Analyst Dojo

You are reading the 2026 curriculum (exams up to November 2026). Sitting in February 2027 or later? Go to the 2027 Quantitative Methods guide.

Learning Module 9 of Quantitative Methods 6–9% of the exam

Parametric and Non-Parametric Tests of Independence

3 lessons. About 20 minutes in cram mode, 90 minutes in full.

Start with lesson 1 Learn with Sensei Read the module summary

Lessons

  1. 1
  2. 2
    Spearman rank correlation: the nonparametric test

    explain 6 min cram, 26 min full

  3. 3

What the exam expects you to do

  1. Explain parametric and nonparametric tests of the hypothesis that the population correlation coefficient equals zero, and determine whether the hypothesis is rejected at a given level of significance.Is a correlation significant? The t-test of correlation
  2. Explain tests of independence based on contingency table data.Chi-square test of independence with a contingency table

Module summary

The whole module on one sheet, organised by lesson: what to memorize, how the ideas fit together, the formulas and the key terms. No paragraphs, just what you need to revise.

How deeply to learn each part
Know the idea
Understand it in general. It frames other topics, but you will not calculate with it.
Use it
Understand it well enough to solve a question with it. Recognize when it applies.
Memorize
Learn it by heart. The exam will not give it to you.
1

Is a correlation significant? The t-test of correlation

Known

MemorizeThe correlation t-test

ItemValue
Null hypothesisH0:ρ=0H_0: \rho = 0 (two-tailed, Ha:ρ≠0H_a: \rho \neq 0)
Statistict=rn−2 / 1−r2t = r\sqrt{n-2} \,/\, \sqrt{1-r^2}
Degrees of freedomn−2n - 2
Decision∣t∣>|t| > critical value → reject H0H_0

Memory hook Same df as a one-variable regression, which estimates two coefficients (intercept and slope): n − 2. The root of 1 − r² is the "noise" under the line.

Know it cold Reject only when |t| beats the critical value the question gives. Never "accept" H₀.

Key terms

  • parametric test of correlation Use it
  • sample correlation coefficient Know the idea
  • t-statistic for a correlation Memorize
  • degrees of freedom Memorize
  • statistically significant Use it
Reading the t-test of a correlation
Compute t from r and nH₀: ρ = 0, df = n − 2
  • |t| greater than the critical value Memorize a large t is far from zero
    • Reject H₀ the correlation is statistically significant
  • |t| at or below the critical value Memorize the sample could easily be noise
    • Fail to reject H₀ not significant, which is not the same as proving ρ = 0

Formulas

t-statistic for a correlation Memorize
t=rn−21−r2,df=n−2t = \dfrac{r\sqrt{n-2}}{\sqrt{1-r^2}},\quad df = n - 2
Quick example

A correlation of 0.52 from 27 observations. t = 0.52 × √25 ÷ √(1 − 0.2704) = 2.60 ÷ 0.854 = 3.04, with 25 degrees of freedom.

On your calculator
  1. 0.52×25√x÷(1−0.52x²)√x= 3.0439

√x square root · x² square

  1. 0.52ENTER25g√x×1ENTER0.52ENTER×−g√x÷ 3.0439

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · √x square root

2

Spearman rank correlation: the nonparametric test

Known

MemorizeSpearman in four steps

StepAction
1Rank XX (best to worst, or the reverse)
2Rank YY the same way; ties share the average rank
3dd = rank XX − rank YY; square and sum
4rs=1−6∑d2/[n(n2−1)]r_s = 1 - 6\sum d^2 / [n(n^2-1)]

Memory hook "1 minus 6 d-squared over n times n-squared-minus-one." If every rank matches, Σd² = 0 and rₛ = 1.

Know it cold Small sample: use the Spearman table the question gives. n > 30: use the t formula with n−2n-2 df.

Key terms

  • Spearman rank correlation coefficient Memorize
  • nonparametric test Use it
  • rank Use it
  • outlier Know the idea
Which test of a correlation?
Test H₀: ρ = 0choose by the data
  • Parametric t-test Memorize Pearson r on the actual values
    • Both variables roughly normal no extreme outliers
    • More powerful when its assumptions hold
  • Spearman rank test Memorize rₛ on the ranks
    • Data are ranks or ordinal fund rankings, credit grades
    • Outliers or non-normal data ranks tame them
    • Critical value small n: Spearman table; n above 30: t formula

Formulas

Spearman rank correlation Memorize
rs=1−6∑di2n (n2−1)r_s = 1 - \dfrac{6\sum d_i^2}{n\,(n^2 - 1)}
Quick example

8 pairs of rankings with Σd² = 14. rₛ = 1 − 6 × 14 ÷ (8 × 63) = 1 − 84 ÷ 504 = 0.833.

On your calculator
  1. 1−(6×14÷(8×(8x²−1)))= 0.8333

x² square

  1. 6ENTER14×8ENTERENTER×1−×÷CHS1+ 0.8333

ENTER enter (on the HP, puts the number on the stack) · CHS change sign

t-statistic for large samples (n > 30) Memorize
t=rsn−21−rs2,df=n−2t = \dfrac{r_s\sqrt{n-2}}{\sqrt{1-r_s^2}},\quad df = n - 2
Quick example

rₛ = 0.45 from 40 pairs. t = 0.45 × √38 ÷ √(1 − 0.2025) = 3.11, against a critical value of 2.024.

On your calculator
  1. 0.45×38√x÷(1−0.45x²)√x= 3.1063

√x square root · x² square

  1. 0.45ENTER38g√x×1ENTER0.45ENTER×−g√x÷ 3.1063

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · √x square root

3

Chi-square test of independence with a contingency table

Known

MemorizeChi-square test of independence

ItemRule
Null hypothesisthe two classifications are independent
Expected countE=row×column÷grand totalE = \text{row} \times \text{column} \div \text{grand total}
Statisticχ2=∑(O−E)2/E\chi^2 = \sum (O-E)^2 / E
Degrees of freedom(r−1)(c−1)(r-1)(c-1)
Tailright tail only: large χ2\chi^2 → reject

Memory hook "Row times column over total." Divide by E, not O: each cell is judged against what independence predicts. If O = E everywhere, χ² = 0.

Know it cold χ2>\chi^2 > critical value → reject independence. Expected counts below about 5 make the test unreliable.

Key terms

  • categorical data Know the idea
  • contingency table Use it
  • test of independence Use it
  • observed frequency Use it
  • expected frequency Memorize
  • chi-square distribution Use it
Which test of independence?
Are two characteristics linked?choose by the type of data
  • Two numerical variables Memorize returns, prices, ratios
    • Parametric t-test of correlation roughly normal data
    • Spearman rank test ranks, outliers, non-normal
  • Two categorical variables Memorize sector, rating, region
    • Chi-square test of independence counts in a contingency table
    • df (r − 1)(c − 1) right-tailed

Formulas

Expected frequency Memorize
Eij=(row i total)×(column j total)grand totalE_{ij} = \dfrac{(\text{row } i \text{ total}) \times (\text{column } j \text{ total})}{\text{grand total}}
Quick example

Row total 90, column total 60, grand total 300. E = 90 × 60 ÷ 300 = 18.

On your calculator
  1. 90×60÷300= 18
  1. 90ENTER60×300÷ 18

ENTER enter (on the HP, puts the number on the stack)

Chi-square statistic Memorize
χ2=∑i,j(Oij−Eij)2Eij\chi^2 = \sum_{i,j} \dfrac{(O_{ij} - E_{ij})^2}{E_{ij}}
Quick example

A 2 × 2 table with observed 40, 60, 60, 40 and every expected count 50. Each cell adds 100 ÷ 50 = 2, so χ² = 8.0.

On your calculator
  1. (40−50)x²÷50×4= 8

x² square

  1. 40ENTER50−ENTER×50÷4× 8

ENTER enter (on the HP, puts the number on the stack)

Degrees of freedom Memorize
df=(r−1)(c−1)df = (r - 1)(c - 1)
Quick example

A table with 3 rows and 4 columns. df = 2 × 3 = 6.

On your calculator
  1. (3−1)×(4−1)= 6
  1. 3ENTER1−4ENTER1−× 6

ENTER enter (on the HP, puts the number on the stack)

Standardized residual Use it
standardized residual=Oij−EijEij\text{standardized residual} = \dfrac{O_{ij} - E_{ij}}{\sqrt{E_{ij}}}
Quick example

A cell with O = 26 and E = 16. (26 − 16) ÷ √16 = +2.5: well above what independence predicts.

On your calculator
  1. 26−16=÷16√x= 2.5

√x square root

  1. 26ENTER16−16g√x÷ 2.5

ENTER enter (on the HP, puts the number on the stack) · g blue shift key · √x square root

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