IB Biology IA statistics toolkit. Table it. Plot it. Test it.

Each test answers one kind of question. Find yours in the first column and check your data match the second. Each test then has a worked example below: the null hypothesis, the working, the result and the sentence for your IA.

The question you are asking	Your data	The test	What it tells you	When not to use it
Do two groups differ?Is stomatal density different on sun and shade leaves?	A measurement, with repeats, in two separate groups	Unpaired t-test	Whether the gap between the two means is bigger than the spread of the repeats would give by chance	Three or more groups (use ANOVA); the same individuals measured twice (use the paired t-test); counts in categories (use chi-squared)
Did the same individuals change?Does each potato cylinder gain mass in water?	Two measurements on each individual or sample: before and after, or under two conditions	Paired t-test	Whether the mean of the differences is different from zero	The two sets come from different individuals, so a value has no partner in the other set
Do three or more groups differ?Does the sugar change how much carbon dioxide yeast gives off?	A measurement, with repeats, in three or more groups	One-way ANOVA, then Tukey’s HSD	ANOVA: whether at least one mean differs. Tukey: which pairs differ	Only two groups (use the t-test); counts in categories (use chi-squared)
Do counts fit an expected ratio?Do seedlings from a cross fit 3 green : 1 albino?	Counts of individuals in categories, and a ratio that predicts them	Chi-squared goodness of fit	Whether the observed counts differ from the expected counts by more than chance would explain	Means, percentages or measurements (it needs the raw counts); any expected count below 5
Are two sets of categories linked?Does each species of plantain favour a different zone beside a path?	Counts of individuals, each sorted two ways into a contingency table	Chi-squared test of association	Whether the two variables are associated, and from the biggest terms, where	As for goodness of fit; and never count one individual twice
Do two measured variables rise or fall together?Does moss cover increase with soil moisture?	Pairs of values measured on the same quadrats or samples, ideally 10 pairs or more	Spearman’s rank correlation	How strongly, and which way, the two are correlated (rs from −1 to +1), and whether that is significant	Comparing groups (use a t-test or ANOVA). A correlation never shows that one variable causes the other

How the numbers are rounded. Every example gives the numbers the calculator gives for the same data, and each has a button that loads them in to check. Means and standard deviations carry one decimal place more than the readings. In the sentences the test statistic has 2 decimal places and p has 3 (or reads p < 0.001), as the calculator’s own sentence does; its results card shows one place more of each. Critical values have 3 decimal places, as tables print them, at the 0.05 significance level the calculator uses. Work with unrounded values and round at the end.

A t-test compares the means of two groups. It asks whether the gap between them is big compared with how much the repeats vary, or small enough to be down to random variation. Use the unpaired t-test for two separate groups, and the paired t-test when the same individuals or samples are measured twice.

Worked example: stomata on sun and shade leaves

Stomata were counted on the lower surface of six leaves from the sunny side of a hedge and six from the shaded side, using nail varnish peels, and each count was converted to stomata per mm².

Null hypothesis (H₀)There is no difference in mean stomatal density between sun leaves and shade leaves.

1. Mean and standard deviation of each group: sun 221.8 ± 15.8, shade 189.3 ± 12.3 stomata per mm² (n = 6 each).
2. Pooled variance: s_p² = [(6 − 1) × 15.82² + (6 − 1) × 12.27²] ÷ (6 + 6 − 2) = 200.4
3. t = (x̄₁ − x̄₂) ÷ √(s_p²(1/n₁ + 1/n₂)) = (221.83 − 189.33) ÷ √(200.4 × (1/6 + 1/6)) = 32.50 ÷ 8.17 = 3.98
4. Degrees of freedom: df = n₁ + n₂ − 2 = 10
5. Critical value of t for df = 10 at p = 0.05 (two-tailed): 2.228. As 3.98 > 2.228, the difference is significant.

t

3.98

df

10

Critical t (p = 0.05)

2.228

p

0.003

The calculator’s t-test pools the two standard deviations (Student’s t-test), which assumes the two groups vary by a similar amount. Here they do: 15.8 and 12.3.

Paired t-test: the same potato cylinders, before and after

Six potato cylinders were weighed, left in distilled water for 30 minutes, blotted dry and weighed again. Each cylinder has its own before and after mass, so the two sets are paired: the test works on each cylinder’s change in mass, which takes the differences between cylinders out of the comparison.

Null hypothesis (H₀)The mean change in mass of the potato cylinders is zero.

1. Change in mass of each cylinder, after − before: the last row of the table.
2. Mean change d̄ = 0.137 g, standard deviation of the changes s_d = 0.027 g, n = 6
3. t = d̄ ÷ (s_d ÷ √n) = 0.1367 ÷ (0.02733 ÷ √6) = 0.1367 ÷ 0.01116 = 12.25
4. Degrees of freedom: df = n − 1 = 5
5. Critical value of t for df = 5 at p = 0.05 (two-tailed): 2.571. As 12.25 > 2.571, the change is significant.

t

12.25

df

5

Critical t (p = 0.05)

2.571

p

< 0.001

Chi-squared works on counts: the number of individuals in each category, never means, percentages or lengths. It adds up how far each observed count (O) is from the count expected (E): χ² = Σ (O − E)² ÷ E. A goodness of fit test takes E from a ratio you predict; a test of association works E out from the table itself. Every expected count should be 5 or more, or the test is not reliable.

Goodness of fit: do seedlings fit a 3 : 1 ratio?

Seeds from a cross between two tobacco plants, both heterozygous for a recessive albino allele, were germinated on damp paper. Of 150 seedlings, 104 were green and 46 albino. A monohybrid cross predicts 3 green : 1 albino.

Null hypothesis (H₀)There is no difference between the observed numbers of green and albino seedlings and the numbers expected from a 3 : 1 ratio.

1. Expected counts from the ratio: 150 × 3/4 = 112.5 green, 150 × 1/4 = 37.5 albino.
2. χ² = (104 − 112.5)² ÷ 112.5 + (46 − 37.5)² ÷ 37.5 = 0.642 + 1.927 = 2.569
3. Degrees of freedom: df = number of categories − 1 = 1
4. Critical value of χ² for df = 1 at p = 0.05: 3.841. As 2.57 < 3.841, the difference is not significant.

χ²

2.57

df

1

Critical χ² (p = 0.05)

3.841

p

0.109

Not significant is a result too: the counts are consistent with a 3 : 1 ratio, though that does not prove it. The calculator’s goodness of fit mode takes the expected counts, so work them out from the ratio first, as in step 1.

Test of association: a contingency table

Beside a footpath, every plantain plant in ten quadrats in each of three zones was identified as greater plantain (Plantago major) or ribwort plantain (Plantago lanceolata). Each plant is counted once, by its species and its zone.

Null hypothesis (H₀)There is no association between the species of plantain and the zone it grows in.

1. Expected count for each cell = row total × column total ÷ grand total. For greater plantain at the path edge: 65 × 45 ÷ 140 = 20.89.
2. Then (O − E)² ÷ E for each of the six cells, as for goodness of fit:

3. χ² = the sum of the six terms = 30.137 (added before rounding)
4. Degrees of freedom: df = (rows − 1) × (columns − 1) = (2 − 1) × (3 − 1) = 2
5. Critical value of χ² for df = 2 at p = 0.05: 5.991. As 30.14 > 5.991, the association is significant.

χ²

30.14

df

2

Critical χ² (p = 0.05)

5.991

p

< 0.001

The biggest terms show where the association lies: the path edge and the grass. The verge counts are almost exactly as expected. In the calculator, choose Test of association and type only the observed counts: it works out the expected counts, every term and df = (rows − 1) × (columns − 1) for you. Run as goodness of fit instead, six cells would wrongly get df = 5.

Spearman’s rank correlation coefficient, r_s, measures how consistently one variable rises (or falls) as the other rises. It works on ranks, not the values themselves, so the relationship does not have to be a straight line. r_s runs from −1 (a perfect negative correlation) through 0 (none) to +1 (a perfect positive correlation).

Worked example: moss cover and soil moisture

Ten quadrats were placed at 2 m intervals along a transect running down a slope to a stream. In each, soil moisture was read with a probe and the percentage cover of moss was estimated.

Null hypothesis (H₀)There is no correlation between soil moisture and the percentage cover of moss.

1. Rank each variable on its own, smallest = 1. Tied values share the mean of their ranks (two values in 3rd and 4th place both get 3.5); the calculator does this for you.
2. For each quadrat, d = the difference between its two ranks. Square each d and add them: Σd² = 24
3. r_s = 1 − (6Σd²) ÷ (n(n² − 1)) = 1 − (6 × 24) ÷ (10 × 99) = 1 − 144 ÷ 990 = 0.8545, so r_s = 0.85
4. Spearman tables are read with n, the number of pairs, rather than df: n = 10
5. Critical value of r_s for n = 10 at p = 0.05 (two-tailed): 0.648. As 0.85 > 0.648, the correlation is significant.

r_s

0.85

n

10

Critical r_s (p = 0.05)

0.648

p (approximate)

0.002

The calculator works p out from an approximation that is less exact below about 10 pairs, so with a small sample compare r_s with the critical value from a table, as in step 5. A significant correlation does not show cause: something else that changes along the transect, such as shade, could be behind both. And the R² that Step 02 gives for a line of best fit says how closely the means follow the line; it describes the fit and is not a test.

A one-way ANOVA compares the means of three or more groups in one test. It works out F: the variation between the group means divided by the variation within the groups. ANOVA only says whether at least one mean differs; Tukey’s honestly significant difference (HSD) test then compares every pair and says which. Running a t-test on every pair instead raises the risk of a false positive: three groups means three tests, each with a 5% chance of a significant result by luck alone.

Worked example: yeast respiring three sugars

Yeast was mixed with a 5% solution of glucose, sucrose or maltose at 30 °C, and the volume of carbon dioxide given off in 10 minutes was measured, five times for each sugar.

Null hypothesis (H₀)There is no difference in the mean volume of carbon dioxide given off by yeast with glucose, sucrose and maltose. For Tukey’s test, each pair has its own: there is no difference between the two means.

1. Grand mean of all 15 values: x̄_grand = 7.86 cm³
2. SS_between = Σ n_i(x̄_i − x̄_grand)² = 6.604, and SS_within = ΣΣ(x_ij − x̄_i)² = 3.012
3. Each mean square is SS ÷ df: MS_between = 6.604 ÷ 2 = 3.302, and MS_within = 3.012 ÷ 12 = 0.251
4. F = MS_between ÷ MS_within = 3.302 ÷ 0.251 = 13.16, with df = 2 (k − 1) and 12 (N − k)
5. Critical value of F for df = 2 and 12 at p = 0.05: 3.885. As 13.16 > 3.885, at least one mean differs. Now Tukey finds which.

6. For each pair, q = |x̄_i − x̄_j| ÷ √(MS_within ÷ n). For glucose and maltose: 1.580 ÷ √(0.251 ÷ 5) = 1.580 ÷ 0.224 = 7.05
7. Critical value of q for 3 groups and df = 12 at p = 0.05: 3.773. A pair whose q is above it differs significantly.

F

13.16

df

2 and 12

Critical F (p = 0.05)

3.885

p

< 0.001