Use this diagnostic to find which Unit 3 topics to revisit: one question per topic.
These are original Marlbridge practice questions with fictional data, not past exam questions. It is not calibrated and gives no predicted score.
How to take it. Allow 30 minutes; answer everything before opening the answers. Graphing calculator allowed; z* is 1.645 (90%), 1.960 (95%), 2.576 (99%).
Question 1 (multiple choice · 3.1)
For p = 0.40, 1,000 simulated random samples give estimator P mean 0.400, standard deviation 0.07, and estimator Q mean 0.430, standard deviation 0.03. Which is correct?
- (A) P appears unbiased; Q appears biased but less variable.
- (B) Q is unbiased, because its values cluster.
- (C) Both are biased: neither standard deviation is 0.
- (D) P is biased: one value could be 0.33.
Answer and explanation
Answer: (A). Bias is about the centre: P’s matches 0.40; Q’s is 0.03 too high.
- (B) confuses bias with variability.
- (C) forgets that unbiased estimators vary.
- (D) judges bias from one value.
If you missed this: 3.1 study guide.
Question 2 (multiple choice · 3.2)
In a fictional region, 4% of adults have a heat pump. For random samples of 150 adults, the sampling distribution of p̂ has:
- (A) Mean 0.04, standard deviation 0.016, approximately normal
- (B) Mean 0.04, standard deviation 0.016, skewed right
- (C) Mean 6, standard deviation 2.4, approximately normal
- (D) Mean 0.04, standard deviation 0.000256, skewed right
Answer and explanation
Answer: (B). σp̂ = √(0.04 × 0.96 ÷ 150) = 0.016, but np = 6 < 10, so the large-counts condition fails: skewed right.
- (A) ignores the failed condition.
- (C) describes the count.
- (D) forgets the square root.
If you missed this: 3.2 study guide.
Question 3 (multiple choice · 3.3)
In a random sample of 640 passengers on a fictional tram network, 208 had a monthly pass. Which is the 95% confidence interval for p, the proportion of all passengers with one?
- (A) (0.289, 0.361)
- (B) (0.295, 0.355)
- (C) (0.324, 0.326)
- (D) (0.277, 0.373)
Answer and explanation
Answer: (A). p̂ = 0.325; SE = √(0.325 × 0.675 ÷ 640) = 0.01851; margin of error = 1.960 × 0.01851 = 0.0363.
- (B) uses 1.645, the 90% value.
- (C) forgets the square root.
- (D) uses 2.576, the 99% value.
If you missed this: 3.3 study guide.
Question 4 (multiple choice · 3.4)
A 95% interval for the proportion of a fictional town’s residents supporting a new market is (0.47, 0.55). A councillor says exactly half do. Which is best?
- (A) The interval proves the claim: 0.50 is inside it.
- (B) 0.50 is plausible, so the claim is not contradicted, but it is not proved.
- (C) The point estimate is 0.51, so a majority supports it.
- (D) 95% of residents’ answers lie between 0.47 and 0.55.
Answer and explanation
Answer: (B). 0.50 is inside the interval, but so are 0.48 and 0.53.
- (A) treats “plausible” as “proven”.
- (C) ignores plausible values below 0.50.
- (D) describes individuals, not p.
If you missed this: 3.4 study guide.
Question 5 (multiple choice · 3.5)
Which situation calls for a one-sample z-test for a population proportion?
- (A) Estimating the proportion of students who own a bike
- (B) Checking a claim that 30% of drivers use winter tyres, suspecting fewer do
- (C) Comparing two schools’ proportions of students who walk
- (D) Asking whether age group and favourite sport are associated
Answer and explanation
Answer: (B). One population, one categorical variable and a claimed value, p₀ = 0.30.
- (A) needs a confidence interval.
- (C) needs a two-sample test.
- (D) needs a chi-square test for independence.
If you missed this: 3.5 study guide.
Question 6 (multiple choice · 3.6)
Random samples X (n = 100) and Y (n = 400) both give p̂ = 0.56 for H₀: p = 0.50, Hₐ: p > 0.50. Which is correct?
- (A) Y gives the smaller p-value.
- (B) The p-values are equal, because p̂ is the same.
- (C) X gives the smaller p-value, because small samples vary more.
- (D) The p-values cannot be compared without α.
Answer and explanation
Answer: (A). For X, z = 0.06 ÷ 0.05 = 1.20 (p-value 0.1151); for Y, z = 0.06 ÷ 0.025 = 2.40 (p-value 0.0082).
- (B) ignores n.
- (C) More variability makes 0.56 less surprising.
- (D) α only matters for the decision.
If you missed this: 3.6 study guide.
Question 7 (short answer · 3.7)
A fictional council says 30% of residents use its leisure centre. A random sample of 400 residents includes 140 users. Is the proportion different from 30%? Conditions are met; α = 0.05.
(a) State the hypotheses, defining the parameter. (b) Find z and the p-value. (c) Conclude in context.
Answer and explanation
(a) p = the true proportion of all residents who use the centre. H₀: p = 0.30; Hₐ: p ≠ 0.30.
(b) p̂ = 0.35. z = (0.35 − 0.30) ÷ √(0.30 × 0.70 ÷ 400) = 2.18. p-value = 2 × P(Z ≥ 2.18) = 0.0291.
(c) 0.0291 < 0.05, so reject H₀. There is convincing statistical evidence that the true proportion of residents using the centre differs from 0.30.
If you missed this: 3.7 study guide.
Question 8 (multiple choice · 3.8)
A test of H₀: p = 0.50 against Hₐ: p > 0.50 uses n = 150 and α = 0.05. Its power is 0.34 if p = 0.55. What could its power be if p = 0.60?
- (A) 0.05
- (B) 0.21
- (C) 0.34
- (D) 0.79
Answer and explanation
Answer: (D). With n and α fixed, a true value further from p₀ is easier to detect, so power rises.
- (A) is α.
- (B) has power falling.
- (C) ignores the true value.
If you missed this: 3.8 study guide.
Question 9 (multiple choice · 3.9)
For independent random samples of 300 from A (pA = 0.62) and 250 from B (pB = 0.48), the sampling distribution of p̂A − p̂B has
- (A) mean 0.14, standard deviation larger than σp̂A and σp̂B
- (B) mean 0.14, standard deviation σp̂A − σp̂B
- (C) mean 0, as the samples are independent
- (D) mean 0.14, standard deviation smaller than σp̂A and σp̂B
Answer and explanation
Answer: (A). Variances add: σp̂A = 0.0280, σp̂B = 0.0316 and σ for the difference = √(0.0280² + 0.0316²) = 0.0422.
- (B) subtracts standard deviations.
- (C) confuses independence with equal proportions.
- (D) Variability adds, not shrinks.
If you missed this: 3.9 study guide.
Question 10 (multiple choice · 3.10)
Independent random samples show 168 of 280 drivers in fictional Town P and 130 of 260 in Town Q use winter tyres. Which is the 90% confidence interval for pP − pQ?
- (A) (0.016, 0.184)
- (B) (0.001, 0.199)
- (C) (0.030, 0.170)
- (D) (−0.170, −0.030)
Answer and explanation
Answer: (C). SE = √(0.60 × 0.40 ÷ 280 + 0.50 × 0.50 ÷ 260) = 0.04265; 0.10 ± 1.645 × 0.04265.
- (A) uses 1.960.
- (B) adds standard errors, not variances.
- (D) estimates pQ − pP.
If you missed this: 3.10 study guide.
Question 11 (multiple choice · 3.11)
For the proportions of students who cycle at fictional East and West colleges, a 95% interval for pE − pW is (−0.21, −0.04). A report says pW exceeds pE by more than 0.05. Which is best?
- (A) Convincing evidence for the report
- (B) Convincing evidence that pW > pE, but not of the report’s claim
- (C) No convincing evidence of a difference, because the interval is negative
- (D) Convincing evidence that pE > pW
Answer and explanation
Answer: (B). All plausible values are negative, so pW > pE; but a difference of 0.04 is plausible.
- (A) ignores differences from 0.04 to 0.05.
- (C) 0 is not in the interval.
- (D) reverses the direction.
If you missed this: 3.11 study guide.
Question 12 (multiple choice · 3.12)
A fictional nursery randomly assigns 120 seedlings to feed A and 160 to feed B; 78 and 88 flower. Which correctly checks the conditions for a two-sample z-test?
- (A) Both groups exceed 30, so normality is met.
- (B) Not met: the seedlings are not a random sample.
- (C) p̂c = (0.65 + 0.55) ÷ 2 = 0.60; 120 is under 10% of all seedlings.
- (D) Random assignment; p̂c ≈ 0.593 gives 71.1, 48.9, 94.9, 65.1, all at least 10.
Answer and explanation
Answer: (D). p̂c = 166 ÷ 280. A test assumes H₀, so normality uses the pooled proportion.
- (A) is a rule for means.
- (B) forgets that random assignment is enough.
- (C) averages unequal groups; experiments need no 10% check.
If you missed this: 3.12 study guide.
Question 13 (short answer · 3.13)
A fictional charity randomly assigns 250 learner cyclists to an on-road course (R) and 250 to a classroom course (C). Within three months, 30 R and 45 C learners have an accident. Does R reduce accidents? Conditions are met; α = 0.05.
(a) State the hypotheses, defining the parameters. (b) Find p̂c, z and the p-value. (c) Conclude. Is cause and effect justified?
Answer and explanation
(a) pR, pC = the true three-month accident proportions for learners like these after each course. H₀: pR = pC; Hₐ: pR < pC.
(b) p̂c = 75 ÷ 500 = 0.15. z = (0.12 − 0.18) ÷ √[0.15 × 0.85 × (1/250 + 1/250)] = −1.88. p-value = P(Z ≤ −1.88) = 0.0301.
(c) 0.0301 < 0.05, so reject H₀: convincing statistical evidence that course R reduces accidents. Random assignment justifies cause and effect, for learners like these.
If you missed this: 3.13 study guide.
Question 14 (multiple choice · 3.14)
A random sample of 600 adults in a fictional region records age group and main transport. Which null hypothesis is correct?
- (A) Age group and transport are associated among the region’s adults
- (B) Each type of transport is used by 25% of the region’s adults
- (C) Age group and transport are independent among the region’s adults
- (D) Age group and transport are independent in the sample
Answer and explanation
Answer: (C). One sample, two categorical variables: a chi-square test for independence.
- (A) is Hₐ.
- (B) is about one variable.
- (D) is about the sample.
If you missed this: 3.14 study guide.
Question 15 (short answer · 3.15)
A random sample of 300 visitors to a fictional park records trip type and main activity.
| Hiking | Cycling | Water sports | Total | |
|---|---|---|---|---|
| Day trip | 102 | 48 | 30 | 180 |
| Overnight | 48 | 42 | 30 | 120 |
| Total | 150 | 90 | 60 | 300 |
(a) Name the test, find the expected counts and check the condition. (b) Find χ², df and the p-value; conclude at α = 0.01.
Answer and explanation
(a) Chi-square test for independence (one sample). Expected = row total × column total ÷ 300: Day trip 90, 54, 36; Overnight 60, 36, 24. All exceed 5.
(b) χ² = 1.60 + 0.67 + 1.00 + 2.40 + 1.00 + 1.50 = 8.17; df = (2 − 1)(3 − 1) = 2; p-value = 0.0169. 0.0169 > 0.01, so fail to reject H₀. There is not convincing statistical evidence of an association between trip type and main activity among park visitors.
If you missed this: 3.15 study guide.
Your next step
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 3.1 | 1 | Estimators |
| 3.2 | 2 | Distribution of p̂ |
| 3.3 | 3 | Interval for p |
| 3.4 | 4 | Interval claims |
| 3.5 | 5 | Test set-up |
| 3.6 | 6 | p-values |
| 3.7 | 7 | One-sample test |
| 3.8 | 8 | Errors and power |
| 3.9 | 9 | Difference of p̂s |
| 3.10 | 10 | Two-sample interval |
| 3.11 | 11 | Two-sample claims |
| 3.12 | 12 | Two-sample test set-up |
| 3.13 | 13 | Two-sample test |
| 3.14 | 14 | Chi-square set-up |
| 3.15 | 15 | Chi-square test |
How to use your result
- Mark each answer right, partly right or wrong. A conclusion without context is only partly right.
- Fix weak topics in order: tests (3.5 to 3.7) build on 3.2; two-sample work repeats one-sample steps.
- Read the guide, do its practice set, then try the Unit 3 mixed review, which combines topics.