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Interpreting data and evaluating developmental research

Edexcel IAL Psychology · WPS03 Applications of Psychology

Unit 3 can ask you to use List A (descriptive statistics) and List B (inferential statistics) in developmental contexts, and to evaluate studies (spec 5.3.4 and 5.3.5). Here they are applied to attachment and language data.

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Choosing the test

Question	Data	Design	Test
Difference	Nominal (categories, frequencies)	Independent groups	Chi-squared
Difference	At least ordinal	Repeated measures	Wilcoxon signed ranks
Relationship	At least ordinal (pairs of scores)	Correlation	Spearman's rank

List B tests

Chi-squared with real attachment data

From van IJzendoorn and Kroonenberg (1988): Germany had 48 avoidant and 88 not avoidant infants; Japan had 5 avoidant and 91 not avoidant.

Expected frequency = row total x column total / grand total. E.g. Germany avoidant: 136 x 53 / 232 = 31.07.
χ² = Σ (O - E)² / E = 28.90, df = (2 - 1) x (2 - 1) = 1.
Critical value (df 1, two-tailed, p ≤ .01) = 6.64. 28.90 ≥ 6.64, so the difference in avoidant attachment between Germany and Japan is significant at p ≤ .01: reject the null hypothesis.

df	One-tailed p ≤ .05	Two-tailed p ≤ .05	Two-tailed p ≤ .01
1	2.71	3.84	6.64
2	4.61	5.99	9.21
3	6.25	7.82	11.35

Extract from Pearson's chi-squared critical values table (observed must be equal to or greater than critical)

Spearman's rank with language data

Invented data for practice: hours of shared reading per week and vocabulary scores for 10 children.

Reading: 2, 5, 1, 7, 4, 6, 3, 8, 0, 5

Vocabulary: 38, 52, 30, 61, 47, 50, 41, 66, 28, 55

Rank each variable (ties share the average rank), find d for each child, square and add: Σd² = 6.5.
rs = 1 - 6Σd² / (n(n² - 1)) = 1 - 6 x 6.5 / (10 x 99) = 0.961.
Critical value (N = 10, two-tailed, p ≤ .05) = 0.648; at p ≤ .01 = 0.794. 0.961 ≥ both, so there is a significant positive correlation. It does not show that reading causes vocabulary growth.

N	One-tailed p ≤ .05	Two-tailed p ≤ .05	Two-tailed p ≤ .01
8	0.643	0.738	0.881
10	0.564	0.648	0.794
12	0.503	0.587	0.727

Extract from Pearson's Spearman's rank critical values table

Evaluating developmental research

Evaluate in the terms the specification uses: reliability, validity, generalisability, credibility, objectivity and subjectivity, ethics and practical application.

Type I error: claiming a significant effect that is really chance (more likely at p ≤ .10). Type II error: missing a real effect (more likely at p ≤ .01).
Descriptive statistics (List A): mean, median, mode, range, standard deviation, frequency tables, bar charts and histograms, percentages and ratios; normal and skewed distributions.
Always sense-check: in a one-tailed test, results must go in the predicted direction.

Worked example

A study finds χ² = 3.20 (df 1) comparing secure attachment in day-care and home-reared infants, with a two-tailed hypothesis. Is it significant at p ≤ .05?

Critical value for df 1, two-tailed, p ≤ .05 is 3.84.
3.20 is less than 3.84.
So it is not significant: keep the null hypothesis.

Answer: Not significant at p ≤ .05.

Key idea: Chi-squared for nominal, independent groups differences; Wilcoxon for repeated measures; Spearman for correlation. Germany v Japan avoidant attachment gives a significant chi-squared. Observed ≥ critical for chi-squared and Spearman; Wilcoxon T ≤ critical. Evaluate with reliability, validity, generalisability, credibility, objectivity, ethics, application.

Check yourself

Using Germany (48 avoidant, 88 not) and Japan (5 avoidant, 91 not), the expected frequency for Japan avoidant is row total x column total / grand total. Calculate it to 2 decimal places.
Show answer
21.93. 96 x 53 / 232 = 21.93.
What are the degrees of freedom for a 2 x 2 chi-squared table?
Show answer
1. (2 - 1) x (2 - 1) = 1.
For the invented reading and vocabulary data, Σd² = 6.5 and n = 10. Calculate Spearman's rs to 3 decimal places.
Show answer
0.961. 1 - 6 x 6.5 / (10 x 99) = 0.961.
A researcher records whether children in two separate nurseries show secure or insecure attachment. Which test should she use?
- Spearman's rank
- Wilcoxon signed ranks
- No test is possible
- Chi-squared
Show answer
Chi-squared. Nominal categories, independent groups, difference.
Using p ≤ .10 instead of p ≤ .05 increases the risk of
- a Type II error
- a Type I error
- attrition
- a cohort effect
Show answer
a Type I error. A lenient level makes false positives more likely.
Most children in a sample score low on a vocabulary test, with a few very high scores. The distribution is
- negatively skewed
- normal
- bimodal and symmetrical
- positively skewed
Show answer
positively skewed. The tail points to the high scores.
Explain why Spearman's rank is the appropriate test for the reading and vocabulary data.
Show answer
The study looks for a relationship (correlation) between two co-variables, each child gives a pair of scores, and the data are at least ordinal.. Hypothesis type plus data.

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