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:
• 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:
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:
• 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.
• 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.
• 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:
$\mathbf\small{a_m =\frac{26.12}{5}=5.224}$
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:
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:
• 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.
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.
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.
So the deviation from the true value is 0.036 cm.
• Table 2.7 below is the table for student B.
So the deviation from the true value is 0.002 cm.
• Table 2.8 below is the table for student C.
So the deviation from the true value is 0.253 cm.
• Table 2.9 below is the table for student D.
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.
So the deviation from the mean value is 0.036 cm.
• Table 2.11 below is the table for student B.
So the deviation from the mean value is 0.002 cm.
• Table 2.12 below is the table for student C.
So the deviation from the mean value is 0.007 cm.
• Table 2.13 below is the table for student D.
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.
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 |
• 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 |
(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.
■ An interesting point:
♦ 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?
• 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 |
• 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 |
• 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 |
• 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.
We will see two solved examples:
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 |
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 |
• Table 2.7 below is the table for student B.
 |
| Table 2.7 |
• Table 2.8 below is the table for student C.
 |
| Table 2.8 |
• Table 2.9 below is the table for student D.
 |
| Table 2.9 |
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 |
• Table 2.11 below is the table for student B.
 |
| Table 2.11 |
• Table 2.12 below is the table for student C.
 |
| Table 2.12 |
• Table 2.13 below is the table for student D.
 |
| Table 2.13 |
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 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.
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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