Frequency	Class Value * Frequency
1	0 ~ 20	10	1	10
2	20 ~ 40	30	2	60
3	40 ~ 60	50	3	150
4	60 ~ 80	70	4	280
5	80 ~ 100	90	3	270
-	TOTAL	-	13	770
-	Mean	-	-	59.2

The method to calculate Mean Value in Frequency Table is as follows.

set Class Values.
set Class Value * Frequency in each Class.
take the sum of product of the Class Value and the Frequency
divide the sum by the Total of Frequency

Then the Mean Value of this Frequency Table results in $59.2$.

Difference from Actual Mean Value

However, the actual Mean Value of this data is calculated as follows.

$Mean$
$= (10 + 20 + 30 + 45 + 50 + 55 + 60 + 65 + 70 + 75 + 85 + 90 + 95) / 13$
$= 57.692307692308$
$≒ 57.7$

There's big difference.

$DIFF = 59.2 - 57.7 = 1.5$

$RATIO_{DIFF} = 1.5 / 57.7 ≒ 0.026 = 2.6\%$

Formula Deviation of Mean Value in Frequency Table

Let the Class Range be the variable $CR$.
Let the Data Range be the variable $DR$.
Let the Number of Classes be the variable $C$.
$$DR = C \times CR \; \Longleftrightarrow \; CR = \frac{DR}{C} \; \Longleftrightarrow \; C = \frac{DR}{CR}$$ Let the Class Value of $k$ th Class be $CV(k)　( 1 \leq k \leq C)$.
$$CV(k) = \frac{CR}{2} + CR \times (k - 1) = (k - \frac{1}{2})CR$$ Let the Frequency of $k$ th Class be $F(k)　( 1 \leq k \leq C)$.
Product of Class Value and Frequency of $k$ th Class is:
$$CV(k) \times F(k) = (k - \frac{1}{2})CR \times F(k)$$ Let the sum of these products be the variable $S$. $$S = \sum^{C}_{k=1}\{(k - \frac{1}{2})CR \times F(k)\} = CR\{\sum^{C}_{k=1}k - \frac{1}{2}\sum^{C}_{k=1}F(k)\}$$ Let the Total of Frequency be the variable $T$.
$$T = \sum^{C}_{k=1}F(k)$$ And,
$$\sum^{C}_{k=1}k = \frac{C(C + 1)}{2}$$ Then,
$$S = CR\{\frac{C(C + 1)}{2} - \frac{T}{2}\} = \frac{CR}{2}\{C(C + 1) - T\}$$ Let the Mean Value in Frequency Table be the variable $MF$.
$$MF = \frac{S}{T} = \frac{CR}{2T}\{C(C + 1) - T\}$$ Let the Number of Data be the variable $N$.
Let each data be the variable $D(k)　(1 \leq k \leq N)$.
Let the Actual Mean Value be the variable $AM$.
$$AM = \frac{1}{T}\sum^{N}_{k=1}D(k)$$ Let the difference of MF and AM be the variable $DIFF$.
$$DIFF = MF - AM = \frac{1}{T}[\frac{CR}{2}\{C(C + 1) - T\} - \sum^{N}_{k=1}D(k)]$$ Let the variable $R_{DIFF}$ be the ratio of $DIFF$ to $AM$.
$$R_{DIFF} = \frac{DIFF}{AM}$$ Let the variable $P_{DIFF}$ be the percentage of $DIFF$ to $AM$.
$$P_{DIFF} = R_{DIFF} \times 100$$ $DIFF$ and $P_{DIFF}$ have complicated behavior.

Examination1: Changing Class Range

The tool used for this examination is macocci7/php-frequency-table.
And PHP Code used for this examination is Examination1_01.php
The result data file is Examination1_01.csv

Gradually change the Class Range from $1$ to $0.001$ in steps of $-0.001$.

From this graph, we can see the $P_{DIFF}$ value tends to converge to $0\%$
as the Class Value approaches $0.001$.
When the Class Value is $0.223$ or less, the $P_{DIFF}$ is mostly with in $0.1\%$.
The only exception is when the Class Value is $0.125$, where $P_{DIFF} = 0.10833333333321\%$.

Examination2: Changing Number of Data

PHP Code used for this examination is Examination2_02.php
The result data file is Examination2_01.csv

Gradually change the Number of Data from $13$ to $1,000$.
Data Range is fixed at $85 ( 10 ～ 95 )$. Class Range is $20$.

$P_{DIFF}$ behaves in complex ways.
However, $P_{DIFF}$ still seems to oscillate around $0.59\%$.
The Mean Value of $P_{DIFF}$ in this case is $0.585637401$.
And $P_{DIFF}$ seems to converge to $0.59\%$.

Conclusion

I have only examined two cases this time, but I can still say something,
so I will draw a conclusion here.

Class Range Greatly affects the accuracy of Mean Value in Frequency Table.
If you want to increase the accuracy of Mean Value in Frequency Table, you should reduce the class value.
If doing so, however, there's no advantage of creating Frequency Table.
An increase in the number of data supresses the oscillation of the error.
However, the error converges to a constant numerical value that is not zero.
Therefore, it cannot be said that an increase in the number of data contributes to the accuracy of the Mean Value in Frequency Table.
Mean Value in Frequency Table is only an approximation.

Document created: 2023/05/29
Document updated: 2023/06/05