Statistical Tests (t, Q & F) — Maths for Analytical Chemistry

Is the difference real?

Confidence intervals and significance tests for analytical chemistry

Every measured value carries uncertainty. Statistical tests let you decide whether an observed difference — between your result and a certified value, or between two methods — is genuine or merely the result of random scatter. The backbone of all these decisions is the confidence interval:

μ = \overset{x}{ˉ} \pm \frac{t s}{n}

where $\overset{x}{ˉ}$ is the sample mean, s the standard deviation, n the number of measurements, and t the critical value from Student's t-distribution for the chosen confidence level and $n - 1$ degrees of freedom. At 95 % confidence with n = 4 (df = 3), t = 3.18; as n grows, t shrinks toward 1.96 (the z-value for a large sample). The interval narrows both because $n$ grows and because t decreases.

Three tests dominate analytical practice. The Student t-test compares a measured mean against a known reference (one-sample form) or two methods against each other (two-sample form). The Q-test (Dixon's Q) provides a quick outlier test: compute the ratio of the suspect point's gap to the overall range and compare to a critical value. The F-test decides whether two sets of measurements have the same precision (variance), which must be checked before pooling them.

Confidence interval

\overset{x}{ˉ} \pm t s / n

use t-table for df = n − 1

One-sample t-test

t_{calc} = \frac{∣ x ˉ - μ _{0} ∣ n}{s}

compare to t-crit at df = n − 1

Two-sample t-test

t = \frac{∣ x ˉ _{1} - x ˉ _{2} ∣}{s _{p} 1/ n _{1} + 1/ n _{2}}

pool variances if F-test passes

Q-test (outlier)

Q = \frac{gap}{range}

reject if Q > Q-crit

F-test (precision)

F = s_{1}^{2} / s_{2}^{2} (s_{1} > s_{2})

reject equal variance if F > F-crit

Critical Q values (95 %)

n = 3: 0.970 | n = 4: 0.829 | n = 5: 0.710
n = 6: 0.625 | n = 7: 0.568 | n = 10: 0.466

Choosing the right test requires reading the experimental design, not just the data. The critical decision tree is:

Paired vs unpaired t-test

Use the paired test when each sample is measured by both methods (removing sample-to-sample variation). Use the unpaired (independent-samples) test when different samples are used in each group. Using the wrong version inflates or deflates the t-statistic.

F-test first rule

Before a two-sample t-test, run an F-test to check equal variances. If F-test passes (equal variances), use the pooled t-test. If it fails, use Welch's t-test (separate variances) with adjusted degrees of freedom.

One-tailed vs two-tailed

Two-tailed: "Do the means differ?" (most common in validation). One-tailed: "Is method A higher than method B?" One-tailed tests are easier to pass — pre-register which you will use.

Q-test limitations

Apply the Q-test only once per dataset (using all n values). Applying it repeatedly to the remaining set inflates the false-positive rejection rate. Never reject more than one outlier with a single Q-test; for more outliers use Grubbs' test.

Common pitfalls

Using an unpaired t-test when the design is paired. Paired tests remove between-sample variation and are far more powerful. Ignoring the pairing can hide a real method difference.
Skipping the F-test before pooling. Pooling variances that are significantly different inflates the t-statistic and leads to false conclusions.
One-tailed vs two-tailed. Choosing the more convenient tail after seeing the data is a form of p-hacking. Decide in advance.
Comparing t_calc to the wrong df row in the table. For a two-sample pooled t-test the df is $n_{1} + n_{2} - 2$ , not $n_{1} - 1$ or $n_{2} - 1$ .
Over-applying the Q-test. Only one Q-test per dataset; the critical Q value is read for the original n, not the reduced set.

The analyst's verdict

Validating a method against a certified reference material, rejecting a bad replicate, or comparing two instruments all rest on these tests. Open the confidence-interval lab and watch precision and sample size shape the answer.

Confidence-interval lab

Adjust the standard deviation and sample size and watch the 95 % CI widen or narrow in real time.

95% confidence interval for a mean of 10.00

μ = \overset{x}{ˉ} \pm \frac{t s}{n}

standard deviation s0.15

number of measurements n4

t (95%, df = 3)3.18

interval half-width± 0.238

95% CI10.00 ± 0.238

Take more measurements and the interval narrows — both because √n grows and because t shrinks with more degrees of freedom. This is the maths behind “is my result consistent with the certified value?”

Worked example 195 % confidence interval for a chloride determination

A student runs four replicate titrations for chloride and obtains: $23.42, 23.51, 23.37, 23.46 mM$ . Calculate the 95 % confidence interval.

Worked example 2One-sample t-test vs a certified reference material

The certified value for a lead standard is $μ_{0} = 10.00 ppm$ . A new method gives five replicates with $\overset{x}{ˉ} = 9.84 ppm$ and $s = 0.14 ppm$ . Does the method show significant bias at 95 % confidence?

Worked examples

Q-test, F-test and a two-sample comparison — try each before revealing the solution.

Worked example 3Q-test for an outlier

Six replicate absorbances are: 0.521, 0.519, 0.524, 0.516, 0.551, 0.520. Is 0.551 an outlier at 95 % confidence? (Critical Q for n = 6 is 0.625.)

Worked example 4F-test comparing two methods' precision

Method A (n = 6) gives $s_{A} = 0.32 ppm$ and Method B (n = 5) gives $s_{B} = 0.14 ppm$ for iron in the same standard. Do the precisions differ significantly at 95 % confidence? (F-crit for df = 5, 4 is 6.26.)

Worked example 5Two-sample t-test comparing two methods' means

Method A: $\overset{x}{ˉ}_{A} = 10.14 ppm, s_{A} = 0.32, n_{A} = 6$ . Method B: $\overset{x}{ˉ}_{B} = 9.98 ppm, s_{B} = 0.14, n_{B} = 5$ . (F-test passed — variances equal.) Do the means differ at 95 % confidence?

Worked example 6Paired t-test for two methods on the same samples

Six soil samples are analysed for arsenic (mg kg⁻¹) by ICP-MS (method A) and hydride-generation AAS (method B):

Sample	ICP-MS	HG-AAS	Difference d
1	12.3	12.0	+0.3
2	18.7	18.2	+0.5
3	9.1	9.4	−0.3
4	24.5	24.0	+0.5
5	15.2	15.5	−0.3
6	20.1	19.7	+0.4

Use a paired t-test at 95 % confidence ( $t_{0.025, 5} = 2.571$ ) to decide whether the methods give different results.

ChallengeChallenge — full method comparison: F-test then appropriate t-test

Two independent labs determine calcium in a reference cement (certified: 24.90 % Ca):

Lab 1 (EDTA titration): $n_{1} = 6, \overset{x}{ˉ}_{1} = 24.75%, s_{1} = 0.18%$
Lab 2 (ICP-OES): $n_{2} = 5, \overset{x}{ˉ}_{2} = 24.88%, s_{2} = 0.08%$

(i) Apply an F-test to compare precisions (F-crit(5,4) = 6.26 at 95 %). (ii) If the F-test passes, apply a pooled t-test to compare means (t-crit df = 9 is 2.262). If it fails, note that Welch's test would be needed. (iii) Test each mean against the certified value using a one-sample t-test (t-crit df = 5 is 2.571; df = 4 is 2.776). State all conclusions at 95 % confidence.

Check yourself

Four quick questions on confidence intervals, the t-test, and the Q-test.

Question 1 of 4 · Score 0

A 95% confidence interval for the mean is calculated as:

\overset{x}{ˉ} \pm \frac{t s}{n}

Choose an answer.