Core Concepts
Practice Quiz

The drug reduces blood pressure → we expect μ to be lower than baseline μ₀. That's a one-tailed left test: H₁: μ < μ₀. Choice D (μ ≠ μ₀) would be a two-tailed test used when we only know the drug changes pressure, not the direction.

p-value (0.03) < α (0.05) → Reject H₀. We never "accept H₁ as proven" — we only say there is sufficient evidence to support H₁. Choice C is wrong because statistical tests don't prove hypotheses, they only provide evidence against H₀.

Type I Error (α) = Rejecting H₀ when H₀ is actually TRUE. Here H₀ = "patient is healthy" is TRUE, but we rejected it — classic false alarm. Memory trick: Type I = False POSITIVE (like a fire alarm going off with no fire).

The correct definition: p-value = P(data this extreme | H₀ is true). It is NOT the probability that H₀ is true (that's a Bayesian posterior). Many students confuse this — the p-value assumes H₀ is true and asks how surprising our data is under that assumption.

For a two-tailed test at α = 0.05, we split α into two tails: 0.025 each side. P(Z < z*) = 0.975 → z* = 1.96. Memory: 90% → 1.65 · 95% → 1.96 · 99% → 2.576. These three values appear on almost every exam.

α = P(Type I Error). Raising α from 0.01 → 0.05 directly increases Type I error probability. The trade-off: a larger α means a wider rejection region → easier to reject H₀ → decreases Type II error (β). These two errors have an inverse trade-off for fixed sample size.

Right-tailed test: reject H₀ if z > z*. Here z = 2.10 > 1.65 = z*, so the test statistic falls in the rejection region → Reject H₀. Note: we never say "accept H₀" — only "fail to reject" or "reject."

z = (78 − 70) / 4 = 8 / 4 = 2.0. The student scored exactly 2 standard deviations above the mean. Since the score is above μ, z is positive. A common mistake: subtracting in the wrong order (70−78 = −8) giving z = −2.0.

μ = 50, σ = 10. The range 40 to 60 = (50 − 10) to (50 + 10) = μ ± 1σ. By the Empirical Rule, approximately 68% of data falls within 1 standard deviation of the mean. 95% would be 30–70 (±2σ), and 99.7% would be 20–80 (±3σ).

P(Z < 1.00) = 0.8413 is a key z-table value to memorize. It comes directly from the Empirical Rule: P(μ − σ < X < μ) = 34%, and P(X < μ) = 50%, so P(X < μ + σ) = 50% + 34% = 84% ≈ 0.8413. Therefore z = 1.00.

Step 1: z = (186 − 170)/8 = 16/8 = 2.0. Step 2: P(Z < 2.0) = 0.9772. Step 3: P(Z > 2.0) = 1 − 0.9772 = 0.0228. Common mistake: forgetting to use the complement and reporting 0.9772 instead. Always ask: "Am I looking for the LEFT or RIGHT tail?"

A larger σ means more spread/variability. The total area under any normal curve = 1 (always). So if it spreads wider, the peak must get shorter to keep total area = 1. Think of it like stretching dough — wider = flatter. μ controls location (left/right shift), σ controls shape (tall/narrow vs. short/wide).

P(Z < z) = 0.84 → z ≈ 1.00 (from z-table, P(Z < 1.00) = 0.8413). Now reverse the z-score formula: x = μ + zσ = 500 + (1.00)(100) = 600. Step: percentile → z-value → x = μ + zσ. This "reverse standardization" is a very common exam question.

σ = √25 = 5. z for 55: (55−60)/5 = −1.0. z for 70: (70−60)/5 = 2.0. P(−1 < Z < 2) = P(Z < 2) − P(Z < −1) = 0.9772 − 0.1587 = 0.8185 ≈ 81.9%. Key trap: using σ = 25 instead of σ = 5. Always check if you're given variance σ² or standard deviation σ!

MoE = (Upper − Lower) / 2 = (58 − 42) / 2 = 16 / 2 = 8. The center (point estimate) = (58 + 42) / 2 = 50. So the interval is 50 ± 8. Many students mistakenly give 16 (the full width) as the margin of error — remember MoE is half the interval width.

The correct interpretation is B. μ is a fixed (unknown) constant — it doesn't have a probability. Once the interval is calculated, either μ IS in it or it ISN'T. The "95%" refers to the procedure: if we repeated sampling many times, 95% of constructed intervals would contain μ. Choice A confuses frequentist intervals with Bayesian credible intervals.

MoE ∝ 1/√n. To halve the MoE (4% → 2%), we need √n to double → n must quadruple. New n = 4 × 600 = 2,400. This is the most commonly tested MoE relationship: halving the error requires 4× the sample size, not 2×. Students almost always choose "double" by intuition — this is the #1 trap on this topic.

SE = √(0.60 × 0.40 / 400) = √(0.24/400) = √0.0006 = 0.02449. MoE = 1.96 × 0.02449 ≈ 0.048. CI = 0.60 ± 0.048 = (0.552, 0.648). Common mistake: using n instead of √n, or forgetting to multiply SE by z*.

MoE = z* × σ/√n. To decrease MoE: ↑n (more data), ↓z* (lower confidence), or ↓σ. Choice C increases n from 100 → 400 (×4), so MoE halves — that's a clear decrease. Choice A increases confidence level → larger z* → increases MoE. Choice B decreases n → increases MoE. Choice D increases σ → increases MoE.

n = (1.96/0.03)² × 0.5 × 0.5 = (65.33)² × 0.25 = 4268.4 × 0.25 = 1067.1 → round UP to 1,068. Always round sample size UP to meet the requirement. p̂ = 0.5 gives maximum variability (p(1−p) = 0.25 is maximum), so this is the conservative (largest) estimate. A very common error is forgetting to square z*/E.