SOP: Testing Significance in Triangle Tests Using the Binomial Distribution

GIFSQ — Global Institute for Food Safety & Quality

Principle

In a triangle test each trial presents three samples, of which one is different and two are the same, so the chance of picking the odd sample correctly by guessing is 1/3. The binomial distribution gives the probability that the odd sample is picked correctly exactly x times in m trials; comparing the observed number of correct picks with the number expected by chance shows whether the panelists truly detected a difference.

Materials

• The triangle-test judgements: m trials per panelist, the number x of correct picks
• The total number of correct replies over all panelists

Procedure

Record for each replicate the number of times x the odd sample was picked correctly out of m trials

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Take the chance probability p = 1/3 for the triangle test

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Compute the probability y of exactly x correct picks in m trials using the binomial expression below

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Multiply the probability by the number of panelists to obtain the expected number of panelists picking the odd sample correctly by guesswork, and compare with the observed numbers

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Compute the expected total number of correct replies (total judgements × p) and the standard deviation of the binomial distribution, and judge whether the observed total exceeds what chance would give

Calculation

Binomial probability:

P(x) = [m! / (x!(m − x)!)] × pˣ × (1 − p)ᵐ⁻ˣ

Mean and standard deviation of the binomial distribution:

μ = np

σ = √[np(1 − p)]

Where,

• m = number of trials
• x = number of trials in which the event occurs (the correct sample is picked)
• m − x = number of trials in which the event does not occur
• p = chance of the event occurring in each trial (1/3 for the triangle test)
• n = total number of judgements
• μ = mean
• σ = standard deviation

Worked example

Eight panelists scored two juice samples by triangle test on 6 occasions (6 replicates), giving 48 judgements in all.

Probability of exactly x = 2 correct picks in m = 6 trials:

P(2) = [6!/(2!4!)] × (1/3)² × (2/3)⁴ = 15 × (1/9) × (16/81) = 240/729 = 0.33

Expected total correct replies by chance: 48 × (1/3) = 16

Standard deviation: σ = √[48 × (1/3) × (2/3)] = 3.27

In the actual test the odd sample was picked correctly 23 times — considerably more than three times the standard deviation — so the correct picking was not due to chance but to an actual difference between the samples.

Note: the book compares the observed total (23) directly with three times the standard deviation (9.81) rather than with the expected total plus three standard deviations; the reasoning is carried as printed.

Precautions

• The binomial distribution applies where one deals with an expected proportion rather than an expected number.
• The comparison is valid only when random effects operate, i.e. the panelists are truly picking by chance under the null hypothesis.

References

Ranganna, S., Handbook of Analysis and Quality Control for Fruit and Vegetable Products, 2nd edition.