Chapter 2.12 - Relative Error

In the previous section, we saw the different types of errors that can occur while taking measurements. In this section, we will see how the 'quantity of errors' can be calculated. In other words, we will see the method to calculate 'how much error occurred', when the experiment was conducted

1. Suppose that, the length of a cylinder is being measured
• Let n trials be done
    ♦ In the first trial, length obtained is a₁ cm 
    ♦ In the second trial, length obtained is a₂ cm 
    ♦ In the third trial, length obtained is a₃ cm
    ♦ . . . .
    ♦ . . . .
    ♦ In the n^th trial, length obtained is a_n cm
2. Each of the above measurements may be different from one another
• We cannot say which among the n values, is the true length
3. In such a situation, the ‘mean of the n values’ will be very close to the true value
• But how can we prove that 'mean is indeed very close to the true value'?
4. For proving it, let us recall the 'basics about mean (average)' that we learned in our previous classes:
• Consider a group of seven students. It is decided to make uniforms for them
• Their heights are: 142 cm, 141 cm, 138 cm , 145 cm, 139 cm, 143 cm and 144 cm
• How much clothes should be bought?
Solution:
(i) Sum of the heights works out to 992 cm
(ii) If we divide this sum by 7, we will get the average height
• So average height = ⁹⁹²⁄₇ = 142 cm
(iii) Assume that all 7 students have a height of 142 m
• Let 2 m of cloth be required for a student of 142 cm height
(iv) Then for the 7 students, (2 × 7) = 14 m of cloth will be required
• This is an easy method for calculating the total quantity of cloth, especially if the number of students in the group or class is large
(v) But doubts arise:
• The student of 138 cm height do not need 2 m of cloth
    ♦ Some cloth will go as waste
    ♦ Same is the case with students of heights less than 142 cm
• Similarly, the student of 145 cm will need more than 2 m
    ♦ His uniform will be incomplete
    ♦ Same is the case with students of heights more than 142 cm
(vi) But in reality, such a wastage or insufficiency do not occur. The reason can be explained using the schematic diagram in fig.2.25 below:

Fig.2.25

• In fig.a, all students stand in ascending order of their heights
• The fourth student have the average height of 142 cm. He is shown in yellow color in fig.b
• The average height is marked by a cyan line in fig.c
• The white rectangles represents the wastage caused by the first 3 students
    ♦ They are below the cyan line
• The magenta rectangles represent the insufficiency suffered by the last 3 students
    ♦ They are above the cyan line
• If the mean is calculated accurately, the wastage will be very nearly equal to the insufficiency
• That is., the total heights of the white rectangles will be equal to the total heights of the magenta rectangles
• In short, the negatives will be cancelled by the positives. This is the advantage of using the mean value
5. When we calculate the mean of the measured values, a similar cancellation occurs. Let us see how:
(i) We have n number of readings
(ii) Half of those readings will be greater than the true value
    ♦ The errors in this case will be all positive
    ♦ Because, Error = Measured value - True value
    ♦ The errors of such over estimated readings are represented by the magenta rectangles in fig.2.25 above 
(iii) The other half will be lesser than the true value
    ♦ The errors in this case will be all negative
    ♦ Because, Error = Measured value - True value
    ♦ The errors of such under estimated readings are represented by the white rectangles in fig.2.25 above
(iv) How can we be sure that exactly half readings are over estimates and the other half are under estimates?
• The answer is that, it is an assumption
• It is a reasonable assumption. Because, 'likely hood of over estimating' is same as the 'likely hood of under estimating'
• Nobody would make all readings greater than the true value  
• Also, nobody would make all readings lesser than the true value  
• Every body wants the readings to be as accurate as possible
• The 'possibility for making positive errors' is same as the 'possibility for making negative errors'
6. So we can confirm that, the mean value is very close to the true value
• We know the formula for calculating the mean value: $\mathbf\small{a_{mean}=\frac{a_1+a_2+a_3+\;.\;.\;.\;+\;a_n}{n}}$
• In short form, this formula is written as: $\mathbf\small{a_{mean}=\frac{\sum\limits_{i=1}^{i=n}{a_i} }{n}}$
7. So we have obtained the mean value. The mean value will be very close to the true value
• So we can now find 'how much error was made in each reading'
• Let 𝚫a₁ be the error made in the first reading
• We have: Error = Measured value - True value
    ♦ So 𝚫a₁ = a₁ - a_mean 
    ♦ 𝚫a₂ = a₂ - a_mean 
    ♦ 𝚫a₃ = a₃ - a_mean
    ♦ . . . .
    ♦ . . . .
    ♦ 𝚫a_n = a_n - a_mean.
8. The above 𝚫a values will be +ve in some cases and -ve in the rest of the cases
• For our present discussion, we do not need the sign
• We need only 'how much error' occurred in each reading 
• So we take absolute values:
    ♦ |𝚫a₁| = |(a₁ - a_mean)|
    ♦ |𝚫a₂| = |(a₂ - a_mean)|
    ♦ |𝚫a₃| = |(a₃ - a_mean)|
    ♦ . . . .
    ♦ . . . .
    ♦ |𝚫a_n| = |(a_n - a_mean)|
9. The mean of the above absolute error values will give us the final absolute error or the mean absolute error
• It is denoted as 𝚫a_mean
• This 𝚫a_mean can be calculated using the same method: $\mathbf\small{\Delta a_{mean}=\frac{|\Delta a_1|+|\Delta a_2|+|\Delta a_3|+\;.\;.\;.\;+\;|\Delta a_n|}{n}}$
• In short form, this formula is written as: $\mathbf\small{\Delta a_{mean}=\frac{\sum\limits_{i=1}^{i=n}{|\Delta a_i|} }{n}}$
10. So now we have two quantities: a_mean and 𝚫a_mean.
■ We must clearly understand the difference between the two
• a_mean is the value which is very close to the true value
• 𝚫a_mean is an indication of 'how much error' occurred when the experiment was conducted
11. So the true value will be within the range: $\mathbf\small{a_{\rm mean}\pm \Delta a_{\rm mean}}$
• That is., (a_mean - 𝚫a_mean) ≤ true value ≤ (a_mean + 𝚫a_mean)

Relative error or Percentage error

1. Instead of mean absolute error, we often use the relative error or percentage error
• Relative error is a ratio. It can be obtained using the relation:
Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
2. This ratio helps us to compare two quantities:
(i) The error (ii) a_mean (which is close to the true value)
3. If the error is very small (when compared to the true value), we will get a small ratio    
• In that case, the error can be considered as negligible
4. If the error is large (when compared to the true value), we will get a large ratio
• In that case, the error cannot be considered as negligible
An example is given below. It is written in 3 steps:
(i) Consider the experiment to find the distance between earth and a star
(ii) If the error made is a few kilometers, it is negligible. We will get a small relative error
(iii) If the error made is thousands of kilometers, it is not negligible. We will get a large relative error
5. Once we calculate the relative error, we can easily convert it into percentage format
• All we need to do is: multiply by 100
• The result thus obtained is called the percentage error. It is denoted as $\mathbf\small{\delta a}$
• So we get: $\mathbf\small{\delta a=\left(\frac{\Delta a_{mean}}{a_{mean}}\right)\times 100 \text{%}}$

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Now we will see two solved examples

Solved example 2.19 
We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s. Calculate the absolute errors, relative error or percentage error
Solution:
The calculations are done in the table below:

1. We get: a_mean = 2.624 s
• This is written in column 3
• This result is got by addition of various numbers
• So the result must not have more decimal places than the least in the given data
• Thus, the result cannot have more than 2 decimal places
• Rounding off, we get: a_mean = 2.62 s
(After addition, we divide the sum by (n = 5). But n is a factor. It has infinite number of significant numbers)   
2. The absolute errors are calculated in the fourth column
• We get: Mean absolute error 𝚫a_mean = 0.107 s
• This is also obtained by addition. Addition of numbers having two decimal places
• So rounding off, we get: 𝚫a_mean = 0.11 s 
3. Now we can write three statements about that pendulum:
(i) The true value of the time period of that pendulum is very close to 2.62 s
(ii) The mean absolute error is 0.11 s
(iii) There is a probability that, the actual time period lies within the range:
(2.62 - 0.11) to (2.62 + 0.11)
• That is., 2.51 ≤ actual time period ≤ 2.73 s
4. Calculation of relative error
• We have: Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
• Substituting the values, we get:
Relative error = $\mathbf\small{\frac{0.11}{2.62}=0.04}$
• Percentage error = Relative error ×100 = 0.04 × 100 = 4%

Solved example 2.20
Diameter of a wire as measured by a screw gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm. Calculate
(i) Mean value of diameter
(ii) absolute error in each measurement
(iii) Mean absolute error
(iv) relative error
(v) Percentage error
(vi) Express the result in terms of percentage error
Solution:
The calculations are done in the table below:

1. We get: a_mean = 2.626 cm
• This is written in column 3
• This is the answer for part (i) 
2. The absolute error of each measurement is written in column 4
• These are the answers for part (ii)
3. The mean absolute error is written in column 5
• This is the answer for part (iii)
4. Calculation of relative error
• We have: Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
• Substituting the values, we get:
Relative error = $\mathbf\small{\frac{0.003}{2.626}=0.001}$
• This is the answer for part (iv)
5. Percentage error = Relative error ×100 = 0.001 × 100 = 0.1%
• This is the answer for part (v)
6. Diameter of the wire can be written as:
d = 2.626 cm ± 0.1%
• This is the answer for part (vi)

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In the next section, we will see combination of errors

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Chapter 2.6 - Accuracy and Precision

In the previous section, we completed a discussion on length. In this section, we will see accuracy and precision.

Accuracy and precision are not related to each other. They are two different items.
We will learn about them by analyzing an example. We will write the analysis in steps:
1. In an experiment, the length of a cylinder is to be determined.
• The following 5 readings were taken by student A:
5.25 cm, 5.19 cm, 5.24 cm, 5.17 cm, 5.27 cm.
• The following 5 readings were taken by student B:
5.32 cm, 5.29 cm, 5.31 cm, 5.30 cm, 5.31 cm.
■ We want to know two things:
(a) Whose readings are more accurate?
(b) Whose readings are more precise?
2. First we will consider question (a).
• To find the answer to that question, the ‘true value’ is a prerequisite.
• That is., 'actual length' of the cylinder should be already known.
• Let this actual length be 5.23 cm.
3. Now we take each reading of student A.
• The difference (deviation) between the true value and the first reading = |(5.23-5.25)| = 0.02.
• Note that, we want the difference only. The sign is not required. That is why we are taking the absolute value of the difference.
• We can call it: Absolute deviation.
    ♦ So absolute deviation of first reading = 0.02.
• The difference between the true value and the second reading = |(5.23-5.19)| = 0.04.
    ♦ That is., absolute deviation of second reading = 0.04.
• The difference between the true value and the third reading = |(5.23-5.24)| = 0.01.
    ♦ That is., absolute deviation of third reading = 0.01.
• The difference between the true value and the fourth reading = |(5.23-5.17)| = 0.06.
    ♦ That is., absolute deviation of fourth reading = 0.06.
• The difference between the true value and the fifth reading = |(5.23-5.27)| = 0.04.
    ♦ That is., absolute deviation of fifth reading = 0.04.
4. The above deviations gives us an important information:
• The fourth reading deviated the most from the true value.
• The third reading deviated the least from the true value.
5. However, all deviations are important. We must calculate their mean. This can be conveniently done in a tabular form as shown in table 2.1 below:

Table 2.1

• The mean of the absolute deviations is denoted as $\mathbf\small{|\Delta a_m|}$
• It can be obtained using the familiar expression for mean (average): $\mathbf\small{|\Delta a_m|=\frac{\sum{|\Delta a|}}{n}}$
• In our present problem, there are 5 readings. So n = 5.
• Thus we get: $\mathbf\small{|\Delta a_m|}$ = 0.034.
6. In a similar way, we must calculate $\mathbf\small{|\Delta a_m|}$ for the readings taken by student B.
The table is shown below:

Table 2.2

7. Now we compare A and B. We see that:
• The readings taken by student A has a mean deviation of 0.034 from the true value.
• The readings taken by student B has a mean deviation of 0.076 from the true value.
8. The readings taken by student A deviates less from the true value.
• 'Less deviation' means 'greater closeness'.
• We can write:
The readings taken by student A is closer to the true value.
• In other words:
The readings taken by student A is more accurate than those taken by student B.
9. A graphical representation will give us a better understanding.
(i) Consider fig.2.9 below:

Fig.2.9

• Imagine that, the students A and B are engaged in an archery competition.
(ii) Both students try to hit the yellow triangle. It is at the true value, which is 5.23.
• The arrows shot by A are indicated by the magenta circles.
• The arrows shot by B are indicated by the cyan circles.
(Note that, for student B, the reading 2.31 occurs twice. So there are two circles, one below the other at 2.31. Thus we see only a total of 4 cyan circles).
(iii) We see this:
• Student A manages to shoot some arrows close to the triangle.
• But arrows shot by B are no where near the triangle.
(iv) We say that, A has 'more closeness to the true value'.
• We need a mathematical tool to express this 'more closeness to true value'.
(v) The tables 2.1 and 2.2 taken together, provides an excellent mathematical tool for this purpose.
• Table 2.1 gives the deviation as 0.034.
• Table 2.2 gives the deviation as 0.076.
• So A has a lesser deviation. 'Lesser deviation' means 'greater closeness'.
• So A has 'more closeness'. That is., readings taken by A are more accurate.

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■ An interesting point:

• We see points on both left and right sides of the yellow triangle.
    ♦ If we consider the points on the left, '(a-T)' will be negative.
    ♦ If we consider the points on the right, '(a-T)' will be positive.
• But sign is not important here.
• What matters is the distance of each point from the yellow triangle.
• That is why we take the absolute value of the difference in each reading.

---

10. Now we take up the question (b): Whose readings are more precise?

• To find the answer to this question, the ‘true value’ is not a prerequisite.
• In fact, the 'true value' is not required at all.
11. Consider the readings obtained by student A.
• We take the mean of those readings. That is., we calculate $\mathbf\small{a_m}$ using the expression: $\mathbf\small{a_m =\frac{\sum{a}}{n}}$
• From the table 2.3 below, we get:
$\mathbf\small{a_m =\frac{26.12}{5}=5.224}$

Table 2.3

12. Now we take each reading of student A.
• The difference between the mean value and the first reading = |(5.224-5.25)| = 0.026.
• Note that, we want the difference only. The sign is not required. That is why we are taking the absolute value of the difference.
• We can call it: Absolute deviation.
    ♦ So absolute deviation of first reading = 0.026.
• The difference between the mean value and the second reading = |(5.224-5.19)| = 0.034.
    ♦ That is., absolute deviation of second reading = 0.034.
• The difference between the mean value and the third reading = |(5.224-5.24)| = 0.016.
    ♦ That is., absolute deviation of third reading = 0.016.
• The difference between the mean value and the fourth reading = |(5.224-5.17)| = 0.054.
    ♦ That is., absolute deviation of fourth reading = 0.054.
• The difference between the mean value and the fifth reading = |(5.224-5.27)| = 0.046.
    ♦ That is., absolute deviation of fifth reading = 0.046.
13. Note that:
• In question (a) we obtain the deviations from the true value.
• In question (b) we obtain the deviations from the mean value.
14. The deviations obtained in (12) gives us an important information:
• The fourth reading deviated most from the mean value.
• The third reading deviated least from the mean value.
15. However, all deviations are important. We must calculate their mean. This is done in table 2.3 above.
• The mean of the absolute deviations is denoted as $\mathbf\small{|\Delta a_m|}$
• It can be obtained using the familiar expression for mean (average): $\mathbf\small{|\Delta a_m|=\frac{\sum{|\Delta a|}}{n}}$.
• In our present problem, there are 5 readings. So n = 5.
• Thus we get: $\mathbf\small{|\Delta a_m|}$ = 0.035.
16. In a similar way, we must calculate $\mathbf\small{|\Delta a_m|}$ for the readings taken by student B.
The table is shown below:

Table 2.4

17. Now we compare A and B. We see that:
• The readings taken by student A has a mean deviation of 0.035 from his own mean value.
• The readings taken by student B has a mean deviation of 0.009 from his own mean value.
18. The readings taken by student B deviates less from the mean value.
• 'Less deviation' means 'greater closeness'.
• We can write:
The readings taken by student B is closer to the mean value.
• In other words:
The readings taken by student B is more precise than those taken by student B.
19. A graphical representation will give us a better understanding.
(i) Consider fig.2.10 below:

Fig.2.10

• Imagine that, the students A and B are engaged in a different type of archery competition.
• This time, 'closeness of the shots' is the criteria.
• A shoots 5 times. Are the shots close to each other?
    ♦ To get an idea, the mean of A's shots is marked as a magenta triangle.
• B shoots 5 times. Are the shots close to each other?
    ♦ To get an idea, the mean of B's shots is marked as a cyan triangle.
(ii) The arrows shot by A are indicated by the magenta circles.
• The arrows shot by B are indicated by the cyan circles.
(iii) We see this:
• Arrows shot by B are close to the cyan triangle.
(Note that, for student B, the reading 2.31 occurs twice. So there are two circles, one below the other at 2.31. Thus we see only a total of 4 cyan circles)
• Arrows shot by A are not that close to the magenta triangle.
(iv) We say that, B has 'more closeness to his mean value'.
• We need a mathematical tool to express this 'more closeness to mean value'.
(v) The tables 2.3 and 2.4 taken together, provides an excellent mathematical tool for this purpose.
• Table 2.3 gives the deviation as 0.035.
• Table 2.4 gives the deviation as 0.009.
• So B has a lesser deviation. 'Lesser deviation' means 'greater closeness'.
• So B has 'more closeness'. That is., readings taken by B are more precise.
■ When we consider accuracy, closeness to the true value of the whole problem is the criteria.
■ When we consider precision, closeness to the mean value of each observer is the criteria.
20. In our present problem, we see this:
■ Observer A has good accuracy but poor precision.
■ Observer B has poor accuracy but good precision.
• In science labs, we see four types of readings:
(i) Readings which are accurate but not precise.
(ii) Readings which are precise but not accurate.
(iii) Readings which are both accurate and precise.
(iv) Readings which are neither accurate nor precise.

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We will see two solved examples:

Solved example 2.10
Table 2.5 below shows the readings obtained by 4 students while trying to find the radius of a curved surface. The true value is 3.043 cm. Analyze the observations with regard to accuracy and precision.

Table 2.5

Solution:
1. First we will consider the accuracy of the readings taken by each student.
Table 2.6 below is the table for student A.

Table.2.6

So the deviation from the true value is 0.036 cm.
• Table 2.7 below is the table for student B.

Table 2.7

So the deviation from the true value is 0.002 cm.
• Table 2.8 below is the table for student C.

Table 2.8

So the deviation from the true value is 0.253 cm.
• Table 2.9 below is the table for student D.

Table 2.9

So the deviation from the true value is 0.384 cm.
2. Writing the deviations in ascending order, we get:
0.002 (B) < 0.036 (A) < 0.253 (C) < 0.384 (D)
• Student B with the least deviation of 0.002 cm has the highest accuracy.
• Student D with the highest deviation of 0.384 cm has the least accuracy.
• Writing the accuracy in ascending order, we get:
D < C < A < B
3. Next we will consider the precision of the readings taken by each student.
• Table 2.10 below is the table for student A.

Table 2.10 

So the deviation from the mean value is 0.036 cm.
• Table 2.11 below is the table for student B.

Table 2.11 

So the deviation from the mean value is 0.002 cm.
• Table 2.12 below is the table for student C.

Table 2.12

So the deviation from the mean value is 0.007 cm.
• Table 2.13 below is the table for student D.

Table 2.13

So the deviation from the mean value is 0.007 cm.
4. Writing the deviations in ascending order, we get:
0.001 (C) < 0.002 (B) < 0.036 (A) < 0.286 (D)
• Student C with the least deviation of 0.001 cm has the highest precision.
• Student D with the highest deviation of 0.286 cm has the least precision.
• Writing the precision in ascending order, we get:
D < A < B < C

Solved example 2.11
Two clocks are being tested against a standard clock located in a national laboratory. At 12:00:00 noon by the standard clock, the readings of the two clocks are given in the table 2.14 below:

If you are doing an experiment that requires precision time interval measurements, which of the two clocks will you prefer ?
Solution:
1. In table 2.15 above, all the times are converted into seconds.
Example: On Wednesday at 12 noon, the clock 1 shows 11:59:08.
In seconds, this is: (11 × 60 × 60) + (59 × 60) + 8 = 43148 s.
2. First we will compare the accuracy.
• The calculations for Clock 1 is shown in table 2.16 below   .
• The calculations for Clock 2 is shown in table 2.17.

• The true value is 12 hours = 43200 s.
• From the tables, we see that the deviation (from the true value) is lesser for clock 1. So clock 1 is more accurate than clock 2.
• The deviation is very large for clock 2. So Cock 2 does not have any accuracy at all. This is evident because, at 12 noon, it is showing around 10:15
3. Next we will compare precision.
• The calculations for Clock 1 is shown in table 2.18 below.
• The calculations for Clock 2 is shown in table 2.19.

• The mean of the readings for clock 1 is 43237.429 s.
• The mean of the readings for clock 2 is 36908.286 s.
• From the tables, we see that the deviation (from the mean value) is lesser for clock 2. So clock 2 is more precise than clock 1.
4. So if we are doing an experiment which requires precision time interval measurements, we must use clock 2.

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Next we have to learn about significant figures and errors. In the next section, we will learn about significant figures. In the section after that, we will see errors.

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