FAD1015: MATHEMATICS III — Exam Focus: Leak Topics

Based on LEAK/EXAM TIPS. Lecturer solved in ~35 minutes. All four topics below are likely Section B (longer, higher-weight questions).

#	Topic	Type	Weight
1	R Language — Matrix Operations	Section B	Heavy
2	R Language — Descriptive Stats Output	Section B	Moderate
3	Permutation & Counting	Section B	Heavy
4	Hypothesis Testing	Section B	Heavy

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R Matrix Operations
R Descriptive Stats Output
Permutation & Counting
Hypothesis Testing
Quick-Reference Tables
Revision Checklists

1. R Language — Matrix Operations (Section B)

1.1 Creating Matrices

`matrix()` — By Column (Default)

# 3x3 matrix filled column-wise by default
A <- matrix(c(1, 2, 3, 4, 5, 6, 7, 8, 9), nrow = 3, ncol = 3)
#      [,1] [,2] [,3]
# [1,]    1    4    7
# [2,]    2    5    8
# [3,]    3    6    9

# Fill row-wise instead
B <- matrix(1:9, nrow = 3, byrow = TRUE)
#      [,1] [,2] [,3]
# [1,]    1    2    3
# [2,]    4    5    6
# [3,]    7    8    9

Exam trap: Remember matrix() fills by column by default! Use byrow = TRUE for row-wise fill.

`rbind()` — Bind Rows

row1 <- c(1, 2, 3)
row2 <- c(4, 5, 6)
row3 <- c(7, 8, 9)
M <- rbind(row1, row2, row3)

`cbind()` — Bind Columns

col1 <- c(1, 4, 7)
col2 <- c(2, 5, 8)
col3 <- c(3, 6, 9)
M <- cbind(col1, col2, col3)

1.2 Naming Rows and Columns

A <- matrix(1:9, nrow = 3)
rownames(A) <- c("Row1", "Row2", "Row3")
colnames(A) <- c("Col1", "Col2", "Col3")

1.3 Subsetting Matrices

A[i, j]      # Element at row i, column j
A[i, ]       # Entire row i
A[, j]       # Entire column j
A[1:2, ]     # First two rows
A[, -2]      # All columns except column 2
A[c(1,3), c(2,3)]  # Rows 1 & 3, columns 2 & 3

Warning: Subsetting a single column/row may return a vector (not a matrix). Use drop = FALSE to preserve matrix structure:
A[, 1, drop = FALSE]  # Returns a 3x1 matrix, not a vector

1.4 Matrix Operations

Operation	R Syntax	Notes
Addition	`A + B`	Element-wise, same dimensions
Subtraction	`A - B`	Element-wise
Scalar multiplication	`2 * A`	Multiplies every element
Matrix multiplication	`A %*% B`	Use `%%`, NOT ``!
Element-wise multiply	`A * B`	NOT matrix multiplication
Transpose	`t(A)`	Switches rows & columns
Determinant	`det(A)`	Only for square matrices
Inverse	`solve(A)`	Only if `det(A) != 0`
Diagonal	`diag(A)`	Extracts diagonal elements
Create diagonal matrix	`diag(c(1,2,3))`	Returns a diagonal matrix
Solve linear system	`solve(A, b)`	Solves $Ax = b$
Eigenvalues	`eigen(A)$values`	For square matrices
Eigenvectors	`eigen(A)$vectors`

Critical distinction — * vs %*%:

A <- matrix(1:4, nrow = 2)   # A = [1 3; 2 4]
B <- matrix(4:1, nrow = 2)   # B = [4 2; 3 1]

A * B       # Element-wise:  [1*4  3*2; 2*3  4*1] = [4  6; 6  4]
A %*% B     # Matrix mult:   [1*4+3*3  1*2+3*1; 2*4+4*3  2*2+4*1] = [13  5; 20  8]

1.5 Dimension Properties

dim(A)      # Returns c(nrow, ncol)
nrow(A)     # Number of rows
ncol(A)     # Number of columns
length(A)   # Total elements (nrow * ncol)

1.6 Likely Exam Question Format

Question type: You are given data on (e.g.) test scores from different groups. Create a matrix in R, compute summary statistics using matrix operations, find the inverse, solve a system, or interpret output.

Example question:

The marks of three students in three subjects are given below:

Student 1: Maths 80, Stats 75, CS 90

Student 2: Maths 65, Stats 85, CS 70

Student 3: Maths 92, Stats 78, CS 88

(a) Create a matrix marks in R using rbind(). Assign row names and column names. (b) Find the total marks for each student using matrix multiplication with a vector of weights (1,1,1). (c) Compute the average marks per subject using colMeans(). (d) Find the determinant and inverse of the marks matrix. What does the determinant tell you?

Model answer:

# (a)
marks <- rbind(c(80, 75, 90),
               c(65, 85, 70),
               c(92, 78, 88))
rownames(marks) <- c("Student1", "Student2", "Student3")
colnames(marks) <- c("Maths", "Stats", "CS")

# (b) — matrix multiplication with weights vector
weights <- c(1, 1, 1)
total_marks <- marks %*% weights  # returns a 3x1 matrix

# (c)
colMeans(marks)

# (d)
det(marks)
solve(marks)  # if det != 0

[!warning] Hypothesis Testing in R NOT Tested Chen Jing confirmed: Hypothesis testing IN R is NOT coming out. All Q5 work is by-hand (Z-table, t-table, formulas). The t.test(), z.test(), and shapiro.test() R functions and their output interpretation will NOT appear. Only summary() descriptive stats output is relevant for R output questions.

2. R Language — Descriptive Stats Output (Section B)

2.1 Interpreting `summary()` Output for Vectors

summary(heights)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  165.0   168.0   170.0   169.9   172.0   175.0

2.2 Common Exam Traps

Trap	Correct Interpretation
"p-value = 0.03 means there is a 3% chance $H_0$ is true"	Wrong. p-value is P(data or more extreme \| $H_0$ true), not P($H_0$ true \| data).
"The 95% CI contains $\mu_0$, so we accept $H_0$"	Better: We "fail to reject $H_0$". We never "accept" $H_0$.
"t = -1.68 means the test is not significant"	Not directly. You must compare p-value to $\alpha$ or compare
"Since $n$ is small, use z-test"	Wrong. Small $n$ + $\sigma$ unknown → t-test.
"p-value > $\alpha$ proves $H_0$ is true"	Wrong. It just means insufficient evidence to reject $H_0$.
Mistaking `` for `%%` in R	`A * B` is element-wise; `A %*% B` is matrix multiplication.

3. Permutation & Counting (Section B)

3.1 Quick Formula Reference

Situation	Formula	When to Use
Permutation (no repetition)	$P(n,r) = \dfrac{n!}{(n-r)!}$	Order matters, pick $r$ from $n$, no reuse
Permutation (with repetition)	$n^r$	Order matters, pick $r$ from $n$, reuse allowed
Permutation (identical objects)	$\dfrac{n!}{n_1!,n_2!,\cdots,n_k!}$	Arranging $n$ objects where some are identical
Circular permutation (different orientations)	$(n-1)!$	Round table, rotations count as same
Circular permutation (same when flipped)	$\dfrac{(n-1)!}{2}$	Ring/necklace, flip is same as rotation
Combination	$C(n,r) = \dbinom{n}{r} = \dfrac{n!}{r!(n-r)!}$	Order does NOT matter
Multiplication rule	$m \times n \times \cdots$	Sequential choices from categories
Complementary counting	Total $-$ Invalid	"At least one", "not together" problems
Grouping method	(block arrangements) $\times$ (within-block arrangements)	Items that must be together

3.2 Decision Flowchart

graph TD
    START([Problem]) --> Q1{Does order matter?}
    Q1 -->|No| COMB[Combination C(n,r)]
    Q1 -->|Yes| Q2{Restrictions?}
    Q2 -->|Items together| GROUP[Grouping method<br/>block × internal]
    Q2 -->|Items apart| COMPL[Total − together]
    Q2 -->|First/last fixed| FIX[Fix positions,<br/>arrange remaining]
    Q2 -->|No restrictions| Q3{All objects used?}
    Q3 -->|Yes| Q4{Identical objects?}
    Q4 -->|Yes| IDENT[n! / n₁! n₂! ...]
    Q4 -->|No| ALL[n!]
    Q3 -->|No, pick r from n| Q5{Repetition allowed?}
    Q5 -->|Yes| REP[n^r]
    Q5 -->|No| NPR[P(n,r)]

    style START fill:#e7f5ff,stroke:#1971c2
    style Q1 fill:#fff4e6,stroke:#e67700
    style COMB fill:#d3f9d8,stroke:#2f9e44
    style GROUP fill:#e5dbff,stroke:#5f3dc4
    style COMPL fill:#e5dbff,stroke:#5f3dc4
    style FIX fill:#e5dbff,stroke:#5f3dc4
    style ALL fill:#ffe8cc,stroke:#d9480f
    style IDENT fill:#ffe8cc,stroke:#d9480f
    style REP fill:#ffe8cc,stroke:#d9480f
    style NPR fill:#ffe8cc,stroke:#d9480f

3.3 Multi-Step Section B Problem Types

Type 1: Arrangements with Restrictions (Classic Section B)

Eight members of a club stand in a line for a photo. Nayla and Dura refuse to stand next to each other. Find the number of possible arrangements.

Solution (complementary counting):

Total arrangements without restriction: $8! = 40,320$
Arrangements with Nayla and Dura together: treat them as one block → $7! \times 2! = 10,080$
Valid arrangements: $40,320 - 10,080 = 30,240$

Type 2: Words with Identical Letters

How many ways can the letters of STATISTICS be arranged? $$\frac{10!}{3! \times 3! \times 2!}$$ (3 S's, 3 T's, 2 I's)

Type 3: Combined Counting + Probability

A committee of 5 is chosen from 6 boys and 4 girls. Find the probability that there are more boys than girls.

Solution:

More boys than girls means: 4B1G or 5B0G
$P = \dfrac{C(6,4)C(4,1) + C(6,5)}{C(10,5)}$

Type 4: Circular Arrangements

In how many ways can 6 people sit at a round table? $$(6-1)! = 5! = 120$$

How many ways can 10 different coloured beads be arranged to form a ring? $$\frac{(10-1)!}{2} = \frac{9!}{2} = 181,440$$

Type 5: Permutation with Repetition + Restrictions

Four-digit numbers are formed from ${1, 2, 3, 4, 5, 6}$. How many are greater than 2000? First digit ∈ {2,3,4,5,6} → 5 choices Remaining 3 digits: $6 \times 6 \times 6$ Total: $5 \times 6^3 = 1080$

3.4 Permutation vs Combination — Quick Test

Phrase	Order Matters?	Use
"arrange in a row"	Yes	Permutation
"select a committee"	No	Combination
"PIN code"	Yes	Permutation
"choose a team"	No	Combination
"sit in a line"	Yes	Permutation
"sit around a table"	Yes (circular)	$(n-1)!$
"number of ways to pick"	Usually No	Combination
"arrange the letters"	Yes	Permutation

3.5 Likely Section B Question Structure

Expect a multi-part question like:

(a) How many arrangements of the word PROBABILITY are possible? [3 marks] (b) In how many of these arrangements do the two B's appear together? [3 marks] (c) If the letters are arranged at random, what is the probability the two B's are separated? [3 marks] (d) How many ways can the letters be arranged around a circle? [3 marks]

4. Hypothesis Testing (Section B)

4.1 The 4-Step Framework (Taught in L23-L24)

graph LR
    S1["Step 1: State H₀ and H₁"] --> S2["Step 2: Compute test statistic<br/>Z or t"]
    S2 --> S3["Step 3: Find p-value or<br/>compare to critical value"]
    S3 --> S4["Step 4: Conclusion<br/>in context"]

Step 1: State the Hypotheses

Test Type	$H_0$	$H_1$	Tail
Two-tailed	$\mu = \mu_0$	$\mu \neq \mu_0$	Both tails
Right-tailed	$\mu \leq \mu_0$	$\mu > \mu_0$	Right tail
Left-tailed	$\mu \geq \mu_0$	$\mu < \mu_0$	Left tail

Step 2: Choose and Compute the Test Statistic

Condition	Test Statistic	Distribution
$\sigma$ known, any $n$	$z = \dfrac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$	$N(0,1)$
$\sigma$ unknown, $n \geq 30$	$z = \dfrac{\bar{x} - \mu_0}{s / \sqrt{n}}$	$N(0,1)$ (CLT)
$\sigma$ unknown, $n < 30$	$t = \dfrac{\bar{x} - \mu_0}{s / \sqrt{n}}$	$t_{n-1}$

Step 3: Decision Rule

Critical Value Method:

Reject $H_0$ if test statistic falls in rejection region
Two-tailed: reject if $|z| > z_{\alpha/2}$ or $|t| > t_{\alpha/2, n-1}$
Right-tailed: reject if $z > z_\alpha$ or $t > t_{\alpha, n-1}$
Left-tailed: reject if $z < -z_\alpha$ or $t < -t_{\alpha, n-1}$

P-value Method:

Two-tailed: $p = 2 \times P(Z > |z|)$ or $2 \times P(T > |t|)$
Right-tailed: $p = P(Z > z)$ or $P(T > t)$
Left-tailed: $p = P(Z < z)$ or $P(T < t)$
Reject $H_0$ if $p \leq \alpha$

Step 4: Conclusion

If reject $H_0$: "At the $\alpha$ significance level, there is sufficient evidence to conclude that [claim in $H_1$]."

If fail to reject $H_0$: "At the $\alpha$ significance level, there is insufficient evidence to conclude that [claim in $H_1$]."

4.2 z-test vs t-test — When to Use

	z-test	t-test
$\sigma$ known?	Yes	No (use $s$)
Sample size	Any (with $\sigma$) or $n \geq 30$	Any, but especially $n < 30$
Distribution	Standard normal $N(0,1)$	$t$ with $df = n-1$

4.3 Full Worked Example (Section B Style)

Problem: The mean cost of a hotel room in KL is said to be RM168 per night. A random sample of 25 hotels resulted in $\bar{x} = \text{RM}172.50$ and $s = \text{RM}15.40$. Test at $\alpha = 0.05$ whether the mean cost differs from RM168. Assume normality.

Step 1: Hypotheses $$H_0: \mu = 168 \quad \text{vs} \quad H_1: \mu \neq 168$$

Step 2: Test Statistic $n = 25$, $\sigma$ unknown → t-test with $df = 24$ $$t = \frac{172.50 - 168}{15.40 / \sqrt{25}} = \frac{4.50}{3.08} = 1.46$$

Step 3: Decision Rule (Critical Value Method) At $\alpha = 0.05$, two-tailed, $df = 24$: Critical value: $t_{0.025, 24} = \pm 2.064$

Since $|1.46| < 2.064$, we fail to reject $H_0$.

Step 3 (Alternative — P-value Method): $$p = 2 \times P(T_{24} > 1.46) = 2 \times 0.0785 = 0.157$$ Since $p = 0.157 > 0.05$, fail to reject $H_0$.

Step 4: Conclusion At the 5% significance level, there is insufficient evidence to conclude that the mean cost of a hotel room in KL differs from RM168.

4.4 P-value Interpretation

p-value ≤ $\alpha$ → Reject $H_0$ (statistically significant result)
p-value > $\alpha$ → Fail to reject $H_0$ (not statistically significant)
Small p-value → Strong evidence against $H_0$
Large p-value → Weak evidence against $H_0$

Common thresholds:

$\alpha$	Interpretation
0.10	Marginal significance
0.05	Standard significance
0.01	Strong significance

4.5 Likely Section B Question Structure