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Chapter 4
Elements of
Scientific Measurement
4.1 Issues in measurement
In most practical situations, a scientist has to measure something. The quantity to be measured could be the average weight of adult male sparrows, the number of microbes in unit volume of a fluid, or the electrical charge of an electron. In the first case the weight of each sparrow is really different from that of another, because of the variation inherent in that species. If one could somehow measure the weight of all adult male sparrows, one could obtain the true answer to the question. But that enterprise is practically impossible, due to constraints of time and money. So one has to obtain a smaller sample and has to obtain the mean. For the scientific question in hand, it could also be important to measure the variability within the species (becase it is the variation that natural selection acts on). Even though the weight of a sparrow is a continuous variable, one always measures upto a definite accuracy: the least count of the instrument used. In the second case, the microbes may not be uniformly dis- tributed in the liquid, and for that reason one would have to take samples from different parts of the liquid. Counting such tiny
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living organisms may also have its problems: each counting may not be accurate. One could miss or overcount. If one does not assume any particular propensity of the experimenter to miss or overcount, the error in counting may be assumed to be random.
In case of the measurement of the charge of an electron, the value to be measured is really a constant. But in the process of measurement, many errors would creep in—systematic as well as random. The scientist would have to remove all chances of systematic errors, but has to learn to live with random errors.
In all these cases there is some objective value of the quantity to be measured, and the scientist has to reach as close as possible to that value, subject to the constraints of time and money, by taking samples. Even in the case of measuring the charge of an electron, in each measurement he is really taking samples from a theoretically infinite number of possible readings.
4.2 Analyzing the sampled data
The first thing one does after recording the data is to obtain the mean given by
Mean : x ¯ =
n
( x 1 + x 2 · · · + xn ) =
n
∑^ n i = 1
xi (4.1)
One would also like to know how much the data points devi- ate from the mean value on an average. Thus one would like to obtain an average of ( xi − x ¯). But this quantity ( xi − x ¯) could be positive as well as negative, although the average of the quantities ( xi − x ¯) is zero. If we really want a measure of the deviation from the mean value, it should be a positive number. To overcome this problem, we take the square of ( xi − x ¯) and obtain its average. Thus, we obtain
Variance : s^2 =
n − 1
∑^ n i = 1
( xi − x ¯)^2 (4.2)
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points fell into each bin. The number of data points in each bin divided by the total number of data points give the normalized frequency of each bin. One then plots the graph of the weight versus the frequency of each weight. An example of such a graph is shown in Fig. 4.1. For different observations the graph may have different shapes.
Figure 4.1: A typical histogram obtained from observational data
If the thing to be measured has an inherent variability (like the variation within a species), the frequency curve may have specific characteristics reflecting the nature of variability. But if the variation is purely due to random errors, one would expect a bell-shaped curve: the so-called ‘normal’ distribution. This curve is given by
f ( x ) =
σ
p 2 π
exp
( (^) x − μ σ
, −∞ < x < ∞ (4.3)
where μ is the mean and σ is the standard deviation. The function is graphed in Fig. 4.2. The position of the standard deviation is shown on the graph. If a set of data has a smaller standard deviation, the graph is narrow and tall and if σ is large, it is broader and of shorter height. One can integrate this curve in different ranges to find the fraction of the population that can be expected to lie in specific ranges:
- Approximately 68.3% of the population are within 1 standard
4.3. Distribution of the data 5
μ
σ
Figure 4.2: The normal distribution.
deviation of the mean (that is, between μ − σ and μ + σ );
- Approximately 95.4% of the population are within 2 standard deviations of the mean (that is, between μ − 2 σ and μ + 2 σ );
- Almost all the population – about 99.7% – are within 3 stan- dard deviations of the mean (that is, between μ − 3 σ and μ + 3 σ ).
One is normally interested in finding the standard deviations within which certain specific percentages of the population lie. These are summarized as follows.
- Approximately 90% of the population are within 1.64 standard deviations of the mean;
- Approximately 95% of the population are within 1.96 standard deviations of the mean;
- Approximately 99% of the population are within 2.58 standard deviations of the mean.
It helps to remember these values. In all observational or experimental situations, there is an objectively existing distribution for the quantity being measured, with ‘true’ values of the mean μ and the standard deviation σ. The challenge is to get as close as possible to these values through a finite number of samplings. Note that, after a scientist publishes his or her results, hundreds of scientists worldwide will repeat the
4.5. Experimental Errors 7
in a sack from similar holes 20 m apart. Then the collected soil is spread over a hard surface and is dried in the sun. After the soil is dry, it is mixed well and spread over a square area. The square is then divided into 4 equal parts. Two of them along the main diagonal are kept for further processing and the other two are discarded. The remaining part is again mixed well, spread over another square area, divided into four parts, two along the diagonal are kept and the rest are thrown away. The process is repeated until the remaining amount is around 2-3 kg. Then the bigger particles are ground, and the material is passed through a 2 mm sieve. The soil that goes through the sieve is considered to be the representative sample of the soil in the field, on which tests are done. Similarly, standard procedures of sampling exist in almost all fields. One has to learn the procedures before proceeding to conduct any experiment. If such time-tested procedures are not available in a field of enquiry, one has to develop a procedure that ensures that the range of variability is aptly represented in the sample. The procedure adopted has to be clearly stated in the paper or scientific report. After the samples are collected, one proceeds to measure the parameters in question.
4.5 Experimental Errors
All measurements are prone to errors, and a scientist has to be conscious of this fact when making measurements. Errors can be divided into two major categories:
Random errors: These are fluctuations in readings around the actual value being measured, caused by thermal and other sources of noise.
Systematic errors: These are consistent deviations of the mea- surement from the value being measured, caused by definite causative factors.
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A researcher has to consciously try to avoid all possibilities of systematic errors. There are two general prescriptions of doing so. First, one has to check the calibration of each measuring instrument, because these may change with time and environ- mental conditions under which an experiment is conducted. Second, to do the measurement of a value in two or more different ways, because it is unlikely that the same systematic error would creep in two different sets of apparatus. Apart from these two prescriptions, there are no other general guidelines, because in different experiments different causative factors may be operative that influence the readings. But there are very definite prescriptions in dealing with ran- dom errors. The first prescription is that, the experiment should be so planned that it is possible to make a large number of measurements of the same quantity, under varying conditions. For example, if one is interested in measuring the electrical resis- tance of a sample, one should make arrangements for applying a variable voltage (which can be done with a potential divider) and to measure the current for each value of the voltage. When the measured values are tabulated, the resistance can be obtained by dividing the voltage across the sample by the current through it. We thus get a large number of measured values of the resistance, say, x 1 , x 2 , x 3 · · · xn which typically will be slightly different from each other due to random error. One can then take the average of the measured values
x ¯ =
x 1 + x 2 + x 3 · · · + xn n
and can hope that the positive errors will cancel out the negative ones, thus getting a mean value close to the actual value being measured. But still many questions remain.
- How many observations need to be taken in order to obtain a confident estimate?
- Which value out of the large number of observations can be
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a few of these insects and will have to weigh them. By the act of catching a few insects, she is actually ‘sampling’ from a pop- ulation of insects, and typically the population will be much larger than the samples chosen. How can she make an objective estimate about the character of the species by taking a relatively small number of samples? Note that the body-weight of the organisms in the insect species might have a distribution that is not a normal distri- bution. But if one could somehow capture all the organisms in that species and could measure them, she could obtain the ‘true’ mean μ and the ‘true’ standard deviation σ. But that is not physically possible. So she would actually measure these quantities from a finite sample. If she captures 10 individuals and measures them, the sample size is 10. From these values she could obtain the sample mean x ¯ and sample standard deviation s. Now, she could again capture another 10 individuals and measure them, i.e., she could again obtain another sample of size 10. Of course she will not get the same value of the sample mean ¯ x and sample standard deviation s. Each time she repeats the experiment and obtains 10 samples, she will get different values. Now if she calculates the distribution of these sample means , what will the distribution be? Similarly, when you are making any measurement (say, the charge of an electron), you are actually ‘sampling’ from an ideally infinite number of possible measurements. Suppose you have taken 10 measurements, and have obtained the mean value. Now if you repeat the experiment and take 10 more readings, will you get the same mean value? No. If you repeat the experiment a number of times (each time taking 10 readings), you will get a scatter of mean values. What will the distribution be? the value of the gravitational constant G by a number of experimental runs; if you are a chemist, think of the task of measuring the specific gravity of a new compound that has been synthesized, etc. It helps to think of a problem from one’s own field. Note that in all these cases, you are taking a small number of samples from a large ‘population’ of possible measurements.
4.6. Specifying the measured value 11
4.6.1 The Central Limit Theorem
The answers to these questions come from the Central Limit Theorem:
For large sample sizes (at least 25), the sampling distribu- tion of the mean for samples of size n from a population with mean μ and standard deviation σ may be approxi- mated by a normal distribution with mean μ and standard deviation σ /
p n.
Thus, the Central limit Theorem says that in all these cases the distribution of the sample means will be approximately a normal distribution. The more the sample size n , the better is the fit to a normal distribution curve. And this is independent of the actual distribution in the population. Now, instead of taking 10 readings in each set, if you had taken 50 readings, the average values that come out in each experiment would be closer to the actual mean value μ , and so you would get a narrower Gaussian function. If you take 100 readings, it will be even narrower, i.e., with a smaller variance. The variance of the distribution of means, σ^2 x ¯ , is thus inversely proportional to the number of readings, i.e.,
σ^2 x ¯ =
σ^2 n
Therefore the standard deviation of the distribution of the sample means is σ (^) x ¯ =
σ p n That is what the Central Limit Theorem says: That the sample means will follow a normal distribution, with the same mean as that of the population (i.e., μ ) and standard deviation σ /
p n. This gives a way of finding out how good will be the measured value x ¯ (using a small number of samples) as an estimate of the mean μ of the actual population. Let us illustrate that with an example.
4.6. Specifying the measured value 13
μ (^) 1.28σ
σ
Figure 4.3: The distribution of the sample means in Example 1
p 50 = 0.099. Therefore we have to find
P
x ¯ lies
= 2.02 standard deviations above the mean
From Table 4.2 we find that 0.9783 fraction of the area lies to the left of this value, and so 1 − 0. 9783 = 0. 0217 fraction lies to the right. This implies that the probability of getting a mean value
beyond 2.2 goes down to 2.17% if she took 50 samples. 2
4.6.2 Standard error of the mean
The standard deviation of the distribution of the sample means, σ (^) x ¯ , is called the “standard error of the mean”, and gives an esti- mate of the error in the mean obtained by taking a finite number of readings. Let us illustrate it with an example.
Example 4.2: Suppose you have measured a quantity 36 times and have obtained a sample mean x ¯ = 112. 0 and sample standard deviation s = 40. What is the probability that the actual mean μ lies in the range [100,124]?
Solution: Here we have a situation where we do not know the actual population mean μ and the population standard deviation σ , i.e., we do not know the actual distribution. But we want a good estimate of the population mean. If we repeated the experiment of taking 36 samples again and again, we would get slightly different values each time which
14 Chapter 4. Elements of Scientific Measurement
will be distributed as a normal distribution, whose mean will be μ and standard deviation will be σ /
p 36 = σ /6. Since we actually do not know the value of σ , the best we can do is to substitute it with what you know: the standard deviation s of the readings actually taken. Thus the sampling distribution will have a standard deviation 40/6=6.67.
−1.8σ μ 1.8σ
σ
Figure 4.4: The distribution of the sample means in Example 2
Now we ask: what is the probability that the population mean μ is within 12 of x ¯ = 112, i.e., μ ∈ [112 − 12 , 112 + 12]? This is the same as asking what is the probability that the quantity x ¯ we have measured is within 12 of the population mean μ? And the quantity 12 is actually 12/6.67=1.8 standard deviations (see Fig. 4.4). To write it mathematically,
P ( μ is within 12 of ¯ x ) = P ( ¯ x is within 12 of μ ) = P ( ¯ x is within 1.8 standard deviations of μ )
From the z -table in Table 4.2 we see that the area under the normal curve below 1.8 standard deviations is 0.9641. The area under the curve from the mean to 1.8 standard deviations is 0. 9641 − 0. 5 = 0. 4641. Thus the area between − 1. 8 standard deviations to +1.8 standard deviations is 0.4641 × 2 = 0.9282. Therefore there is 92.82% chance that the actual population
mean lies within ±12 of the measured value. 2
Now let us consider the question: how many data points are necessary for a confident estimate of the population mean and
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Here the error is expressed in absolute magnitude, and so it has a unit. The error can also be expressed as a percentage, i.e., a measurement of a length x can also be expressed as
x ¯ cm ±
δx x ¯
× 100%
where ¯ x is the mean and δx is the standard error of the mean.
In this case the error is expressed as a fraction and will not have any unit.
Example 4.3: You have conducted an experiment to measure two values x and y , and have obtained the following data.
x 5.23 4.97 4.78 5.05 5.34 4.78 4.96 5. 5.15 5.26 4.92 4.78 4.98 5.01 5.19 5. 5.15 4.94 4.92 4.89 5.10 5.22 4.80 5.
y 3.43 3.45 3.85 3.29 3.96 3.10 3.11 3. 3.43 3.24 3.29 3.24 3.16 3.45 3.23 3. 3.29 3.37 3.42 3.10 3.29 3.27 3.17 3.
How will you describe the result scientifically?
Solution:
The mean values can be easily obtained as
x ¯ =
n
∑^ n i = 1
xi = 5.
y ¯ =
n
∑^ n i = 1
yi = 3.
The standard deviations are
σsx =
n − 1
∑^ n i = 1
( xi − x ¯)^2 = 0.
4.7. Estimating with confidence 17
σs y =
n − 1
∑^ n i = 1
( yi − y ¯)^2 = 0.
Using the approximation σ ≈ s , we get
Standard error in x =
σ p n
Standard error in y = σ p n
Therefore, the results are to be specified as
x = 5.018 ± 0. y = 3.335 ± 0.
4.7 Estimating with confidence
Thus, the usual experimental procedure is to obtain samples from a population, to obtain the mean and the standard deviation from the data, and to use these to estimate the characteristics of the population. The question is: How good is the sample mean as an estimate of the population mean? One normally approaches this question by finding an interval of values within which one can be fairly confident that the population mean lies. We have seen earlier that the sampling distribution of the mean for samples of size n has mean μ and
Standard error SE =
σ p n
Since this distribution is approximately normal, 95% of the sam- ples will lie within 1. 96 ×SE of the population mean. So, for ap- proximately 95% of samples of size n , the difference between the sample mean x ¯ and the population mean μ is less than 1. 96 σ /
p n.
4.7. Estimating with confidence 19
Only one in 3.5 million data points lie outside this range, i.e., if the results were due to chance (not caused by the phenomenon in question) then the obtained result can occur to most once in 3.5 million repetitions of the experiment. Still a question may remain in your mind: Was it a good idea to replace σ by s? Doesn’t it introduce error in our estimate of the 95% confidence interval? There is actually an intuitive justification for it. You have seen that the sample mean varies from sample to sample less for large sample sizes than for small ones. In a similar way, it can be shown that the sample standard deviation also varies from sample to sample less for large sample sizes than for small ones. Thus, the larger the sample size, the better s is as an estimate of σ. Moreover, due to the division by p n , for large sample sizes s / p n would not be much different from σ / p n , and the confidence interval thus calculated would really contain the population mean μ in 95% of the cases. Now notice a few things.
- Since the 95% confidence interval is proportional to 1/
p n , you would need to take samples four times as large in order to halve the widths of confidence intervals.
- The calculation of a 95% confidence interval does not depend on the size of the population. The only assumption is that the population size is much larger than the sample size. The interval will remain the same if you have drawn a sample of 100 from a population of 10,000 or 10^7.
- The calculation of this interval does depend on the size of the sample, because we have assumed that the sampling distribu- tion follows a normal curve. This is true only if the sample size is at least 25. Moreover, for small n the replacement of σ by s becomes questionable. But still, a scientist may encounter situations where it is impossible (or expensive) to draw a large number of samples ( n < 25). In that case the distribution of the sample means cannot be assumed to be normal, and
20 Chapter 4. Elements of Scientific Measurement
one has to fit it to some other distribution (generally the t - distribution). We shall come to this issue later.
- Following a similar line of argument, any other confidence in- terval can also be calculated. For example, a 99% confidence interval would be given by [ x ¯ − 2.58 × s p n
, x ¯ + 2.58 × s p n
]
because 99% of the area under the normal distribution curve lies between −2.58 to +2.58 standard deviations.
Example 4.4: For the data in Example 3, define a range of x in which the “true” value of x must lie with 99% probability.
Solution: From the data, we get the standard error of x as
SE x = σ p n
Therefore the true value of x will lie in the range
[ ¯ x − 2.58 SE x , ¯ x + 2.58 SE x ] = [5.018 − 2.58 × 0.032, 5.018 + 2.58 × 0.032] = [4.937, 5.102]
4.8 When the data size is small
The above procedures are applicable to situations where the data size is reasonably large (at least 25), without which the Central Limit Theorem would not be applicable. But there are situations in which it is difficult (or very expensive) to obtain many data points. What to do in such cases?
