How to Solve: IB AI HL/SL – Statistical Tests (t-test, Chi-squared, Spearman’s Rank, p-Values)

How to Solve: IB AI HL/SL – Statistical Tests (t-test, Chi-squared, Spearman’s Rank, p-Values)

Introduction

"Mastering statistical tests in IB Math AI HL/SL can earn you 10-15% of your final exam score—and help you answer real-world questions like: ‘Is this new drug effective?’ or ‘Is there a link between study time and grades?’ Today, you’ll learn the exact steps to ace t-tests, Chi-squared, Spearman’s Rank, and p-values—so you can walk into your exam confident and prepared."

What You Need To Know First

Before diving in, ensure you understand: 1. Hypothesis Testing Basics – Null hypothesis (H₀) vs. alternative hypothesis (H₁), significance level (α). 2. Normal Distribution & Critical Values – How to read statistical tables (t-distribution, Chi-squared, Spearman’s). 3. Basic Probability – p-values, Type I/II errors.

Key Vocabulary

Formulas To Know

1. t-test (Independent Samples)

Formula: [ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} ] Variables: - (\bar{x}_1, \bar{x}_2) = sample means - (s_1^2, s_2^2) = sample variances - (n_1, n_2) = sample sizes MEMORISE THIS (but check if given on your exam sheet).

Degrees of Freedom (df): [ df = n_1 + n_2 - 2 ]

2. Chi-squared (χ²) Test for Independence

Formula: [ \chi^2 = \sum \frac{(O - E)^2}{E} ] Variables: - (O) = observed frequency - (E) = expected frequency (row total × column total ÷ grand total) GIVEN ON EXAM SHEET (but know how to use it).

Degrees of Freedom (df): [ df = (r - 1)(c - 1) ] (where (r) = rows, (c) = columns)

3. Spearman’s Rank Correlation Coefficient

Formula: [ r_s = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)} ] Variables: - (d_i) = difference between ranks - (n) = number of pairs MEMORISE THIS (but check exam sheet).

Interpretation: - (r_s = 1) → perfect positive correlation - (r_s = -1) → perfect negative correlation - (r_s = 0) → no correlation

4. p-Value Interpretation

• If p < α → Reject H₀ (statistically significant).
• If p ≥ α → Fail to reject H₀ (not significant). MEMORISE THIS LOGIC (not a formula, but critical).

Step-by-Step Method

General Hypothesis Testing Steps (Applies to All Tests)

1. State Hypotheses
2. Write H₀ (no effect) and H₁ (effect).

3. Example: H₀: μ₁ = μ₂ vs. H₁: μ₁ ≠ μ₂ (two-tailed t-test).

4. Choose Significance Level (α)

5. Usually α = 0.05 (given in exam).

6. Calculate Test Statistic

7. Use the correct formula (t, χ², or rₛ).

8. Find Critical Value or p-Value

9. Critical Value Approach: Compare test statistic to table value.

10. p-Value Approach: Compare p-value to α.

11. Make Decision

12. If test statistic > critical value OR p < α → Reject H₀.

13. Otherwise → Fail to reject H₀.

14. Write Conclusion in Context

15. Example: "There is sufficient evidence to conclude that the drug has an effect."

Worked Example: t-test (Independent Samples)

Question: Two groups of students took a math test. Group A (n=10) had a mean score of 75 with s=8. Group B (n=12) had a mean of 70 with s=6. Test at α=0.05 if Group A performed better.

Step-by-Step Solution:

1. State Hypotheses
2. H₀: μ₁ = μ₂ (no difference)

3. H₁: μ₁ > μ₂ (Group A is better) → One-tailed test

4. Significance Level

5. α = 0.05

6. Calculate t-statistic
   [ t = \frac{75 - 70}{\sqrt{\frac{8^2}{10} + \frac{6^2}{12}}} = \frac{5}{\sqrt{6.4 + 3}} = \frac{5}{3.07} ≈ 1.63 ]

7. Degrees of Freedom
   [ df = 10 + 12 - 2 = 20 ]

8. Find Critical Value

9. One-tailed t-table (df=20, α=0.05) → 1.725

10. Compare t to Critical Value

11. 1.63 < 1.725 → Fail to reject H₀

12. Conclusion
    "There is insufficient evidence to conclude that Group A performed better."

What we did and why: - Used a one-tailed t-test because the question asked "better" (directional). - Compared t to critical value (not p-value) since the exam often expects this. - Failed to reject H₀ because the test statistic was less extreme than the critical value.

Worked Examples

Example 1 – Basic: Chi-squared Test

Question: A survey asked 100 students if they prefer online or in-person learning. Test at α=0.05 if preference is independent of grade level.

Solution:

1. Hypotheses
2. H₀: Preference is independent of grade level.

3. H₁: Preference depends on grade level.

4. Expected Frequencies (E)

5. E (Grade 11, Online) = (50 × 55) / 100 = 27.5
6. E (Grade 11, In-Person) = (50 × 45) / 100 = 22.5
7. E (Grade 12, Online) = 27.5

8. E (Grade 12, In-Person) = 22.5

9. Calculate χ²
   [ \chi^2 = \frac{(30-27.5)^2}{27.5} + \frac{(20-22.5)^2}{22.5} + \frac{(25-27.5)^2}{27.5} + \frac{(25-22.5)^2}{22.5} ]
   [ = \frac{6.25}{27.5} + \frac{6.25}{22.5} + \frac{6.25}{27.5} + \frac{6.25}{22.5} ≈ 0.227 + 0.278 + 0.227 + 0.278 = 1.01 ]

10. Degrees of Freedom
    [ df = (2-1)(2-1) = 1 ]

11. Critical Value (α=0.05, df=1)

12. χ² table → 3.841

13. Decision

14. 1.01 < 3.841 → Fail to reject H₀

15. Conclusion
    "There is no evidence that learning preference depends on grade level."

Example 2 – Medium: Spearman’s Rank

Question: 6 students ranked their math and science difficulty (1=easiest, 6=hardest). Test if there’s a correlation at α=0.05.

Solution:

1. Hypotheses
2. H₀: No correlation between math and science difficulty.

3. H₁: There is a correlation (two-tailed).

4. Calculate Differences (d) and d²
   | Student | Math (x) | Science (y) | d = x-y | d² |
   |---------|----------|-------------|---------|----|
   | A | 2 | 3 | -1 | 1 |
   | B | 1 | 1 | 0 | 0 |
   | C | 4 | 5 | -1 | 1 |
   | D | 3 | 2 | 1 | 1 |
   | E | 6 | 6 | 0 | 0 |
   | F | 5 | 4 | 1 | 1 |
   Σd² = 4

5. Calculate rₛ
   [ r_s = 1 - \frac{6 \times 4}{6(36 - 1)} = 1 - \frac{24}{210} ≈ 0.886 ]

6. Critical Value (n=6, α=0.05, two-tailed)

7. Spearman’s table → 0.886

8. Decision

9. 0.886 ≥ 0.886 → Reject H₀

10. Conclusion
    "There is a significant positive correlation between math and science difficulty rankings."

Example 3 – Exam-Style: p-Value Interpretation

Question: A t-test gives p=0.03. The significance level is α=0.05. What conclusion can you draw?

Solution:

1. Compare p to α

2. p = 0.03 < 0.05 → Reject H₀

3. Conclusion
   "There is sufficient evidence to reject the null hypothesis at the 5% significance level."

What we did and why: - Used p-value approach (common in exams). - Rejected H₀ because p was less than α, meaning the result is statistically significant.

Common Mistakes

Exam Traps

1-Minute Recap

"Here’s your last-minute cheat sheet for statistical tests in IB Math AI:

1. t-test: Compare means of two groups. Formula: ((\bar{x}_1 - \bar{x}_2) / \text{SE}). Degrees of freedom = n₁ + n₂ – 2.
2. Chi-squared: Test independence in tables. Formula: (\sum (O-E)^2 / E). Degrees of freedom = (rows-1)(columns-1).
3. Spearman’s Rank: Correlation for ranked data. Formula: (1 - 6 \sum d^2 / n(n^2-1)).
4. p-value: If p < α, reject H₀. If p ≥ α, fail to reject.
5. Always state hypotheses, check assumptions, and write a conclusion in context.

Now go ace that exam!