Summarizing data with a few numbers

Same mean, two very different varieties

You tested two wheat varieties on five plots each. Both had a mean yield of 500 kg/da:

	Plot yields (kg/da)	Mean	Standard deviation
Variety A	490, 505, 498, 512, 495	500	8.6
Variety B	310, 690, 455, 620, 425	500	153.5

Someone who looks only at the mean thinks the two are the same. But A yields around 500 on every plot, while B gives 310 in one place and 690 in another. A risk-averse farmer picks A.

That is why describing a data set takes at least two numbers:

• A measure of location (centre): around which number do the values cluster? Mean, median, mode.
• A measure of spread (dispersion): how far do the values stray from that centre? Standard deviation, range, coefficient of variation.

“A mean of 500 ± 9 kg/da” says far more than “a mean of 500” on its own.

Try it with your own data

Click a measure to see how it was calculated, step by step, on the right. Below is a plain reading of the results.

Your data

Sample data:Student heightsPull-upsMidterm scoresWheat yield

188, 192, 170, 165, 155, 152, 155, 178, 195, 150

10 numbers read.

Trimming5%10%20%Quartile methodStandard (same as Excel)Course notesDecimals234

Centre Which value does the data cluster around?

• Arithmetic mean$\bar{x}$170

  The most widely used measure of centre. It uses every value, so it is affected by extreme values.

  1. Add up all values and divide by the number of values.

     $$\bar{x} = \dfrac{\sum x_i}{n}$$
  2. $$\bar{x} = \dfrac{188 + 192 + 170 + 165 + \cdots + 195 + 150}{10}$$
  3. $$\bar{x} = \dfrac{1700}{10} = 170$$
• Median$\tilde{x}$167.5
• Mode$Mo$155
• Geometric mean$G$169.22
• Harmonic mean$H$168.45
• Trimmed mean$\bar{x}_{tr}$169.38
• Winsorized mean$\bar{x}_{w}$169.9

Spread How different are the values from each other?

• Range$R$45
• Variance$s^2$299.56
• Standard deviation$s$17.31
• Coefficient of variation$CV$%10.18
• 1st quartile$Q_1$154.25
• 3rd quartile$Q_3$189
• Interquartile range$Q_3-Q_1$34.75

Shape Is the distribution symmetric or piled to one side?

• Skewness$Sk_P$0.43
• Kurtosis$\beta_2$1.51

What do these numbers say?

• Values range from 150 to 195 cm; the mean is 170 cm.
• There is a small gap between the mean (170) and the median (167.5): the distribution is slightly right-skewed.
• The standard deviation is 17.31 cm. Roughly speaking, most values lie between 152.69 and 187.31 cm.
• The coefficient of variation is 10.2%, so variability is moderate. Because the CV has no unit, it lets you compare different measurements (e.g. height and weight).

Which one should I use?

Situation	Suggested measure	Why?
Data are symmetric, no extreme values	Arithmetic mean and standard deviation	Uses all the data; every later test is built on them
A few very high or very low values	Median and quartiles	Not affected by extreme values
You suspect the extreme values are measurement errors	Trimmed or winsorized mean	Weakens the extremes without ignoring them entirely
Categorical data (variety, disease type)	Mode	The others can’t be calculated
Annual growth rates	Geometric mean	Suits the multiplicative nature of rates
Speeds, amounts per unit	Harmonic mean	The correct average for values given “per something”
Comparing variability across different units or magnitudes	Coefficient of variation	Unit-free

Short notes

Why is the mean so sensitive to extreme values?

Say five plots yield 480, 495, 500, 510 and 515 kg/da. The mean is 500 and so is the median. If wild boar trample one plot and its yield drops to 120, the mean falls to 424 while the median becomes 495. The median still describes “a typical plot”; the mean no longer describes any plot.

Why divide by n − 1 rather than n for the standard deviation?

The values in a sample are closer to their own mean than to the population mean, because $\bar{x}$ was calculated from those very values. Dividing by $n$ would underestimate the spread; dividing by $n - 1$ corrects that bias. This $n - 1$ is called the degrees of freedom, and you will meet it again and again in t, χ² and F tests.

$$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}$$

What does the coefficient of variation compare?

The standard deviation of plant height is 12 cm, that of kernel weight 3 g. Which is more variable? With different units you can’t compare the standard deviations. The coefficient of variation expresses the standard deviation as a percentage of the mean and makes that comparison possible:

$$CV = \frac{s}{\bar{x}} \times 100$$

In field trials the CV is also reported as a sign of how carefully the trial was run. A very high CV suggests there were differences between plots other than the treatments (soil, irrigation, measurement error).