FAD1015: Rapid-Fire Drill Pack — Leak Topics

Objective: Master the 4 leak-highlighted topics for the final exam via mechanical repetition.
Target: 2–4 min per problem. If you stall >5 minutes, skip and mark it.
Total problems: 40
Estimated time: ~90 min

Cheat Sheet (Memorize First)

Permutation & Counting

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

Hypothesis Testing

4-Step Framework:

Step Action
Step 1 State $H_0$ and $H_1$
Step 2 Compute test statistic: $z = \dfrac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$ or $t = \dfrac{\bar{x} - \mu_0}{s / \sqrt{n}}$
Step 3 Decision: compare $p$ to $\alpha$, or statistic to critical value
Step 4 Conclusion in context

Test Selection:

Condition Test
$\sigma$ known z-test
$\sigma$ unknown, $n \ge 30$ z-test with $s$ (CLT)
$\sigma$ unknown, $n < 30$ t-test with $df = n-1$

P-value Decision: $p \le \alpha \Rightarrow$ Reject $H_0$; $p > \alpha \Rightarrow$ Fail to reject $H_0$.

R Matrix Operations

Task R Code
Create matrix (col-wise) matrix(data, nrow, ncol)
Create matrix (row-wise) matrix(data, nrow, ncol, byrow=TRUE)
Bind rows rbind(row1, row2)
Bind columns cbind(col1, col2)
Matrix multiplication A %*% B
Element-wise multiply A * B
Transpose t(A)
Determinant det(A)
Inverse solve(A)
Diagonal diag(A)
Row/col names rownames(A), colnames(A)
Dimensions dim(A), nrow(A), ncol(A)

R Output Interpretation

t.test() Field Meaning
t = ... Test statistic: $t = (\bar{x} - \mu_0) / (s / \sqrt{n})$
df = ... Degrees of freedom $= n-1$
p-value = ... Probability of observing result this extreme if $H_0$ true
95% CI [...] 95% confidence interval for true mean
mean of x Sample mean $\bar{x}$
z.test() Field Meaning
z = ... Test statistic: $z = (\bar{x} - \mu_0) / (\sigma / \sqrt{n})$
p-value = ... Same interpretation as t-test
95% CI [...] Same as t-test

Decision Rule: If $\mu_0$ lies inside the CI, fail to reject $H_0$. If $\mu_0$ lies outside the CI, reject $H_0$.

Part A: Permutation & Counting (Section B)

Target: 2–3 min per problem.

Set A1 — Formula Identification (5 problems)

Identify which formula to use. Do NOT compute the final number.

  1. A 4-digit PIN code is formed using digits 0–9, repetition allowed. How many possible PINs exist? (Identify: formula only)
  2. In how many ways can 6 different books be arranged on a shelf? (Identify: formula only)
  3. A committee of 3 is selected from 10 students. How many different committees are possible? (Identify: formula only)
  4. How many distinct arrangements of the letters in MISSISSIPPI are there? (Identify: formula only)
  5. In how many ways can 8 people sit around a circular table where rotations are considered the same? (Identify: formula only)

Score: ___/5

Set A2 — Basic Calculations (5 problems)

Compute the exact value.

  1. Compute $P(8,3)$ — the number of ways to arrange 3 objects chosen from 8 distinct objects.
  2. Compute $C(10,4)$ — the number of ways to select 4 objects from 10 distinct objects.
  3. A password consists of 5 characters, each chosen from 26 letters (A–Z), repetition allowed. How many possible passwords are there? Express as a power.
  4. In how many ways can 7 distinct keys be arranged on a circular key ring? (Flipping the ring yields the same arrangement.)
  5. How many distinct arrangements of the word BANANA are possible?

Score: ___/5

Set A3 — Section B Style Multi-Step (5 problems)

Word problems requiring multiple counting steps — restrictions, complementary counting, and combined counting + probability.

  1. Six people (A, B, C, D, E, F) stand in a line for a photo. How many arrangements are possible if A and B refuse to stand next to each other?
  2. Five boys and four girls are to be arranged in a row. Find the number of arrangements where all the boys stand together.
  3. A committee of 4 is chosen from 7 men and 5 women. Find the probability that the committee has at least 3 women.
  4. In how many ways can the letters of the word MATHEMATICS be arranged? (M: 2, A: 2, T: 2, H: 1, E: 1, I: 1, C: 1, S: 1)
  5. Four-digit numbers are formed from digits ${1, 2, 3, 4, 5, 6, 7}$ without repetition. How many such numbers are greater than 4000?

Score: ___/5

Part B: Hypothesis Testing (Section B)

Target: 3–5 min per problem (B1), 4–6 min per problem (B2, B3).

Set B1 — Test Selection (4 problems)

For each scenario: (i) Choose z-test or t-test. (ii) State $H_0$ and $H_1$ in symbols. (iii) State one-tailed or two-tailed.

  1. A researcher claims the mean IQ of students is greater than 100. A sample of 25 students has $\bar{x} = 105$, $s = 15$. Population $\sigma$ is unknown. Test at $\alpha = 0.05$.
  2. A factory claims its cereal boxes contain 500 g on average. A sample of 50 boxes has $\bar{x} = 495$ g, and the population standard deviation is known to be $\sigma = 10$ g. Test at $\alpha = 0.01$.
  3. A medical researcher wants to test whether a new drug lowers blood pressure. A sample of 15 patients is measured before and after treatment. The population $\sigma$ is unknown. (Assume one-sample test against 0 change.)
  4. A car manufacturer claims the mean fuel efficiency of a model is at least 18 km/L. A sample of 40 cars has $\bar{x} = 17.5$ km/L, $s = 2.5$ km/L. Test at $\alpha = 0.05$.

Score: ___/4

Set B2 — Full Hypothesis Tests (5 problems)

Apply the 4-step framework: $H_0/H_1 \to$ test statistic $\to$ decision (critical value or p-value) $\to$ conclusion.

  1. A lightbulb manufacturer claims the mean lifetime of its bulbs is 1500 hours. A sample of 30 bulbs has $\bar{x} = 1460$ hours, $s = 120$ hours. Test at $\alpha = 0.05$ whether the mean lifetime differs from 1500 hours. (Use p-value method.)

    Given: $P(T_{29} > 1.83) \approx 0.0385$, so two-tailed $p \approx 0.077$.

  2. The mean score on a standardized test is known to be 500 ($\sigma = 100$). A sample of 36 students from a new teaching program has $\bar{x} = 530$. Test at $\alpha = 0.05$ whether the program produces a higher mean score. Use the critical value method.

    Given: $z_{0.05} = 1.645$.

  3. A dietitian claims that a new diet reduces mean weight by at least 5 kg. A sample of 20 participants loses an average of 4.2 kg with $s = 2.8$ kg. Test at $\alpha = 0.01$ whether the mean weight loss is less than 5 kg.

    Given: Critical value $t_{0.01, 19} = -2.539$.

  4. A company claims its batteries last 48 hours on average. A sample of 25 batteries has $\bar{x} = 46.2$ hours, $s = 4.5$ hours. Test at $\alpha = 0.10$ whether the mean is less than 48 hours.

    Given: Critical value $t_{0.10, 24} = -1.318$.

  5. The average monthly rent for a studio in a city is claimed to be RM 800. A sample of 40 studios has $\bar{x} = \text{RM } 825$, $s = \text{RM } 95$ (use $z$ approximation since $n \ge 30$). Test at $\alpha = 0.05$ whether the mean differs from RM 800. Use critical value method.

    Given: $z_{0.025} = 1.96$.

Score: ___/5

Set B3 — R Output Interpretation (4 problems)

Given R console output, answer the interpretation questions.

  1. Output:

            One Sample t-test
data:  weights
t = 2.1345, df = 14, p-value = 0.0508
alternative hypothesis: true mean is not equal to 50
95 percent confidence interval:
 49.987 54.013
sample estimates:
mean of x
 52.0

    (a) What is the sample size?
    (b) What is the test statistic value?
    (c) At $\alpha = 0.05$, do you reject $H_0$? Why?
    (d) Does the 95% CI contain $\mu_0 = 50$? What does this imply?

  2. Output:

            One-sample z-Test
data:  scores
z = 2.856, p-value = 0.0043
alternative hypothesis: true mean is greater than 70
95 percent confidence interval:
 71.234   Inf
sample estimates:
mean of x
 74.8

    (a) What type of test was performed (one-tailed or two-tailed)?
    (b) At $\alpha = 0.01$, do you reject $H_0$? Explain.
    (c) Interpret the p-value in plain language.

  3. Output:

            Shapiro-Wilk normality test
data:  sample_data
W = 0.9214, p-value = 0.0342

    (a) What is the null hypothesis of the Shapiro-Wilk test?
    (b) At $\alpha = 0.05$, what do you conclude about the normality of the data?
    (c) What implication does this have for using a t-test on this data?

  4. Output:

            One Sample t-test
data:  heights
t = -1.296, df = 19, p-value = 0.2103
alternative hypothesis: true mean is less than 170
95 percent confidence interval:
    -Inf 169.45
sample estimates:
mean of x
 168.2

    (a) What is the sample mean?
    (b) Is this a left-tailed, right-tailed, or two-tailed test?
    (c) At $\alpha = 0.05$, is there sufficient evidence that the mean height is less than 170? Explain.

Score: ___/4

Part C: R Matrix Operations

Target: 2–4 min per problem.

Set C1 — Matrix Construction (3 problems)

Write the R code to create each matrix.

  1. Create a $3 \times 3$ matrix named A using matrix() with data 1:9 filled row-wise.

         [,1] [,2] [,3]
[1,]    1    2    3
[2,]    4    5    6
[3,]    7    8    9

  2. Using rbind(), create a matrix M with rows:

    • Row 1: c(10, 20, 30)
    • Row 2: c(40, 50, 60)
    • Row 3: c(70, 80, 90) Then assign row names "R1", "R2", "R3" and column names "A", "B", "C".

  3. Create a $2 \times 4$ matrix P where the first row is c(1, 3, 5, 7) and the second row is c(2, 4, 6, 8) using rbind(). Then use dim(P) to check the dimensions.

Score: ___/3

Set C2 — Matrix Operations (4 problems)

Given matrices, compute by hand or write R code. Use: $$A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 0 \ 1 & 3 \end{pmatrix}, \quad C = \begin{pmatrix} 1 & 0 & 2 \ 3 & 1 & 0 \ 0 & 2 & 1 \end{pmatrix}$$

  1. Compute $A %*% B$ (matrix multiplication). Show the resulting $2 \times 2$ matrix.
  2. Compute det(A) and det(C). Is C invertible?
  3. Write R code to: compute t(A), the transpose of $A$, and solve(A), the inverse of $A$. Then verify that A %*% solve(A) gives the identity matrix.
  4. Let $v = \mathtt{diag}(C)$. What is $v$? Show the vector.

Score: ___/4

Set C3 — R Output from Matrix Code (3 problems)

Given R code, predict the exact R output or the resulting matrix.

  1. X <- matrix(c(1, 0, 0, 1), nrow = 2)
Y <- matrix(c(2, 3, 4, 5), nrow = 2, byrow = TRUE)
X %*% Y

    What is the output?

  2. A <- rbind(c(1, 2, 3), c(4, 5, 6))
B <- A[, c(1, 3)]

    What is the matrix B?

  3. M <- matrix(c(3, 1, 2, 4), nrow = 2)
det(M)
M_inv <- solve(M)
M %*% M_inv

    What does det(M) return? What does the last line M %*% M_inv return approximately?

Score: ___/3

Final Scorecard

Part Sets Problems Raw Score
A — Permutation & Counting A1, A2, A3 15 ___/15
B — Hypothesis Testing B1, B2, B3 13 ___/13
C — R Matrix Operations C1, C2, C3 10 ___/10
TOTAL 40 ___/40

Proficiency Benchmarks

  • 28/40 (70%) — Proficient. You can handle standard exam problems.
  • 34/40 (85%) — Solid. Fast and accurate on most patterns.
  • 38/40 (95%) — Exam-ready. Any mistake is a careless slip.

Speed Benchmarks

  • <60 min: Excellent mechanical fluency.
  • 60–90 min: Good. Review missed patterns.
  • >90 min: Drill the specific sets you scored lowest on again tomorrow.

Error Log Template

After grading, list every wrong problem number with a one-word reason:

Problem Reason
e.g. 12 forget grouping factorial

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Set A1

  1. $n^r = 10^4$ (permutation with repetition)
  2. $6! = 720$ (permutation of distinct objects)
  3. $C(10,3) = 120$ (combination, order does not matter)
  4. $\dfrac{11!}{4!,4!,2!}$ (permutation with identical objects: 4 S, 4 I, 2 P)
  5. $(8-1)! = 7!$ (circular permutation, rotations same)

Set A2

  1. $P(8,3) = 8 \times 7 \times 6 = 336$
  2. $C(10,4) = \dfrac{10!}{4!,6!} = 210$
  3. $26^5$
  4. $\dfrac{(7-1)!}{2} = \dfrac{6!}{2} = 360$
  5. $\dfrac{6!}{3!,2!} = \dfrac{720}{6 \times 2} = 60$

Set A3

  1. Total: $6! = 720$. Together (A,B as block): $5! \times 2! = 240$. Valid: $720 - 240 = 480$
  2. Treat 5 boys as one block: $5!$ ways to arrange within block. Then block + 4 girls $= 5$ items: $5!$ arrangements. Total: $5! \times 5! = 120 \times 120 = 14,400$
  3. At least 3 women = 3W1M or 4W0M. $\displaystyle \frac{C(5,3)C(7,1) + C(5,4)C(7,0)}{C(12,4)} = \frac{10 \times 7 + 5 \times 1}{495} = \frac{75}{495} = \frac{5}{33}$
  4. $\dfrac{11!}{2!,2!,2!} = \dfrac{39,916,800}{8} = 4,989,600$
  5. First digit $\ge 4 \Rightarrow$ choices: ${4,5,6,7}$ = 4 choices. Remaining 3 digits from remaining 6: $P(6,3) = 120$. Total: $4 \times 120 = 480$

Set B1

  1. (i) t-test ($\sigma$ unknown, $n < 30$). (ii) $H_0: \mu \le 100$, $H_1: \mu > 100$. (iii) Right-tailed.
  2. (i) z-test ($\sigma$ known). (ii) $H_0: \mu = 500$, $H_1: \mu \neq 500$. (iii) Two-tailed.
  3. (i) t-test ($\sigma$ unknown, $n < 30$). (ii) $H_0: \mu \ge 0$, $H_1: \mu < 0$. (iii) Left-tailed.
  4. (i) z-test (use $s$, $n \ge 30$ by CLT). (ii) $H_0: \mu \ge 18$, $H_1: \mu < 18$. (iii) Left-tailed.

Set B2

  1. Step 1: $H_0: \mu = 1500$, $H_1: \mu \neq 1500$ (two-tailed).
    Step 2: $t = \dfrac{1460 - 1500}{120 / \sqrt{30}} = \dfrac{-40}{21.91} = -1.826$, $df = 29$.
    Step 3: $p \approx 0.077 > 0.05$ (given two-tailed $p \approx 0.077$). Fail to reject $H_0$.
    Step 4: At $\alpha = 0.05$, insufficient evidence that mean lifetime differs from 1500 hours.

  2. Step 1: $H_0: \mu \le 500$, $H_1: \mu > 500$ (right-tailed).
    Step 2: $z = \dfrac{530 - 500}{100 / \sqrt{36}} = \dfrac{30}{16.67} = 1.80$.
    Step 3: Critical value $z_{0.05} = 1.645$. Since $1.80 > 1.645$, reject $H_0$.
    Step 4: At $\alpha = 0.05$, sufficient evidence that the program produces a higher mean score.

  3. Step 1: $H_0: \mu \ge 5$, $H_1: \mu < 5$ (left-tailed).
    Step 2: $t = \dfrac{4.2 - 5}{2.8 / \sqrt{20}} = \dfrac{-0.8}{0.626} = -1.278$, $df = 19$.
    Step 3: Critical value $t_{0.01, 19} = -2.539$. Since $-1.278 > -2.539$, fail to reject $H_0$.
    Step 4: At $\alpha = 0.01$, insufficient evidence that mean weight loss is less than 5 kg.

  4. Step 1: $H_0: \mu \ge 48$, $H_1: \mu < 48$ (left-tailed).
    Step 2: $t = \dfrac{46.2 - 48}{4.5 / \sqrt{25}} = \dfrac{-1.8}{0.9} = -2.00$, $df = 24$.
    Step 3: Critical value $t_{0.10, 24} = -1.318$. Since $-2.00 < -1.318$, reject $H_0$.
    Step 4: At $\alpha = 0.10$, sufficient evidence that mean battery life is less than 48 hours.

  5. Step 1: $H_0: \mu = 800$, $H_1: \mu \neq 800$ (two-tailed).
    Step 2: $z = \dfrac{825 - 800}{95 / \sqrt{40}} = \dfrac{25}{15.02} = 1.664$.
    Step 3: Critical values $\pm z_{0.025} = \pm 1.96$. Since $\vert 1.664 \vert < 1.96$, fail to reject $H_0$.
    Step 4: At $\alpha = 0.05$, insufficient evidence that mean rent differs from RM 800.

Set B3

  1. (a) $n = df + 1 = 15$. (b) $t = 2.1345$. (c) $p = 0.0508 > 0.05$, so fail to reject $H_0$. (d) Yes, the CI $[49.987, 54.013]$ contains 50. This is consistent with failing to reject $H_0$.
  2. (a) Right-tailed ($H_1$: true mean is greater than 70). (b) $p = 0.0043 < 0.01$, so reject $H_0$. (c) If the null hypothesis were true, there is only a 0.43% chance of observing a sample mean this extreme or more.
  3. (a) $H_0$: Data is normally distributed. (b) $p = 0.0342 < 0.05$, so reject normality — data is not normally distributed. (c) A t-test assumes normality; a non-parametric test may be more appropriate, or check for large enough sample size for CLT to apply.
  4. (a) $\bar{x} = 168.2$. (b) Left-tailed ($H_1$: true mean is less than 170). (c) $p = 0.2103 > 0.05$, so fail to reject $H_0$. Insufficient evidence that mean height is less than 170.

Set C1

  1. A <- matrix(1:9, nrow = 3, byrow = TRUE)
  2. M <- rbind(c(10, 20, 30), c(40, 50, 60), c(70, 80, 90))
rownames(M) <- c("R1", "R2", "R3")
colnames(M) <- c("A", "B", "C")

  3. P <- rbind(c(1, 3, 5, 7), c(2, 4, 6, 8))
dim(P)   # returns 2 4

Set C2

  1. $A %*% B = \begin{pmatrix} 1(2)+2(1) & 1(0)+2(3) \ 3(2)+4(1) & 3(0)+4(3) \end{pmatrix} = \begin{pmatrix} 4 & 6 \ 10 & 12 \end{pmatrix}$
  2. $\det(A) = 1(4) - 2(3) = 4 - 6 = -2$. $\det(C) = 1(1\cdot1 - 0\cdot2) - 0 + 2(3\cdot2 - 1\cdot0) = 1(1) + 2(6) = 13$. Since $\det(C) \neq 0$, $C$ is invertible.
  3. t(A)        # returns matrix [1 3; 2 4]
solve(A)    # returns matrix [-2  1; 1.5 -0.5]
A %*% solve(A)  # returns identity matrix approx diag(1,1)
  4. $\mathtt{diag}(C) = \mathtt{c(1, 1, 1)}$ — extracts diagonal elements $[1, 1, 1]$.

Set C3

  1. $X = \begin{pmatrix}1&0\0&1\end{pmatrix}$ (identity). $Y = \begin{pmatrix}2&4\3&5\end{pmatrix}$. $X %*% Y = Y = \begin{pmatrix}2&4\3&5\end{pmatrix}$.
  2. $A = \begin{pmatrix}1&2&3\4&5&6\end{pmatrix}$, $B = A[, c(1,3)] = \begin{pmatrix}1&3\4&6\end{pmatrix}$.
  3. $M = \begin{pmatrix}3&2\1&4\end{pmatrix}$. $\det(M) = 3(4) - 2(1) = 10$. $M %*% M_{\text{inv}}$ returns $\begin{pmatrix}1&0\0&1\end{pmatrix}$ (identity matrix, up to rounding).

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