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Student's t-test

Edexcel IAL Biology · WBI16 Practical Skills in Biology II

The t-test decides whether the difference between two means is significant. It uses the means, the standard deviations and the sample sizes.

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The formula

t = (x̄₁ - x̄₂) / √( s₁²/n₁ + s₂²/n₂ )

x̄₁, x̄₂ = means of the two groups; s₁, s₂ = standard deviations; n₁, n₂ = numbers of values.
Degrees of freedom = n₁ + n₂ - 2 (shown in Pearson papers as (n₁ - 1) + (n₂ - 1)).
Ignore a negative sign: compare the size of t (the October 2023 mark scheme: 'minus signs are ignored').

A large t comes from a big difference between means, small standard deviations and large samples.

Degrees of freedom	4	6	8	10	12	14	16	18
Critical t (p = 0.05)	2.78	2.45	2.31	2.23	2.18	2.14	2.12	2.10

Two-tailed critical values of t at p = 0.05, as printed in Pearson papers

Step by step

1. Write H₀: no significant difference between the two means.

2. Calculate each mean and standard deviation.

3. Calculate s²/n for each group and add them.

4. Square root, then divide the difference in means by it.

5. Degrees of freedom = n₁ + n₂ - 2; read the critical value at p = 0.05.

6. Compare, decide and conclude in context.

	Antibiotic A	Antibiotic B
Zone diameters / mm	18, 20, 19, 21, 17, 19	15, 16, 14, 17, 15, 16
Mean / mm	19.0	15.5
s / mm	1.414	1.049
s²/n	0.3333	0.1833

CP14 example data

Interpreting

For the CP14 data: t = (19.0 - 15.5) / √(0.3333 + 0.1833) = 4.87; degrees of freedom = 6 + 6 - 2 = 10; critical value 2.23.

4.87 > 2.23, so reject H₀: there is a significant difference between the mean zones of inhibition; antibiotic A inhibits growth more under these conditions.

The January 2024 report found most candidates calculated t correctly and compared it with the critical value: the marks are there for careful working.

Worked example

CP10: oxygen collected (cm³ in 5 min) at high light: 8.2, 7.6, 8.8, 8.0, 7.9; low light: 5.1, 6.0, 5.6, 4.8, 5.5. Carry out a t-test.

Means: 8.10 and 5.40; s: 0.447 and 0.464
s²/n: 0.0400 + 0.0430 = 0.0830
t = 2.70 / √0.0830 = 9.37
df = 5 + 5 - 2 = 8; critical value 2.31
9.37 > 2.31: reject H₀

Answer: t = 9.4; significant difference: more oxygen was produced at high light intensity.

Key idea: t = (x̄₁ - x̄₂) / √(s₁²/n₁ + s₂²/n₂); df = n₁ + n₂ - 2; compare |t| with the p = 0.05 critical value. t ≥ critical: reject H₀ (significant difference).

Check yourself

Two groups have 8 and 7 values. How many degrees of freedom are used for the t-test?
Show answer
13. 8 + 7 - 2 = 13.
Locusts ate leaves of two varieties. Variety A: mean 17.1 g, s² = 0.084, n = 6. Variety B: mean 17.7 g, s² = 0.032, n = 6. Calculate t to 2 significant figures.
Show answer
4.3. s²/n: 0.084 / 6 = 0.0140 and 0.032 / 6 = 0.00533; √0.01933 = 0.139; t = 0.6 / 0.139 = 4.32, matching the October 2023 mark scheme (t about 4.3).
Calculate t for these data. Group A: 12, 15, 11, 14, 13. Group B: 13, 16, 12, 15, 14.
Show answer
1.0. Means 13 and 14; both s = 1.581, so s²/n = 0.5 each; t = 1 / √1.0 = 1.0.
Look up the critical value of t at p = 0.05 for 10 degrees of freedom, from the table in this lesson.
Show answer
2.23. df 10 gives 2.23, the value used in the October 2023 mark scheme.
For the Group A and B data above, t = 1.0 with 8 degrees of freedom (critical value 2.31). What is the conclusion?
- Reject the null hypothesis: a significant difference
- Accept the null hypothesis: no significant difference between the means
- There is a significant correlation
- The t-test cannot be used
Show answer
Accept the null hypothesis: no significant difference between the means. 1.0 is less than 2.31, so the difference may be due to chance.
Which change would make a calculated t value larger, if the means stay the same?
- Larger standard deviations
- Fewer values in each group
- Rounding the means
- Smaller standard deviations
Show answer
Smaller standard deviations. Less spread makes the denominator smaller.
A student calculates t = -4.3. What should they do?
- Conclude t is less than any critical value
- Compare 4.3 with the critical value, since the sign only shows which mean was larger
- Repeat the experiment
- Use chi-squared instead
Show answer
Compare 4.3 with the critical value, since the sign only shows which mean was larger. The October 2023 mark scheme ignores minus signs; the January 2024 report credited correct reasoning with negative values.
When is a t-test appropriate?
- Comparing the means of two groups of continuous data
- Testing for correlation between two variables
- Comparing observed and expected counts
- Comparing six groups at once
Show answer
Comparing the means of two groups of continuous data. Two means.

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Last updated 8 November 2026. Spotted a mistake? Tell us.