Accuracy, Precision and Significant Figures
Science is based on experimentation that requires good measurements. The validity of a measurement can be described in terms of its accuracy and its precision (see Figure 1.19 and Figure 1.20). Accuracy is how close a measurement is to the correct value for that measurement. For example, let us say that you are measuring the length of standard piece of printer paper. The packaging in which you purchased the paper states that it is 11 inches long, and suppose this stated value is correct. You measure the length of the paper three times and obtain the following measurements: 11.1 inches, 11.2 inches, and 10.9 inches. These measurements are quite accurate because they are very close to the correct value of 11.0 inches. In contrast, if you had obtained a measurement of 12 inches, your measurement would not be very accurate. This is why measuring instruments are calibrated based on a known measurement. If the instrument consistently returns the correct value of the known measurement, it is safe for use in finding unknown values.
Precision states how well repeated measurements of something generate the same or similar results. Therefore, the precision of measurements refers to how close together the measurements are when you measure the same thing several times. One way to analyze the precision of measurements would be to determine the range, or difference between the lowest and the highest measured values. In the case of the printer paper measurements, the lowest value was 10.9 inches and the highest value was 11.2 inches. Thus, the measured values deviated from each other by, at most, 0.3 inches. These measurements were reasonably precise because they varied by only a fraction of an inch. However, if the measured values had been 10.9 inches, 11.1 inches, and 11.9 inches, then the measurements would not be very precise because there is a lot of variation from one measurement to another.
The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not precise, or they are precise but not accurate. Let us consider a GPS system that is attempting to locate the position of a restaurant in a city. Think of the restaurant location as existing at the center of a bull’seye target. Then think of each GPS attempt to locate the restaurant as a black dot on the bull’s eye.
In Figure 1.21, you can see that the GPS measurements are spread far apart from each other, but they are all relatively close to the actual location of the restaurant at the center of the target. This indicates a low precision, high accuracy measuring system. However, in Figure 1.22, the GPS measurements are concentrated quite closely to one another, but they are far away from the target location. This indicates a high precision, low accuracy measuring system. Finally, in Figure 1.23, the GPS is both precise and accurate, allowing the restaurant to be located.
Uncertainty
The accuracy and precision of a measuring system determine the uncertainty of its measurements. Uncertainty is a way to describe how much your measured value deviates from the actual value that the object has. If your measurements are not very accurate or precise, then the uncertainty of your values will be very high. In more general terms, uncertainty can be thought of as a disclaimer for your measured values. For example, if someone asked you to provide the mileage on your car, you might say that it is 45,000 miles, plus or minus 500 miles. The plus or minus amount is the uncertainty in your value. That is, you are indicating that the actual mileage of your car might be as low as 44,500 miles or as high as 45,500 miles, or anywhere in between. All measurements contain some amount of uncertainty. In our example of measuring the length of the paper, we might say that the length of the paper is 11 inches plus or minus 0.2 inches or 11.0 ± 0.2 inches. The uncertainty in a measurement, A, is often denoted as δA ("delta A"),
The factors contributing to uncertainty in a measurement include the following:
 Limitations of the measuring device
 The skill of the person making the measurement
 Irregularities in the object being measured
 Any other factors that affect the outcome (highly dependent on the situation)
In the printer paper example uncertainty could be caused by: the fact that the smallest division on the ruler is 0.1 inches, the person using the ruler has bad eyesight, or uncertainty caused by the paper cutting machine (e.g., one side of the paper is slightly longer than the other.) It is good practice to carefully consider all possible sources of uncertainty in a measurement and reduce or eliminate them,
Percent Uncertainty
One method of expressing uncertainty is as a percent of the measured value. If a measurement, A, is expressed with uncertainty, δA, the percent uncertainty is
1.2 $$\text{\%uncertainty=}\frac{\delta \text{A}}{\text{A}}\text{\xd7100\%}\text{.}$$
Worked Example
Calculating Percent Uncertainty: A Bag of Apples
A grocery store sells 5lb bags of apples. You purchase four bags over the course of a month and weigh the apples each time. You obtain the following measurements:
 Week 1 weight: $4.\text{8}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{lb}$
 Week 2 weight: $5.3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{lb}$
 Week 3 weight: $4.\text{9}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{lb}$
 Week 4 weight: $5.4\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{lb}$
You determine that the weight of the 5 lb bag has an uncertainty of ±0.4 lb. What is the percent uncertainty of the bag’s weight?
Strategy
First, observe that the expected value of the bag’s weight, $A$ , is 5 lb. The uncertainty in this value, $\delta A$ , is 0.4 lb. We can use the following equation to determine the percent uncertainty of the weight
$$\text{\%uncertainty=}\frac{\delta \text{A}}{\text{A}}\text{\xd7100\%}\text{.}$$
Solution
Plug the known values into the equation
$$\%\text{uncertainty=}\frac{0.4\text{lb}}{5\text{lb}}\times 100\%=8\%.$$
Discussion
We can conclude that the weight of the apple bag is 5 lb ± 8 percent. Consider how this percent uncertainty would change if the bag of apples were half as heavy, but the uncertainty in the weight remained the same. Hint for future calculations: when calculating percent uncertainty, always remember that you must multiply the fraction by 100 percent. If you do not do this, you will have a decimal quantity, not a percent value.
Uncertainty in Calculations
There is an uncertainty in anything calculated from measured quantities. For example, the area of a floor calculated from measurements of its length and width has an uncertainty because the both the length and width have uncertainties. How big is the uncertainty in something you calculate by multiplication or division? If the measurements in the calculation have small uncertainties (a few percent or less), then the method of adding percents can be used. This method says that the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent uncertainties in the items used to make the calculation. For example, if a floor has a length of 4.00 m and a width of 3.00 m, with uncertainties of 2 percent and 1 percent, respectively, then the area of the floor is 12.0 m^{2} and has an uncertainty of 3 percent (expressed as an area this is 0.36 m^{2}, which we round to 0.4 m^{2} since the area of the floor is given to a tenth of a square meter).
For a quick demonstration of the accuracy, precision, and uncertainty of measurements based upon the units of measurement, try this simulation. You will have the opportunity to measure the length and weight of a desk, using milli versus centi units. Which do you think will provide greater accuracy, precision and uncertainty when measuring the desk and the notepad in the simulation? Consider how the nature of the hypothesis or research question might influence how precise of a measuring tool you need to collect data.
Precision of Measuring Tools and Significant Figures
An important factor in the accuracy and precision of measurements is the precision of the measuring tool. In general, a precise measuring tool is one that can measure values in very small increments. For example, consider measuring the thickness of a coin. A standard ruler can measure thickness to the nearest millimeter, while a micrometer can measure the thickness to the nearest 0.005 millimeter. The micrometer is a more precise measuring tool because it can measure extremely small differences in thickness. The more precise the measuring tool, the more precise and accurate the measurements can be.
When we express measured values, we can only list as many digits as we initially measured with our measuring tool (such as the rulers shown in Figure 1.24). For example, if you use a standard ruler to measure the length of a stick, you may measure it with a decimeter ruler as 3.6 cm. You could not express this value as 3.65 cm because your measuring tool was not precise enough to measure a hundredth of a centimeter. It should be noted that the last digit in a measured value has been estimated in some way by the person performing the measurement. For example, the person measuring the length of a stick with a ruler notices that the stick length seems to be somewhere in between 36 mm and 37 mm. He or she must estimate the value of the last digit. The rule is that the last digit written down in a measurement is the first digit with some uncertainty. For example, the last measured value 36.5 mm has three digits, or three significant figures. The number of significant figures in a measurement indicates the precision of the measuring tool. The more precise a measuring tool is, the greater the number of significant figures it can report.
Zeros
Special consideration is given to zeros when counting significant figures. For example, the zeros in 0.053 are not significant because they are only placeholders that locate the decimal point. There are two significant figures in 0.053—the 5 and the 3. However, if the zero occurs between other significant figures, the zeros are significant. For example, both zeros in 10.053 are significant, as these zeros were actually measured. Therefore, the 10.053 placeholder has five significant figures. The zeros in 1300 may or may not be significant, depending on the style of writing numbers. They could mean the number is known to the last zero, or the zeros could be placeholders. So 1300 could have two, three, or four significant figures. To avoid this ambiguity, write 1300 in scientific notation as 1.3 × 10^{3}. Only significant figures are given in the x factor for a number in scientific notation (in the form $x\times {10}^{y}$). Therefore, we know that 1 and 3 are the only significant digits in this number. In summary, zeros are significant except when they serve only as placeholders. Table 1.4 provides examples of the number of significant figures in various numbers.
Number 
Significant Figures 
Rationale 

1.657 
4 
There are no zeros and all nonzero numbers are always significant. 
0.4578 
4 
The first zero is only a placeholder for the decimal point. 
0.000458 
3 
The first four zeros are placeholders needed to report the data to the tenthousandths place. 
2000.56 
6 
The three zeros are significant here because they occur between other significant figures. 
45,600 
3 
With no underlines or scientific notation, we assume that the last two zeros are placeholders and are not significant. 
15895000 
7 
The two underlined zeros are significant, while the last zero is not, as it is not underlined. 
5.457 $\times $ 10^{13} 
4 
In scientific notation, all numbers reported in front of the multiplication sign are significant 
6.520 $\times $ 10^{–23} 
4 
In scientific notation, all numbers reported in front of the multiplication sign are significant, including zeros. 
Table 1.4
Significant Figures in Calculations
When combining measurements with different degrees of accuracy and precision, the number of significant digits in the final answer can be no greater than the number of significant digits in the least precise measured value. There are two different rules, one for multiplication and division and another rule for addition and subtraction, as discussed below.

For multiplication and division: The answer should have the same number of significant figures as the starting value with the fewest significant figures. For example, the area of a circle can be calculated from its radius using $A=\pi {r}^{2}$. Let us see how many significant figures the area will have if the radius has only two significant figures, for example, r = 2.0 m. Then, using a calculator that keeps eight significant figures, you would get
$$A=\text{\pi}{r}^{2}=\text{}\left(\mathrm{3.1415927...}\right)\text{}\times \text{}{\left(2.0\text{m}\right)}^{2}=\text{}4.5238934{\text{m}}^{2}.$$
But because the radius has only two significant figures, the area calculated is meaningful only to two significant figures or
$$A=\text{}4.5{\text{m}}^{2}$$
even though the value of $\pi $ is meaningful to at least eight digits.

For addition and subtraction: The answer should have the same number places (e.g. tens place, ones place, tenths place, etc.) as the leastprecise starting value. Suppose that you buy 7.56 kg of potatoes in a grocery store as measured with a scale having a precision of 0.01 kg. Then you drop off 6.052 kg of potatoes at your laboratory as measured by a scale with a precision of 0.001 kg. Finally, you go home and add 13.7 kg of potatoes as measured by a bathroom scale with a precision of 0.1 kg. How many kilograms of potatoes do you now have, and how many significant figures are appropriate in the answer? The mass is found by simple addition and subtraction:
$$\begin{array}{l}\underset{\_}{\begin{array}{cc}\hfill 7.56& \text{kg}\hfill \\ \hfill \mathrm{6.052}& \text{kg}\hfill \\ \hfill +13.7& \text{kg}\hfill \end{array}}\\ \begin{array}{cc}\hfill 15.208& \text{kg}\hfill \end{array}\end{array}$$
The least precise measurement is 13.7 kg. This measurement is expressed to the 0.1 decimal place, so our final answer must also be expressed to the 0.1 decimal place. Thus, the answer should be rounded to the tenths place, giving 15.2 kg. The same is true for nondecimal numbers. For example,
$$6527.23\text{}+\text{}2\text{}=\text{}6528.23\text{}=\text{}6528\text{.}$$
We cannot report the decimal places in the answer because 2 has no decimal places that would be significant. Therefore, we can only report to the ones place.
It is a good idea to keep extra significant figures while calculating, and to round off to the correct number of significant figures only in the final answers. The reason is that small errors from rounding while calculating can sometimes produce significant errors in the final answer. As an example, try calculating $\mathrm{5,098}\left(5.000\right)\times \left(\mathrm{1,010}\right)$ to obtain a final answer to only two significant figures. Keeping all significant during the calculation gives 48. Rounding to two significant figures in the middle of the calculation changes it to $\text{5,100\u2013(5}\text{.000)\xd7(1,000)=100,}$ which is way off. You would similarly avoid rounding in the middle of the calculation in counting and in doing accounting, where many small numbers need to be added and subtracted accurately to give possibly much larger final numbers.
Significant Figures in this Text
In this textbook, most numbers are assumed to have three significant figures. Furthermore, consistent numbers of significant figures are used in all worked examples. You will note that an answer given to three digits is based on input good to at least three digits. If the input has fewer significant figures, the answer will also have fewer significant figures. Care is also taken that the number of significant figures is reasonable for the situation posed. In some topics, such as optics, more than three significant figures will be used. Finally, if a number is exact, such as the 2 in the formula, $c=2\pi r$, it does not affect the number of significant figures in a calculation.
Worked Example
Approximating Vast Numbers: a Trillion Dollars
The U.S. federal deficit in the 2008 fiscal year was a little greater than $10 trillion. Most of us do not have any concept of how much even one trillion actually is. Suppose that you were given a trillion dollars in $100 bills. If you made 100bill stacks, like that shown in Figure 1.25, and used them to evenly cover a football field (between the end zones), make an approximation of how high the money pile would become. (We will use feet/inches rather than meters here because football fields are measured in yards.) One of your friends says 3 in., while another says 10 ft. What do you think?
Strategy
When you imagine the situation, you probably envision thousands of small stacks of 100 wrapped $100 bills, such as you might see in movies or at a bank. Since this is an easytoapproximate quantity, let us start there. We can find the volume of a stack of 100 bills, find out how many stacks make up one trillion dollars, and then set this volume equal to the area of the football field multiplied by the unknown height.
Solution
 Calculate the volume of a stack of 100 bills. The dimensions of a single bill are approximately 3 in. by 6 in. A stack of 100 of these is about 0.5 in. thick. So the total volume of a stack of 100 bills is
$$\begin{array}{l}\text{volumeofstack}=\text{length}\times \text{width}\times \text{height,}\\ \text{volumeofstack}=\text{6in}\text{.}\times \text{3in}\text{.}\times \text{0}\text{.5in}\text{.,}\\ \text{volumeofstack}=\text{9in}{\text{.}}^{\text{3}}.\end{array}$$

Calculate the number of stacks. Note that a trillion dollars is equal to $\text{\$}1\times {10}^{12}$, and a stack of onehundred $\text{\$}100$ bills is equal to $\text{\$}10,000,$ or $\text{\$}1\times {10}^{4}$. The number of stacks you will have is
1.3$${\text{\$1\xd710}}^{\text{12}}{\text{(atrilliondollars)/\$1\xd710}}^{\text{4}}{\text{perstack=1\xd710}}^{\text{8}}\text{stacks}\text{.}$$

Calculate the area of a football field in square inches. The area of a football field is $100\text{yd}\times \text{50}\text{\hspace{0.17em}}\text{yd}$, which gives $5,000{\text{yd}}^{\text{2}}$. Because we are working in inches, we need to convert square yards to square inches
$$\begin{array}{l}{\text{Area=5,000yd}}^{\text{2}}\text{\xd7}\frac{\text{3ft}}{\text{1yd}}\text{\xd7}\frac{\text{3ft}}{\text{1yd}}\text{\xd7}\frac{\text{12in}\text{.}}{\text{1foot}}\text{\xd7}\frac{\text{12in}\text{.}}{\text{1foot}}\text{=6,480,000in}{\text{.}}^{\text{2}}\text{,}\hfill \\ \text{Area}\approx {\text{6\xd710}}^{\text{6}}\text{in}{\text{.}}^{\text{2}}\text{.}\hfill \end{array}$$
This conversion gives us ${\text{6\xd710}}^{\text{6}}\text{in}{\text{.}}^{\text{2}}$ for the area of the field. (Note that we are using only one significant figure in these calculations.)
 Calculate the total volume of the bills. The volume of all the $100bill stacks is $$\text{9in}{\text{.}}^{\text{3}}{\text{/stack\xd710}}^{\text{8}}{\text{stacks=9\xd710}}^{\text{8}}\text{in}{\text{.}}^{\text{3}}$$
 Calculate the height. To determine the height of the bills, use the following equation
$$\begin{array}{ccc}\text{volumeofbills}& \text{=}& \text{areaoffield\xd7heightofmoney}\\ \text{Heightofmoney}& \text{=}& \frac{\text{volumeofbills}}{\text{areaoffield}}\\ \text{Heightofmoney}& \text{=}& \frac{{\text{9\xd710}}^{\text{8}}\text{in}{\text{.}}^{\text{3}}}{{\text{6\xd710}}^{\text{6}}\text{in}{\text{.}}^{\text{2}}}\text{=1}{\text{.33\xd710}}^{\text{2}}\text{in}\text{.}\\ \text{Heightofmoney}& \text{=}& {\text{1\xd710}}^{\text{2}}\text{in}\text{.=100in}\text{.}\end{array}$$
The height of the money will be about 100 in. high. Converting this value to feet gives
$$100\text{in}\text{.}\times \frac{\text{1ft}}{\text{12in}\text{.}}=8.33\text{ft}\approx \text{8ft}\text{.}$$
Discussion
The final approximate value is much higher than the early estimate of 3 in., but the other early estimate of 10 ft (120 in.) was roughly correct. How did the approximation measure up to your first guess? What can this exercise tell you in terms of rough guesstimates versus carefully calculated approximations?
In the example above, the final approximate value is much higher than the first friend’s early estimate of 3 in. However, the other friend’s early estimate of 10 ft. (120 in.) was roughly correct. How did the approximation measure up to your first guess? What can this exercise suggest about the value of rough guesstimates versus carefully calculated approximations?