1.3 - Central tendency, ratios and percentages

Calculating the arithmetic mean

A measure of central tendency summarises the centre or typical value of a dataset.

The arithmetic mean divides the sum of all scores by the number of scores. Each observation contributes to the total, so changing one score changes the mean.

In an original illustrative recall task, five participants remember 2, 4, 4, 6 and 9 words. The sum is 25, so the mean is $25/5=5$ words. The mean need not equal anyone's actual score. Keep the score units and count participants rather than dividing by the number of different score values.

When a table gives frequencies, each score must contribute once for every occurrence. Scores 2, 4, 6 and 9 with frequencies 1, 2, 1 and 1 give $(2\times1+4\times2+6\times1+9\times1)/5=5$. Averaging the four different values would wrongly underweight the repeated score.

Finding the median and mode

The median uses ordered positions, while the mode uses frequency.

The median is the middle score after arranging values in order. For an odd number of numerical scores, its position is $(N+1)/2$. In 2, 4, 4, 6, 9, the third score is 4, so the median is 4 words. For an even number of equal-unit numerical scores, average the two central scores: 3, 5, 8, 8 gives $(5+8)/2=6.5$ words. For ordinal category labels, report the central ordered category or categories without inventing a halfway psychological quantity.

The mode is the most frequent value or category. The recall dataset 2, 4, 4, 6, 9 has mode 4 words. A dataset may have two equally most frequent values, or no useful unique mode when every value occurs equally often. A favourite study setting can have a mode such as “quiet”; a mean of its numerical category codes would have no meaningful interpretation.

The median is determined by central position rather than how extreme an outlier is. The mode is easy to identify and works for nominal data, but may summarise little of the remaining distribution. Neither uses every score's numerical distance from the centre.

Choosing a useful typical value

Choose a summary that matches the measurement and represents the distribution fairly.

Use a mode for nominal categories, since they have no numerical order or equal intervals. A median is appropriate for ordered ratings and often useful for skewed numerical data. A mean is useful for equal-unit numerical scores without extreme values because it incorporates every score and supports further calculations.

In the illustrative scores 2, 4, 4, 6, 9, the mean is 5 and the median is 4. Replacing 9 with 29 raises the mean to $45/5=9$, while the median stays 4. Reporting only the new mean would give a misleading impression of a typical participant; most recalled far fewer than nine. Report the distribution or an appropriate spread measure alongside the centre. Do not remove an extreme score solely to make the mean more convenient: investigate recording errors and apply a justified rule.

A higher mean in one condition describes this sample's performance. It does not itself demonstrate a statistically significant difference, a causal effect or generalisation to a population. The research design and later inferential analysis address those questions.

Fractions, ratios and percentage checks

Proportions allow fair comparisons when groups have different totals.

If 18 of 30 participants answer yes and 12 answer no, the fraction answering yes is $18/30=3/5$. The yes:no ratio is $18:12=3:2$, dividing both counts by 6. The ratio compares yes with no; the fraction compares yes with the whole sample. They are not interchangeable denominators.

Here yes is $(18/30)\times100=60\%$. In a second group, 24 of 60 answer yes: its larger yes count is only $40\%$ of that group. Compare percentages rather than counts when assessing the relative prevalence of yes responses. Include group totals so the reader knows the amount of evidence behind each percentage.

Use estimation to catch errors. Nineteen of 80 is about one quarter, so an answer near 25% is plausible; 237.5% is not a possible proportion of this whole sample. Exact calculation gives $23.75\%$, or $24\%$ to two significant figures. Retain unrounded working and round at the requested final stage.

An order of magnitude check compares powers of ten. If about $1.2\times10^5$ total responses are recorded from $3\times10^3$ participants, the mean is about $4\times10^1=40$ responses per person, of order tens rather than thousands. An estimate checks scale; it does not replace an exact result when one is requested.