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- Measuring With Precision
- Accuracy vs. Precision

We have learned that physical measurement involves error and that every physical measurement is an approximation. This leads us to a new question: How much error is involved in any given measurement? The terms precision and accuracy relate to how good an approximation is. For example, how precise were our measurements of the sides of the right triangles, and how accurate were our measurements of the distance from Mars to the Sun?

Since measurements are approximate, the most meaningful way of interpreting a measurement is as an interval with a lower bound and an upper bound. Imagine that we have measured a line segment, using a ruler divided into centimeters, and found the length to be 5 cm. To be more precise, we can state the measure as an interval — either in words, 5 cm to the nearest 0.5 cm, or using notation, such as 5 cm 0.5 cm (read “5 cm plus or minus 0.5 cm”). Either presentation gives the center of the interval and the distance of the upper and lower bounds from this center (5 0.5 implies a lower bound of 4.5 and an upper bound of 5.5). We can also state that the maximum possible error for this measure is 0.5 cm (which is half the size of the measurement unit). **Note 12**

In summary, the precision of a measurement depends on the size of the smallest measuring unit — whether the measurement is, for example, to the nearest 10 feet, to the nearest foot, or to the nearest tenth of a foot. The smaller the interval, the more we have “narrowed it down,” and thus the more precise the measurement.

Video SegmentWatch this video segment to see the participants discuss whether a measurement is an approximate or an exact value. They also discuss the role partitioning plays in answering this question.Can you think of any other reasons that would support their conjecture?You can find this segment on the session video approximately 6 minutes and 4 seconds after the Annenberg Media logo. |

**Problem C1**

In Part A, we discussed how a unit can be partitioned into smaller subunits. How are partitioning and precision related?

Tip: Think about what you would do to a measuring instrument if you wanted to measure something more precisely.

**Problem C2**

When measuring length, the precision unit is determined by the smallest unit being repeated on the measuring tool (the smallest hash mark). Examine the rulers below (not drawn to scale), and identify the precision unit:

**Problem C3**

The maximum possible error of a measurement is always half the size of the precision unit. For example, if the precision unit is 1 cm, the maximum possible error is 0.5 cm; if the precision unit is 4 cm, the maximum possible error is 2 cm, etc. What is the maximum possible error for each ruler above?

**Problem C4**

In Problem B6, you measured isosceles right triangles and found that the hypotenuse of the triangle with legs of 3.0 cm was 4.2 cm. What is the precision unit? Give the interval that shows a more accurate measure of the hypotenuse.

In everyday language, we use the terms accuracy and precision interchangeably. In mathematical terminology, however, the accuracy of a measure (an approximate number) is defined as the ratio of the size of the maximum possible error to the size of the number.

This ratio is called the relative error. We express the accuracy as a percent, by converting the relative error to a decimal and subtracting it from 1 (and writing the resulting decimal as a percent). The smaller the relative error, the more accurate the measure.

Here’s how it works. We can measure two different items to the nearest centimeter: a desk and a notepad. The desk is 120 cm long, and the notepad is 12 cm long. The maximum possible error in each case is 0.5 cm. Both measures are to the same level of precision, but the relative error differs:

Relative Error = ^{Maximum Error
} Size of Number

So the relative errors for the desk and the notepad are

0.5 _{=} 0.0042 0.5 = 0.0042

120 12

Therefore, though the measurements of 120 cm and 12 cm are equally precise, they are not equally accurate. The measurement of 120 cm is more accurate, because it has the smaller relative error. You can record the accuracy as a percentage by subtracting the relative error from one and writing the resulting decimal as a percentage. So, in the first instance, the accuracy is 1 – 0.0042 = 0.9958, or 99.58%, and in the second instance, it is 1 – 0.042 = 0.958, or 95.8%. Clearly, the accuracy of the two measurements differs significantly. **Note 13**

**Problem C5**

Calculate the relative error for the measurement you made on the isosceles right triangle with side 3, and then one or two others.

**Note 12**

Another way to think of an interval is as a boundary; namely, that the measurement is going to fall within a range given by an upper and lower boundary. The range is determined first by the smallest measuring unit on the instruments being used (the precision unit) and then by the maximum possible error for that precision unit. Discuss or reflect on why the maximum possible error cannot be more than half the precision unit.

**Note 13**

If you are working in a group, be sure to discuss the idea of accuracy. How do maximum possible error and relative error differ? Why do you think relative error is calculated as a ratio? How might we improve our approximations when measuring?

**Problem C1**

The precision of a measurement is directly related to the partitioning of the measuring instrument. We can more precisely measure something when the partitions are closer together. For example, a beaker with milliliter partitions gives a more precise liquid measure than a measuring cup with 100-mL partitions.

**Problem C2**

- The precision unit is 1 ft.
- The precision unit is 1 mm.
- The precision unit is 1 cm.
- The precision unit is 1 in.

**Problem C3**

- The maximum possible error is 0.5 ft.
- The maximum possible error is 0.5 mm.
- The maximum possible error is 0.5 cm.
- The maximum possible error is 0.5 in

Problem C4

The length of the hypotenuse mentioned in this problem is 4.2 cm, so the precision unit is 0.1 cm, because each centimeter is divided into tenths. The maximum possible error is 0.05 cm. The side length of the hypotenuse could be written as 4.2 cm 0.05 cm, or between 4.15 and 4.25 cm.

If you expressed the length of the hypotenuse as 42 mm, then the precision unit would be 1 mm, and the maximum possible error would be 0.5 mm (in either direction). The side length of the hypotenuse could be written as 42 mm 0.5 mm, or between 41.5 and 42.5 mm.

**Problem C5**

Since each centimeter is divided into tenths, the precision unit is 0.1 cm, and the absolute error for the calculations is 0.05 cm; therefore, the relative error is 0.05 divided by the measurement. For example, when the hypotenuse was measured as 4.2 cm, the relative error was 0.05/4.2, which gives you 0.012 or 1.2/100 (about 1.2%). Larger measurements with the same absolute error will have lower relative error.