Research Methodology II: Central Tendency and Dispersion, the Normal Probability Curve, the t-Test and the Non-Parametric Tests, Power Analysis and Effect Size — Worked
2. The normal probability curve
The normal probability curve is bell-shaped, symmetrical and unimodal, with mean = median = mode at the centre; its tails are asymptotic (they approach but never touch the baseline); its skewness is zero and its kurtosis mesokurtic (β₂ = 3). A more peaked curve is leptokurtic (β₂ > 3), a flatter one platykurtic (β₂ < 3). Areas: 68.26 per cent of cases lie within ±1σ of the mean, 95.44 per cent within ±2σ and 99.74 per cent within ±3σ; 34.13 per cent lie between the mean and +1σ, and 1.96σ on either side encloses exactly 95 per cent.
A raw score becomes a z-score by z = (X − M) ÷ σ, which has mean 0 and SD 1; the T-score removes negatives and decimals as T = 50 + 10z. Worked: on a test with M = 50 and σ = 10, a score of 65 gives z = (65 − 50) ÷ 10 = 1.5 and T = 50 + 15 = 65; a score of 40 gives z = −1 and T = 40. Since 84.13 per cent of the curve lies below +1σ, a z of +1 is roughly the 84th percentile. Uses of the curve in psychology: converting scores to percentiles and standard scores, normalising distributions, dividing a group into ability categories, and determining the relative difficulty of items.
3. Hypothesis testing and the t-test
A test decides whether to reject H₀. Rejecting a true H₀ is a Type I error, with probability α (the significance level, usually 0.05); retaining a false H₀ is a Type II error, with probability β. The t-test compares means when σ is estimated from the sample. It assumes interval or ratio data, approximate normality, independence of observations and, for independent groups, homogeneity of variance. The independent-samples t has df = n₁ + n₂ − 2; the paired (related) t has df = n − 1 pairs; the one-sample t has df = n − 1.
| Step | Worked example: two groups of 10, M₁ = 24, M₂ = 20, s₁ = s₂ = 4 |
|---|---|
| Pooled SD | sₚ = √{[(9 × 16) + (9 × 16)] ÷ 18} = √16 = 4 |
| Standard error of the difference | SE = sₚ √(1/n₁ + 1/n₂) = 4 × √0.2 ≈ 4 × 0.447 = 1.79 |
| t | t = (24 − 20) ÷ 1.79 ≈ 2.24, with df = 10 + 10 − 2 = 18 |
| Decision | Critical t (two-tailed, 0.05, df 18) = 2.10; 2.24 > 2.10, so reject H₀ |
| Effect size | d = (24 − 20) ÷ 4 = 1.0, a large effect |
4. The non-parametric tests: which replaces which
Non-parametric (distribution-free) tests make no assumption about the shape of the population, work on ranks or signs, and suit ordinal data, small samples and badly skewed distributions. They are somewhat less powerful than their parametric counterparts when the parametric assumptions actually hold. The key exam skill is matching each to the parametric test it replaces, which depends on two things: how many groups, and whether they are independent (different people) or related (the same or matched people).
| Design | Parametric test | Non-parametric replacement | What it uses |
|---|---|---|---|
| Two related samples | Paired t-test | Sign test | Only the direction (+ or −) of each difference; binomial test |
| Two related samples | Paired t-test | Wilcoxon signed-rank test | Direction and the rank of the size of each difference; T = the smaller sum of signed ranks — more powerful than the sign test |
| Two independent samples | Independent-samples t-test | Mann–Whitney U test | Ranks of all scores pooled; U = n₁n₂ + n₁(n₁ + 1) ÷ 2 − R₁ |
| Three or more independent samples | One-way ANOVA | Kruskal–Wallis H test | Ranks of all scores pooled; H is referred to chi-square with k − 1 df |
| Three or more related samples | Repeated-measures ANOVA | Friedman test | Ranks within each participant across conditions |
Mann–Whitney worked. Group A has n₁ = 4 and group B n₂ = 5; when all nine scores are ranked together, A's ranks sum to R₁ = 12. Then U₁ = (4 × 5) + (4 × 5) ÷ 2 − 12 = 20 + 10 − 12 = 18, and U₂ = n₁n₂ − U₁ = 20 − 18 = 2. The test statistic is the smaller, U = 2, which is compared with the tabled critical value (smaller U means a larger difference). Check: the ranks 1 to 9 sum to 45, so R₂ = 33, and U₂ = 20 + (5 × 6) ÷ 2 − 33 = 20 + 15 − 33 = 2.
5. Power analysis and effect size
Power is the probability of rejecting H₀ when it is false: power = 1 − β. Jacob Cohen recommended a power of at least 0.80 (so β = 0.20). Power rises with (1) a larger effect size, (2) a larger sample size, (3) a more lenient α (0.05 rather than 0.01), (4) a one-tailed test in the predicted direction, and (5) lower error variance (reliable measures, repeated-measures designs). Power analysis is used before a study (a priori) to find the sample size needed to detect an expected effect with a given α and power, and after it to interpret a non-significant result.
Effect size measures how large a difference or relationship is, independent of sample size — a very large sample can make a trivial difference "significant". Cohen's d = (M₁ − M₂) ÷ sₚ, the mean difference in standard-deviation units. Cohen's benchmarks: d = 0.2 small, 0.5 medium, 0.8 large; for r, 0.1, 0.3 and 0.5; for η² (proportion of variance explained in ANOVA), 0.01, 0.06 and 0.14. Worked: a therapy group scores 108 and a control group 100 on a scale with a pooled SD of 16, so d = 8 ÷ 16 = 0.5, a medium effect.
Key takeaways
- In positive skew mean > median > mode; Mode ≈ 3 Median − 2 Mean; CV = (σ ÷ M) × 100 compares variability across scales.
- The normal curve places 68.26, 95.44 and 99.74 per cent within ±1, ±2 and ±3σ; z = (X − M) ÷ σ and T = 50 + 10z.
- Independent t has df = n₁ + n₂ − 2 and paired t df = n − 1; Type I error is α, Type II is β.
- Mann–Whitney replaces the independent t, Wilcoxon (and the sign test) the paired t, Kruskal–Wallis one-way ANOVA, and Friedman repeated-measures ANOVA.
- Power = 1 − β, conventionally 0.80, and rises with effect size, sample size, α and one-tailed testing; Cohen's d = mean difference ÷ pooled SD, with 0.2, 0.5 and 0.8 as small, medium and large.
Practice questions (10)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
On a test with a mean of 50 and a standard deviation of 10, a student scores 65. What is the student's z-score? Type a number.
Numerical answer — type the value.
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Answer: 1.5
z = (X − M) ÷ σ = (65 − 50) ÷ 10 = 15 ÷ 10 = 1.5. The student is one and a half standard deviations above the mean, and the equivalent T-score is 50 + 10 × 1.5 = 65.A therapy group has a mean of 108 and a control group a mean of 100 on a depression-recovery scale; the pooled standard deviation is 16. What is Cohen's d? Type a number.
Numerical answer — type the value.
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Answer: 0.5
d = (M₁ − M₂) ÷ sₚ = (108 − 100) ÷ 16 = 8 ÷ 16 = 0.5, which Cohen classes as a medium effect (0.2 small, 0.5 medium, 0.8 large).A researcher compares anxiety ratings (ordinal) of the SAME participants before and after training, and the difference scores are badly skewed. The most appropriate test is:
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Answer: D — Wilcoxon signed-rank test
Two related measurements on ordinal, skewed data call for the non-parametric replacement of the paired t-test, which is the Wilcoxon signed-rank test. Mann–Whitney is for two independent groups and Kruskal–Wallis for three or more independent groups.The Friedman test is the non-parametric counterpart of:
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Answer: B — Repeated-measures (related-samples) ANOVA
Friedman ranks scores within each participant across three or more conditions, replacing repeated-measures ANOVA. Kruskal–Wallis replaces one-way independent ANOVA, and Mann–Whitney the independent t.In a Mann–Whitney test, group A (n₁ = 4) and group B (n₂ = 5) are ranked together, and the ranks of group A sum to 12. What is the value of U (the smaller of U₁ and U₂)? Type a number.
Numerical answer — type the value.
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Answer: 2
U₁ = n₁n₂ + n₁(n₁ + 1) ÷ 2 − R₁ = 20 + 10 − 12 = 18, and U₂ = n₁n₂ − U₁ = 20 − 18 = 2. The statistic is the smaller value, U = 2. As a check, R₂ = 45 − 12 = 33 and U₂ = 20 + 15 − 33 = 2.An independent-samples t-test compares a group of 12 with a group of 15. How many degrees of freedom does it have? Type a number.
Numerical answer — type the value.
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Answer: 25
For independent samples, df = n₁ + n₂ − 2 = 12 + 15 − 2 = 25. One degree of freedom is lost for each group mean estimated.Other things being equal, which of the following INCREASE the statistical power of a test? Select all that apply.
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Answer: A — A larger sample size; C — A larger true effect size
Power grows with sample size and effect size. A stricter α (0.01) makes H₀ harder to reject and lowers power, and greater error variance blurs the effect and also lowers power.In a positively skewed distribution, the usual order of the three averages is:
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Answer: A — Mean > median > mode
The long right tail pulls the mean up most, the median less, and the mode (the peak) least, giving mean > median > mode. The order reverses in negative skew, and the three are equal only in a symmetrical distribution.What percentage of cases in a normal distribution lies between the mean and +1 standard deviation? Type the number to two decimal places.
Numerical answer — type the value.
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Answer: 34.13
68.26 per cent of cases lie within ±1σ, and the curve is symmetrical, so half of that — 34.13 per cent — lies between the mean and +1σ.Assertion (A): A statistically significant result may nonetheless reflect a trivially small effect. Reason (R): With a very large sample, the standard error becomes so small that even a tiny difference between means yields a significant test statistic. Choose the correct option.
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Answer: A — Both A and R are true, and R is the correct explanation of A
Both are true and R explains A: significance depends on sample size through the standard error, while effect size does not — which is why an effect size such as d must be reported with every test.