Lesson Explainer: Measurement Error Physics • 9th Grade

In this explainer, we will learn how to define and calculate the absolute and relative errors of measured values.

When measuring a value, it is important to be able to know how accurate the measurement is. When determining such accuracy, the value must be compared to some other value that is deemed to be correct, the accepted value.

An accepted value, also called the actual value, is a measured value obtained by an error-free measurement process. It is what all other measured values are compared to. Accepted values are typically constants, such as the gravitational constant or charge of an electron.

Measurement error is when the measurement of a value differs from the accepted value. If we know that the mass of a block of cheese is 1 kg, but a scale says it is 1.2 kg, this is an example of measurement error.

Whatever the source of the error is, there are two different ways to quantify it. Let’s first look at absolute error.

Absolute error is the absolute difference between the accepted value and the measured value. When expressed as an equation, it looks as follows: absoluteerroracceptedvaluemeasuredvalue=||.

The lines on the right side of the equation indicate that the difference is an absolute value. An absolute value only cares about the magnitude of the number, meaning it will always be positive, even if the measured value is larger than the accepted value.

For the cheese, the accepted value is 1 kg, and the measured value is 1.2 kg. Substituting these values into the equation gives |11.2|=0.2.kgkgkg

So, even though 11.2 results in a negative 0.2, because it is an absolute value, it becomes positive. The cheese has an absolute error of 0.2 kg.

Let’s have a look at an example.

Example 1: Calculating the Absolute Error in the Measurement of an Accepted Value

In an experiment, the acceleration due to gravity at the surface of Earth is measured to be 9.90 m/s2. Find the absolute error in the measurement using an accepted value of 9.81 m/s2.

Answer

To find the absolute error of the measurement value of 9.90 m/s2, we must find the difference between it and the accepted value of 9.81 m/s2, as shown in the equation for absolute error. Recall that the equation for absolute error is absoluteerroracceptedvaluemeasuredvalue=||.

The accepted value is 9.81 m/s2, and the measured value is 9.90 m/s2, so substituting these into the equation for absolute error gives ||9.81/9.90/||=0.09/.msmsms

Absolute error is an absolute value, and so it will always be positive, even though 9.819.90 results in a negative number. The absolute error is thus 0.09 m/s2.

Absolute error is not always helpful in determining the accuracy of a measurement though. Say that we have a colossal cheese wheel with an accepted value of mass of 1‎ ‎000 kg. When the cheese wheel is put on a scale, it has a measured mass of 1‎ ‎000.2 kg.

Using these values, we see that when putting them into the equation for absolute error |10001000.2|=0.2,kgkgkg we have the same value of absolute error for the colossal 1‎ ‎000 kg cheese wheel as we had for the considerably smaller 1 kg block of cheese. The 0.2 kg matters more for smaller masses than larger ones, and there is a way to express this, relative error.

Relative error is a way of showing the error proportional to the accepted value. It is found by taking the absolute error and dividing it by the accepted value 𝑟=Δ𝑥𝑥, where 𝑟 is the relative error, Δ𝑥 is the absolute error, and 𝑥 is the accepted value.

Both the colossal wheel of cheese and the block have the same value of absolute error, 0.2 kg. Since the colossal wheel of cheese has a much larger accepted value, we should expect the relative error to be smaller than the single block of cheese. The relative error for the wheel is 0.21000=0.0002,kgkg and the relative error for the block is 0.21=0.2.kgkg

Note that because the units are the same for both the numerator and denominator of the equation, they cancel, making the relative error unitless.

Let’s have a look at some examples.

Example 2: Calculating an Absolute Error from a Relative Error

If the relative error in measuring an area of 320 m2 was 0.03, calculate the absolute error for that measurement.

Answer

We are given two values initially, the relative error of 0.03 and the accepted value of 320 m2. We need to find the absolute error, which we can do by looking at the equation for relative error. Recall that the equation for relative error is 𝑟=Δ𝑥𝑥, where 𝑟 is the relative error, Δ𝑥 is the absolute error, and 𝑥 is the accepted value.

To isolate the absolute error, Δ𝑥, we need to think algebraically. Let’s multiply both sides of the equation by the accepted value, 𝑥𝑟×𝑥=Δ𝑥𝑥×𝑥, which cancels the accepted value on the right side of the equation, giving 𝑟×𝑥=Δ𝑥.

Using this modified equation, we can now substitute in the given values. Relative error is 0.03, and the accepted value is 320 m2: 0.03×320=9.6.mm

Relative error is unitless, so the multiplication inherits the units of m2. Our value of absolute error is thus 9.6 m2.

Example 3: Identifying the Measurement That Has the Greatest Accuracy

Which of the following measurements of time is the most accurate?

  1. 3.4±0.1 s
  2. 5.2±0.01 s
  3. 7.3±0.2 s
  4. 4.1±0.2 s

Answer

The ± symbol means plus or minus a particular value, with the number following it being the absolute error. To determine which measurement of time is most accurate, we will need to find the relative error, as the measurement that has the lowest relative error is the most accurate. Recall that the relative error equation is absolute error over the accepted value, 𝑟=Δ𝑥𝑥.

In this problem, the absolute error is the number after the ± and the accepted value is before it. Let’s look at each potential answer individually, starting with A: 0.13.4=0.029.ss

Subsequently, the relative error for B is 0.015.2=0.002,ss the relative error for C is 0.27.3=0.027,ss and the relative error for D is 0.24.1=0.049.ss

We see from these that answer B has the smallest relative error, of only 0.002. We could also have determined this by looking at the absolute errors for each option: much smaller absolute errors would also give smaller relative errors.

Relative error is often expressed using a slight modification, making it a percentage.

Percent relative error is relative error expressed as a percentage, which is calculated by multiplying the value by 100%: 𝑟×100%=𝑟,% where 𝑟% is the percent relative error.

Looking back at the cheese, the smaller block of cheese had a relative error of 0.2. The percent relative error is thus 0.2×100%=20%, so the block of cheese has a percent relative error of 20%, or the measurement was off by 20%.

The colossal wheel of cheese has a much smaller percent relative error: 0.0002×100%=0.02%.

This larger proportional difference in percentage error for the smaller blocks of cheese means that the errors in measurement will stack up much faster. If, for instance, you are tasked with measuring out 1‎ ‎000 kg of cheese, choosing the single colossal wheel of 1‎ ‎000 kg will result in an accuracy of 0.02%. If you were to instead choose 1‎ ‎000 of the smaller blocks, the percent relative error would use the much higher 20%.

To get the actual value of how much cheese in kilograms the percent relative error will result in, divide the percent relative error by 100% to convert back to the relative error. Comparing the two, the colossal wheel’s is 1000×0.02%100%=0.2,kgkg while the smaller block of cheese’s is 1000×20%100%=200.kgkg

So, while the colossal wheel’s mass will only vary by 0.2 kg, choosing to instead use the stack of 1‎ ‎000 smaller cheese blocks will have their mass vary by a full 200 kg. Bringing anywhere between 800 and 1‎ ‎200 kg of cheese when you were supposed to have 1‎ ‎000 kg is a big mistake to make.

Since relative error is based on absolute error and the accepted value, the equation for percent relative error, 𝑟% is written as 𝑟=Δ𝑥𝑥×100%,% where Δ𝑥 is the absolute error and 𝑥 is the accepted value.

Let’s look at some examples using the percent relative error.

Example 4: Calculating the Relative Error in a Measurement of an Accepted Value

In an experiment, the speed of sound waves on Earth at sea level at a temperature of 21C is 333 m/s. Find the percent relative error in the measurement using an accepted value of 344 m/s. Give your answer to one decimal place.

Answer

In this problem, the given values are the measured value of 333 m/s and the accepted value of 344 m/s. Recall the percent relative error equation 𝑟=Δ𝑥𝑥×100%,% where Δ𝑥 is the absolute error and 𝑥 is the accepted value.

The absolute error is needed, which is found by taking the difference between the measured and accepted values: 344/333/=11/.msmsms

The relative error is then calculated by dividing the absolute error, 11 m/s, by the accepted value of 344 m/s: Δ𝑥𝑥=11/344/11/344/=0.03197,msmsmsms making the relative error 0.03197. The answer should eventually be to one decimal place, but it is not rounded until the end of the problem for maximum accuracy. To get the percent relative error, this value is then multiplied by 100%: 0.03197×100%=3.197%.

Now that the answer is in its final form, it can be rounded off to one decimal place, making the percent relative error 3.2%.

Example 5: Determining a Value from Its Absolute and Relative Error

The relative and absolute errors in measuring the mass of some box are found to be 1.6% and 0.4 kg respectively. Calculate the actual value of the mass.

Answer

The actual value is the accepted value, and it can be found by using the extended equation for percent relative error Δ𝑥𝑥×100%=𝑟,% where Δ𝑥 is the absolute error and 𝑥 is the accepted value.

The accepted value, 𝑥, needs to be isolated, which can be done algebraically. Let’s start by multiplying both sides by the accepted value: Δ𝑥𝑥×100%×𝑥=𝑟×𝑥.%

This causes the accepted values on the left to cancel out, leaving behind Δ𝑥×100%=𝑟×𝑥.%

Both sides can then be divided by the percent relative error to give Δ𝑥×100%𝑟=𝑟×𝑥𝑟,%%% making the percent relative error cancel on the right, which forms an equation with an isolated accepted value: Δ𝑥×100%𝑟=𝑥.%

Now, the values of absolute error, 0.4 kg, and percent relative error of 1.6% can be substituted in 0.4×100%1.6%=25,kgkg causing the percentage signs to cancel, leaving behind the accepted value of the mass as 25 kg.

Let’s now summarize what we learned in this explainer.

Key Points

  • The accepted value is the actual value that is considered correct.
  • Measurement error is when the measured value differs from the accepted value.
  • Absolute error is the difference between the accepted value and measured value, and it is in the same units as the values.
  • Relative error is the proportion of absolute error and the accepted value, and it is unitless.
  • Percentage relative error is relative error expressed as a percent.

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