# 4.1: Characterizing Measurements and Results

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Let’s begin by choosing a simple quantitative problem that requires a single measurement: What is the mass of a penny? You probably recognize that our statement of the problem is too broad. For example, are we interested in the mass of a United States penny or of a Canadian penny, or is the difference relevant? Because a penny’s composition and size may differ from country to country, let’s narrow our problem to pennies from the United States.

There are other concerns we might consider. For example, the United States Mint produces pennies at two locations (Figure 4.1.1 ). Because it seems unlikely that a penny’s mass depends on where it is minted, we will ignore this concern. Another concern is whether the mass of a newly minted penny is different from the mass of a circulating penny. Because the answer this time is not obvious, let’s further narrow our question and ask “What is the mass of a circulating United States Penny?”

A good way to begin our analysis is to gather some preliminary data. Table 4.1.1 shows masses for seven pennies collected from my change jar. In examining this data we see that our question does not have a simple answer. That is, we can not use the mass of a single penny to draw a specific conclusion about the mass of any other penny (although we might reasonably conclude that all pennies weigh at least 3 g). We can, however, characterize this data by reporting the spread of the individual measurements around a central value.

Penny | Mass (g) |
---|---|

1 | 3.080 |

2 | 3.094 |

3 | 3.107 |

4 | 3.056 |

5 | 3.112 |

6 | 3.174 |

7 | 3.198 |

## Measures of Central Tendency

One way to characterize the data in Table 4.1.1 is to assume that the masses of individual pennies are scattered randomly around a central value that is the best estimate of a penny’s expected, or “true” mass. There are two common ways to estimate central tendency: the mean and the median.

### Mean

The * mean*, \(\overline{X}\), is the numerical average for a data set. We calculate the mean by dividing the sum of the individual values by the size of the data set

\[\overline{X} = \frac {\sum_{i = 1}^n X_i} {n} \nonumber\]

where \(X_i\) is the *i*^{th} measurement, and *n* is the size of the data set.

What is the mean for the data in Table 4.1.1 ?

*Solution*

To calculate the mean we add together the results for all measurements

\[3.080 + 3.094 + 3.107 + 3.056 + 3.112 + 3.174 + 3.198 = 21.821 \text{ g} \nonumber\]

and divide by the number of measurements

\[\overline{X} = \frac {21.821 \text{ g}} {7} = 3.117 \text{ g} \nonumber\]

The mean is the most common estimate of central tendency. It is not a robust estimate, however, because a single extreme value—one much larger or much smaller than the remainder of the data—influences strongly the mean’s value [Rousseeuw, P. J. *J. Chemom. ***1991**, *5*, 1–20]. For example, if we accidently record the third penny’s mass as 31.07 g instead of 3.107 g, the mean changes from 3.117 g to 7.112 g!

An estimate for a statistical parameter is robust if its value is not affected too much by an unusually large or an unusually small measurement.

### Median

The * median*,

*\(\widetilde{X}\), is the middle value when we order our data from the smallest to the largest value. When the data has an odd number of values, the median is the middle value. For an even number of values, the median is the average of the*

*n*/2 and the (

*n*/2) + 1 values, where

*n*is the size of the data set.

When *n *= 5, the median is the third value in the ordered data set; for *n *= 6, the median is the average of the third and fourth members of the ordered data set.

What is the median for the data in Table 4.1.1 ?

*Solution*

To determine the median we order the measurements from the smallest to the largest value

\(3.056 \quad 3.080 \quad 3.094 \quad 3.107 \quad 3.112 \quad 3.174 \quad 3.198\)

Because there are seven measurements, the median is the fourth value in the ordered data; thus, the median is 3.107 g.

As shown by Example 4.1.1 and Example 4.1.2 , the mean and the median provide similar estimates of central tendency when all measurements are comparable in magnitude. The median, however, is a more robust estimate of central tendency because it is less sensitive to measurements with extreme values. For example, if we accidently record the third penny’s mass as 31.07 g instead of 3.107 g, the median’s value changes from 3.107 g to 3.112 g.

## Measures of Spread

If the mean or the median provides an estimate of a penny’s expected mass, then the spread of individual measurements about the mean or median provides an estimate of the difference in mass among pennies or of the uncertainty in measuring mass with a balance. Although we often define the spread relative to a specific measure of central tendency, its magnitude is independent of the central value. Although shifting all measurements in the same direction by adding or subtracting a constant value changes the mean or median, it does not change the spread. There are three common measures of spread: the range, the standard deviation, and the variance.

Problem 13 at the end of the chapter asks you to show that this is true.

### Range

The * range*,

*w*, is the difference between a data set’s largest and smallest values.

\[w = X_\text{largest} - X_\text{smallest} \nonumber\]

The range provides information about the total variability in the data set, but does not provide information about the distribution of individual values. The range for the data in Table 4.1.1 is

\[w = 3.198 \text{ g} - 3.056 \text{ g} = 0.142 \text{ g} \nonumber\]

### Standard Deviation

The * standard deviation*,

*s*, describes the spread of individual values about their mean, and is given as

\[s = \sqrt{\frac {\sum_{i = 1}^{n} (X_i - \overline{X})^{2}} {n - 1}} \label{4.1}\]

where \(X_i\) is one of the *n* individual values in the data set, and \(\overline{X}\) is the data set's mean value. Frequently, we report the relative standard deviation, *s _{r}*, instead of the absolute standard deviation.

\[s_r = \frac {s} {\overline{X}} \nonumber\]

The percent relative standard deviation, %*s _{r}*, is \(s_r \times 100\).

The relative standard deviation is important because it allows for a more meaningful comparison between data sets when the individual measurements differ significantly in magnitude. Consider again the data in Table 4.1.1 . If we multiply each value by 10, the absolute standard deviation will increase by 10 as well; the relative standard deviation, however, is the same.

Report the standard deviation, the relative standard deviation, and the percent relative standard deviation for the data in Table 4.1.1 ?

*Solution*

To calculate the standard deviation we first calculate the difference between each measurement and the data set’s mean value (3.117), square the resulting differences, and add them together to find the numerator of Equation \ref{4.1}

\[\begin{align*}

(3.080-3.117)^2 = (-0.037)^2 = 0.001369\\

(3.094-3.117)^2 = (-0.023)^2 = 0.000529\\

(3.107-3.117)^2 = (-0.010)^2 = 0.000100\\

(3.056-3.117)^2 = (-0.061)^2 = 0.003721\\

(3.112-3.117)^2 = (-0.005)^2 = 0.000025\\

(3.174-3.117)^2 = (+0.057)^2 = 0.003249\\

(3.198-3.117)^2 = (+0.081)^2 = \underline{0.006561}\\

0.015554

\end{align*}\]

For obvious reasons, the numerator of Equation \ref{4.1} is called a sum of squares. Next, we divide this sum of squares by *n *– 1, where *n *is the number of measurements, and take the square root.

\[s = \sqrt{\frac {0.015554} {7 - 1}} = 0.051 \text{ g} \nonumber\]

Finally, the relative standard deviation and percent relative standard deviation are

\[s_r = \frac {0.051 \text{ g}} {3.117 \text{ g}} = 0.016 \nonumber\]

\[\% s_r = (0.016) \times 100 = 1.6 \% \nonumber\]

It is much easier to determine the standard deviation using a scientific calculator with built in statistical functions.

Many scientific calculators include two keys for calculating the standard deviation. One key calculates the standard deviation for a data set of *n *samples drawn from a larger collection of possible samples, which corresponds to Equation \ref{4.1}. The other key calculates the standard deviation for all possible samples. The latter is known as the population’s standard deviation, which we will cover later in this chapter. Your calculator’s manual will help you determine the appropriate key for each.

### Variance

Another common measure of spread is the * variance*, which is the square of the standard deviation. We usually report a data set’s standard deviation, rather than its variance, because the mean value and the standard deviation share the same unit. As we will see shortly, the variance is a useful measure of spread because its values are additive.

What is the variance for the data in Table 4.1.1 ?

Solution

The variance is the square of the absolute standard deviation. Using the standard deviation from Example 4.1.3 gives the variance as

\[s^2 = (0.051)^2 = 0.0026 \nonumber\]

The following data were collected as part of a quality control study for the analysis of sodium in serum; results are concentrations of Na^{+} in mmol/L.

\(140 \quad 143 \quad 141 \quad 137 \quad 132 \quad 157 \quad 143 \quad 149 \quad 118 \quad 145\)

Report the mean, the median, the range, the standard deviation, and the variance for this data. This data is a portion of a larger data set from Andrew, D. F.; Herzberg, A. M. *Data: A Collection of Problems for the Student and Research Worker*, Springer-Verlag:New York, 1985, pp. 151–155.

**Answer**-
*Mean*: To find the mean we add together the individual measurements and divide by the number of measurements. The sum of the 10 concentrations is 1405. Dividing the sum by 10, gives the mean as 140.5, or \(1.40 \times 10^2\) mmol/L.*Median*: To find the median we arrange the 10 measurements from the smallest concentration to the largest concentration; thus\(118 \quad 132 \quad 137 \quad 140 \quad 141 \quad 143 \quad 143 \quad 145 \quad 149 \quad 157\)

The median for a data set with 10 members is the average of the fifth and sixth values; thus, the median is (141 + 143)/2, or 142 mmol/L.

*Range*: The range is the difference between the largest value and the smallest value; thus, the range is 157 – 118 = 39 mmol/L.*Standard Deviation*: To calculate the standard deviation we first calculate the absolute difference between each measurement and the mean value (140.5), square the resulting differences, and add them together. The differences are\(–0.5 \quad 2.5 \quad 0.5 \quad –3.5 \quad –8.5 \quad 16.5 \quad 2.5 \quad 8.5 \quad –22.5 \quad 4.5\)

and the squared differences are

\(0.25 \quad 6.25 \quad 0.25 \quad 12.25 \quad 72.25 \quad 272.25 \quad 6.25 \quad 72.25 \quad 506.25 \quad 20.25\)

The total sum of squares, which is the numerator of Equation \ref{4.1}, is 968.50. The standard deviation is

\[s = \sqrt{\frac {968.50} {10 - 1}} = 10.37 \approx 10.4 \nonumber\]

*Variance*: The variance is the square of the standard deviation, or 108.