Neither student 2 nor student 4 has average values close to the accepted value; neither determination is very accurate. However, student 4 has values that agree closely with each other; the precision is good. This student probably had a consistent error in his or her measuring technique. Student 2 had neither good accuracy nor precision. The accuracy and precision of the four students is summarized below.
Usually, measurements with a high degree of precision are also somewhat accurate. Because the scientists or students don’t know the accepted value beforehand, they strive for high precision and hope that the accuracy will also be high. This was not the case for student 4.
So remember, accuracy and precision are not the same thing:
✔ Accuracy: Accuracy describes how closely a measurement approaches an actual, true value.
✔ Precision: Precision, which we discuss more in the next section, describes how close repeated measurements are to one another, regardless of how close those measurements are to the actual value. The bigger the difference between the largest and smallest values of a repeated measurement, the less precision you have.
The two most common measurements related to accuracy are error and percent error:
✔ Error: Error measures accuracy, the difference between a measured value and the actual value:
✔
Percent error: Percent error compares error to the size of the thing being measured:
Being off by 1 meter isn’t such a big deal when measuring the altitude of a mountain, but it’s a shameful amount of error when measuring the height of an individual mountain climber.
Examples
Q. A police officer uses a radar gun to clock a passing Ferrari at 131 miles per hour (mph). The Ferrari was really speeding at 127 mph. Calculate the error in the officer’s measurement.
A. –4 mph. First, determine which value is the actual value and which is the measured value:
✔ Actual value = 127 mph
✔ Measured value = 131 mph
Then calculate the error by subtracting the measured value from the actual value:
Q. Calculate the percent error in the officer’s measurement of the Ferrari’s speed.
A. 3.15 %. First, divide the error’s absolute value (the size, as a positive number) by the actual value:
Next, multiply the result by 100 to obtain the percent error:
Practice Questions
1. Two people, Reginald and Dagmar, measure their weight in the morning by using typical bathroom scales, instruments that are famously unreliable. The scale reports that Reginald weighs 237 pounds, though he actually weighs 256 pounds. Dagmar’s scale reports her weight as 117 pounds, though she really weighs 129 pounds. Whose measurement incurred the greater error? Who incurred a greater percent error?
2. Two jewelers were asked to measure the mass of a gold nugget. The true mass of the nugget is 0.856 grams (g). Each jeweler took three measurements. The average of the three measurements was reported as the “official” measurement with the following results:
✔ Jeweler A: 0.863 g, 0.869 g, 0.859 g
✔ Jeweler B: 0.875 g, 0.834 g, 0.858 g
Which jeweler’s official measurement was more accurate? Which jeweler’s measurements were more precise? In each case, what was the error and percent error in the official measurement?
Practice Answers
1. Reginald’s measurement incurred the greater magnitude of error, and Dagmar’s measurement incurred the greater percent error. Reginald’s scale reported with an error of
2. Jeweler A’s official average measurement was 0.864 grams, and Jeweler B’s official measurement was 0.856 grams. You determine these averages by adding up each jeweler’s measurements and then dividing by the total number of measurements, in this case three. Based on these averages, Jeweler B’s official measurement is more accurate because it’s closer to the actual value of 0.856 grams.
However, Jeweler A’s measurements were more precise because the differences between A’s measurements were much smaller than the differences between B’s measurements. Despite the fact that Jeweler B’s average measurement was closer to the actual value, the range of his measurements (that is, the difference between the largest and the smallest measurements) was 0.041 grams (
This example shows how low-precision measurements can yield highly accurate results through averaging of repeated measurements. In the case of Jeweler A, the error in the official measurement was
Identifying Significant Figures
Significant figures (no, we’re not talking about supermodels) are the number of digits that you report in the final answer of the mathematical problem you’re calculating. If we told you that one student determined the density of an object to be 2.3 g/mL and another student figured the density of the same object to be 2.272589 g/mL, we bet that you’d believe that the second figure was the result of a more accurate experiment. You may be right, but then again, you may be wrong. You have no way of knowing whether the second student’s experiment was more accurate unless both students obeyed the significant figure convention.
If we ask you to count the number of automobiles that you and your family own, you can do it without any guesswork involved. Your answer may be 0, 1, 2, or 10, but you know exactly how many autos you have. Those numbers are what are called counted numbers. If we ask you how many inches are in a foot, your answer will be 12. That number is an exact number – it’s exact by definition. Another exact number is the number of centimeters per inch, 2.54. In both exact and counted numbers, you have no doubt what the answer is. When you work with these types of numbers, you don’t have to worry about significant figures.
Now suppose that we ask you and four of your friends to individually