Two-out-of-five Test

Last modified at 2/2/2021 4:12 PM by Maren Johnson
Among the discriminative testing methods is the two-out-of-five test. Like other discriminative methods, the two-out-of-five test can be used to determine if a significant difference exists between two products. Additionally, the test can be applied to screen the sensitivity of potential descriptive panelists. The two-out-of-five test is used less often than the Triangle Test, Duo-Trio Test, or the paired-comparison test1.

Principle of the Test

In the two-out-of-five test, the panelist is given five samples. The panelist is instructed to identify the two samples that are different from the other three. Like the Triangle Test, the presentation order of the samples is randomized in order to avoid presentation order bias


Compared to the triangle test, in which the chance of choosing the correct sample is 1/3, the two-out-of-five test is more statistically efficient because the probability of a panelist choosing the two correct products by chance is 1/102. This probability is formed by the product of 2/5, the chances of getting the first sample right, and 1/4, the chances of getting the second sample right (if the first one has already been chosen correctly)3.


The two-out-of-five test is more appropriate for testing non-food consumer products because of the possibility of sensory fatigue. If a panelist were to taste five food samples, his or her discriminative performance may suffer from sensory fatigue and/or non-clarity of memory4. However, the test is suitable for non-food products such as tissues or lotions.

Data Analysis

The chi-square and binomial statistics are suitable for analyzing most discriminative tests including the two-out-of-five test5. The chi-square statistic takes the form of X2=Σ (|O-E|)2/E, where E=expected and O=observed. The total number of expected correct answers is the product of the number of panelists (n) and the probability that an individual would identify the two correct samples by chance (1/10), which takes the form, E=n*1/10. The binomial statistic takes the form, P(x greater than or equal to c) = Sum(N! / (C!(N-C)!) (p^n)(1-p^(n-c))), where n=number of panelists, c=correct responses, and p=probability of each panelist choosing the two correct samples by chance (1/10).


Research has suggested that the two-out-of-five test is preferred over the Triangle Test when measuring perceptible differences in color of a cosmetic product6.
The two-out-of-five test has also been employed when researchers wished to determine if a distillation procedure significantly changed champagne extract. The authors concluded that the pre- and post- distillation extracts were not significantly different7.


1 Kilcast, D. (1999). Sensory techniques to study food texture. In Food texture: measurement and perception. Editor: Rosenthal, A.J. Aspen Publishers: Maryland.
2 Meilgaard, M., Civille, G.V., Carr, T.C. (2007). Sensory Evaluation Techniques. Boca Raton: CRC Press.
3 Stone, H., Bleibaum, R.N. (2009). Sensory Evaluation. In Food Science and Technology. Editor: Campbell-Platt, G. Wiley Blackwell: United Kingdom.
4 Bower, J.A. (1996). Statistics for food science III: sensory evaluation data. Part B – discrimination tests. Nutrition and Food Science, 96, 16-22.
5 Smith, G.L. (1981). Statistical properties of simple sensory difference tests: confidence limits and significance tests. Journal of the science of food and agriculture, 32, 513-520.
6 Whiting R., Murray, S., Caintic, Z., Ellison, K. (2004). The use of sensory difference tests to investigate perceptible colour-difference in a cosmetic product. Color research & application, 29, 299-304.
7 Priser, C., Etiévant, P.X., Nicklaus, S., Brun, O. (1997). Representative champagne extracts for gas chromatography olfactometry analysis. Journal of agriculture and food chemistry, 45, 3511-3514.
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