SOP: Testing the Mean Against a Specified Value by the t-Test

GIFSQ — Global Institute for Food Safety & Quality

Principle

The t-test is used to judge whether the mean of the lot from which a sample has been drawn differs significantly from a specified value. The null hypothesis states that there is no real difference between the lot mean and the specified value; the calculated t-value is compared with tabulated t-values at the 5% and 1% significance levels, and the null hypothesis is rejected only if the calculated value equals or exceeds the tabulated value.

Materials

• The sample of n observations x₁, x₂, …, xₙ from the lot
• The specified value μ₀
• The book’s t-distribution table for (n − 1) degrees of freedom

Procedure

Take a sample of size n from the lot and record the observations

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Compute the sample mean x̄ and the sum of squares Σx²

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Compute the sample standard deviation s using the expression below

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Compute the t-value using the expression below

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Decide whether a one-tailed or a two-tailed test applies: when only values on one side of the specified value are of interest (for example, only low drained weights matter in law), the test is one-sided and the one-tailed critical values are used

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Compare the calculated t-value (irrespective of its sign) with the tabulated t-values at the 5% and 1% levels for (n − 1) degrees of freedom; if it is greater than or equal to the tabulated value, reject the null hypothesis

Calculation

Sample standard deviation:

s = √{[Σx² − (Σx)²/n] / (n − 1)}

t-value:

t = (x̄ − μ₀) / (s/√n)

Where,

• x̄ = sample mean
• μ₀ = specified value
• s = standard deviation of the sample
• n = number of observations
• The t-value has (n − 1) degrees of freedom

Worked example 1 — drained weight of canned fruit

Specified limit: 55%. Thirteen drained-weight observations: 54.5, 53.2, 54.1, 55.2, 55.0, 56.2, 54.5, 53.8, 55.5, 55.6, 55.0, 53.0, 54.8.

Σx = 710.4; Σx² = 38886.62 (as printed); x̄ = 710.4/13 = 54.646

s = √[(38886.62 − 710.4²/13)/12] = 2.345

t = (54.646 − 55)/(2.345/√13) = (−0.354)/0.65 = −0.354 (as printed)

With 12 degrees of freedom the one-tailed critical values are 1.78 (5%) and 2.68 (1%). The calculated t-value of 0.354 is less than the tabulated value, so the null hypothesis is not doubted and the lot mean is taken to be in agreement with the specified 55%.

Notes: (1) the book’s printed Σx² (38886.62) differs by 55.5 from the sum recomputed from the printed x values (38831.12); (2) the printed t-value (−0.354) does not recompute from the printed intermediate values (s = 2.345, n = 13 give t ≈ −0.54); both are carried as printed, and the conclusion — not significant against the one-tailed critical value 1.78 — is unaffected either way.

Worked example 2 — copper content in tomato ketchup

Maximum specified limit: 15 ppm. Thirteen copper determinations (ppm): 14.8, 15.5, 16.2, 16.5, 15.0, 15.1, 14.9, 15.4, 16.5, 14.7, 16.0, 15.8, 15.0.

Σx = 201.4; Σx² = 3125.14; x̄ = 201.4/13 = 15.4923

s = √[(3125.14 − 201.4²/13)/12] = 0.6448

t = (15.49 − 15)/(0.6448/√13) = 2.75

The calculated t-value of 2.75 is higher than the tabulated values of 1.78 and 2.68 at the 5% and 1% levels with 12 degrees of freedom; hence the copper content of the ketchup samples is significantly more than the specified limit.

Precautions

• The t-test assumes that the distribution is normal and the samples have been taken at random.
• Except on rare occasions when the one-tailed test is applicable, and when in doubt, the two-tailed test is recommended as it only marginally affects the conclusion.
• If the probability is less than 5% but more than 1%, the difference is said to be significant; if less than 1%, it is said to be highly significant.
• The t-test and the F-test should not be used for acceptance of lots.

References

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