Statistics -- Diagnostic Tests

Statistics is like a lens for seeing patterns in data — it separates signal from noise, revealing what data actually tells us: Statistical inference lets us draw conclusions about entire populations from carefully chosen samples, bridging observation and generalization

Why it matters: From election polling to clinical trials to quality control, statistics is the science of learning from data

The key insight: Statistical inference lets us draw conclusions about entire populations from carefully chosen samples, bridging observation and generalization

Tests edge cases, boundary conditions, and common misconceptions for statistics.

Question:

For a dataset, the quartiles are $Q_1 = 42$, $Q_2 = 55$ And $Q_3 = 70$.

(a) Determine whether the data is positively skewed, negatively skewed, or symmetric.

(b) A student argues: “Since $Q_2 - Q_1 = 13$ and $Q_3 - Q_2 = 15$The data is positively skewed because $Q_3 - Q_2 \gt Q_2 - Q_1$.” Is this reasoning correct?

(c) If the interquartile range is $IQR = 28$State the outlier boundaries using the $1.5 \times IQR$ rule.

[Difficulty: hard. Tests interpretation of quartile positions to identify skewness and outlier detection.]

Solution:

(a) The distances from the median are:

• $Q_2 - Q_1 = 55 - 42 = 13$
• $Q_3 - Q_2 = 70 - 55 = 15$

Since $Q_3 - Q_2 \gt Q_2 - Q_1$The right tail is longer than the left tail, indicating positive skew.

(b) The student”s reasoning is correct in principle: positive skew means the right tail is longer. However, the student should note that this is a heuristic — formal skewness is measured by the moment coefficient $\frac{1}{n}\sum\left(\frac{x_i - \bar{x}}{s}\right)^3$Not just quartile differences. The quartile-based test is a quick check, not definitive proof.

(c) Lower fence: $Q_1 - 1.5 \times IQR = 42 - 42 = 0$. Upper fence: $Q_3 + 1.5 \times IQR = 70 + 42 = 112$.

Outliers are values below $0$ or above $112$.

Question:

A dataset has the following coded values. The coding is $y = \frac{x - 50}{10}$:

$\sum y = 45, \quad \sum y^2 = 285, \quad n = 9$

(a) Find $\bar{x}$ and $s_x$ (the standard deviation of $x$).

(b) A student computes $s_y = \sqrt{\frac{285}{9} - 25} = \sqrt{31.67 - 25} = \sqrt{6.67}$ and concludes $s_x = s_y$. Explain why this is wrong.

[Difficulty: hard. Tests coded data transformations and the effect on mean and standard deviation.]

Solution:

(a) $\bar{y} = \frac{45}{9} = 5$.

Since $y = \frac{x - 50}{10}$We have $x = 10y + 50$:

$\bar{x} = 10\bar{y} + 50 = 10(5) + 50 = 100$

For the standard deviation: $s_x = 10s_y$.

$s_y = \sqrt{\frac{\sum y^2}{n} - \bar{y}^2} = \sqrt{\frac{285}{9} - 25} = \sqrt{31.67 - 25} = \sqrt{6.67} \approx 2.58$

$s_x = 10 \times 2.58 = 25.8$

(b) The student’s error is concluding $s_x = s_y$. The coding $y = \frac{x-50}{10}$ scales by a factor of $\frac{1}{10}$ and shifts by $50$. Scaling by $c$ multiplies the standard deviation by $|c|$ So $s_x = 10s_y$Not $s_y$. The student forgot to account for the scaling factor. Additionally, the student used $\frac{285}{9} \approx 31.67$ and then subtracted $25$ (where $25 = 5^2$), which is correct for computing $s_y$ But then incorrectly applied the result to $s_x$.

Tests synthesis of statistics with other topics.

Question:

Given five data points $(x_i, y_i)$ with \sum x_i = 15$$\sum y_i = 20$$\sum x_i^2 = 55$$\sum x_iy_i = 68 And $\sum y_i^2 = 90$:

(a) Find the equation of the least squares regression line of $y$ on $x$ in the form $y = a + bx$.

(b) Find PMCC (Pearson product-moment correlation coefficient).

(c) Predict the value of $y$ when $x = 5$.

[Difficulty: hard. Combines regression computation with correlation analysis.]

Solution:

(a)

$b = \frac{n\sum x_iy_i - \sum x_i \sum y_i}{n\sum x_i^2 - (\sum x_i)^2} = \frac{5(68) - 15(20)}{5(55) - 225} = \frac{340 - 300}{275 - 225} = \frac{40}{50} = 0.8$

$a = \bar{y} - b\bar{x} = \frac{20}{5} - 0.8 \times \frac{15}{5} = 4 - 2.4 = 1.6$

Regression line: $y = 1.6 + 0.8x$.

(b)

$r = \frac{n\sum x_iy_i - \sum x_i\sum y_i}{\sqrt{\big[n\sum x_i^2 - (\sum x_i)^2\big]\big[n\sum y_i^2 - (\sum y_i)^2\big]}}$

$= \frac{340 - 300}{\sqrt{(275 - 225)(450 - 400)}} = \frac{40}{\sqrt{50 \times 50}} = \frac{40}{50} = 0.8$

(c) When $x = 5$: $y = 1.6 + 0.8(5) = 1.6 + 4 = 5.6$.

This content page provides comprehensive coverage of Maths content for the Ib qualification, with detailed explanations, worked examples, and practice questions aligned to the specification.

This page includes:

• Key Definitions: Precise explanations of essential concepts
• Core Concepts: Detailed treatment of fundamental principles
• Worked Examples: Step-by-step solutions demonstrating application
• Practice Questions: Examination-style questions with mark schemes
• Common Pitfalls: Frequent errors and how to avoid them
• Exam Tips: Strategies for maximising marks

1. Read through the introductory material to establish context
2. Study the definitions and core concepts carefully
3. Work through the worked examples, following each step
4. Attempt the practice questions independently
5. Review your answers against the provided solutions
6. Note any areas requiring further revision

flowchart TD
    A[Diag Statistics] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

• Foundational definitions and terminology
• Application of principles to examination contexts
• Connections to related topics within the specification
• Assessment objective alignment

• Active Recall: Test yourself on the material rather than passively re-reading
• Spaced Repetition: Review this content at increasing intervals
• Interleaving: Mix this topic with others during study sessions
• Elaborative Interrogation: Ask yourself why each concept works

Practise applying these concepts under timed conditions. Focus on understanding what each question is asking and how marks are allocated. Review examiner reports to learn from common mistakes made by other students.

• Diagnostics
• Differentiation — Diagnostic Tests
• Integration — Diagnostic Tests