Pearson Edexcel International Advanced Level Mathematics: Samples and hypothesis tests (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set tests section 4 (topics 4.1 to 4.6, “Hypothesis tests”) of Unit S2: Statistics 2 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics, Further Mathematics and Pure Mathematics specification (YMA01), Issue 3, April 2019. The specification lists S2 as an optional unit for IAS Further Mathematics and for IAL Mathematics and Further Mathematics. Needed cumulative probabilities are given.

Method: study guide and revision notes.

Questions

1. A sports club with 480 members keeps a list of every member. It wants opinions on a new kit from a sample of members.

(a) State the population and the sampling unit. [2] (b) Identify a suitable sampling frame. [1]

2. A company tests each fuse in a batch by passing a current until it blows.

(a) Explain why a census of the batch is not sensible. [2] (b) Give one advantage of using a sample. [1] (c) Give one disadvantage of using a sample survey. [1]

3. A population has values 2, 4 and 6, each equally likely. Samples of size 2 are taken with replacement. T is the smaller sample value.

(a) Find the sampling distribution of T. [3] (b) Find E(T). [2] (c) Explain why T is a statistic. [1]

4. A website says 25% of visitors sign up for its newsletter. After a redesign the owner thinks the proportion has fallen. In a random sample of 28 visitors, 3 sign up. Let X ~ B(28, p) be the number who sign up. Use P(X ≤ 1) = 0.0033, P(X ≤ 2) = 0.0166, P(X ≤ 3) = 0.0551 for p = 0.25.

(a) State suitable hypotheses. [1] (b) Find the critical region for a test at the 5% level. [3] (c) State the conclusion of the test, in context. [1]

5. A machine makes components. Of its components, 10% are defective. After servicing, an engineer suspects the proportion defective has risen. In a random sample of 40 components, 9 are defective. Test at the 1% level. For X ~ B(40, 0.1): P(X ≥ 8) = 0.0419, P(X ≥ 9) = 0.0155, P(X ≥ 10) = 0.0051.

(a) State hypotheses. [1] (b) Show that the critical region is X ≥ 10 and state the actual significance level. [4] (c) State the conclusion in context. [1]

6. A survey says 45% of commuters cycle weekly. A researcher tests whether this has changed, using 18 commuters, at the 5% level (2.5% in each tail). For X ~ B(18, 0.45): P(X ≤ 3) = 0.0120, P(X ≤ 4) = 0.0411, P(X ≥ 12) = 0.0537, P(X ≥ 13) = 0.0183.

(a) State hypotheses. [1] (b) Find the critical region. [4] (c) Find the actual significance level. [1] (d) In the sample, 13 commuters cycle weekly. State the conclusion in context. [1]

7. A call centre receives complaints at a mean rate of 3.2 per day. After a promotion the manager expects an increase. Over 5 days there are 25 complaints. Test at the 5% level. For Po(16): P(X ≥ 23) = 0.0582, P(X ≥ 24) = 0.0367.

(a) State hypotheses and the distribution of the number of complaints in 5 days under H₀. [2] (b) Find the critical region. [2] (c) State the conclusion in context. [1]

8. A server records failed logins at a mean of 0.8 per minute. After an update the administrator thinks the rate has changed. In 10 minutes there are 2. Test at the 10% level (5% in each tail). For Po(8): P(X ≤ 3) = 0.0424, P(X ≤ 4) = 0.0996, P(X ≥ 13) = 0.0638, P(X ≥ 14) = 0.0342.

(a) State hypotheses and the distribution of X under H₀. [2] (b) Find the critical region. [4] (c) State the conclusion in context. [1]

9. A theatre says 15% of bookings are cancelled. After introducing a deposit, the manager thinks fewer are. Of 200 bookings, 22 are cancelled. Test at the 5% level using a normal approximation.

(a) State hypotheses and the normal distribution that approximates X, the number cancelled, under H₀. [3] (b) Calculate the probability, using a continuity correction, that X is at most 22. [3] (c) State the conclusion in context. [1]

10. A fitness app says 25% of users complete a 7-day streak. After adding reminders, the developer thinks this has risen. In a random sample of 12 users, 6 complete a streak. Test at the 5% level. For X ~ B(12, 0.25): P(X ≥ 6) = 0.0544, P(X ≥ 7) = 0.0143.

(a) State hypotheses. [1] (b) Find the critical region. [3] (c) State the conclusion in context. [2] (d) State the actual significance level. [1] (e) Give one comment on the sample size and one assumption the binomial model needs. [2]

11. A water company logs blockages at a mean of 2.5 per hour. An engineer tests whether a new filter has reduced the rate, using 4 hourly periods chosen at random from the 720 hours in a month.

(a) State the sampling unit and a suitable sampling frame. [2] (b) State hypotheses and the distribution of the total count over the 4 hours under H₀. [2] (c) For a 5% level, find the critical region, given P(X ≤ 4) = 0.0293 and P(X ≤ 5) = 0.0671. [3] (d) The total over the 4 hours is 3. State the conclusion in context. [1]

Answers

1. (a) Population: all 480 members [1]. Sampling unit: one member [1]. (b) The secretary’s list of all members [1]. [3] Examiner insight: A frame must list every unit.

2. (a) Testing destroys each fuse [1], so a census would leave nothing to sell [1]. (b) Cheaper or quicker than testing every fuse [1]. (c) Results vary from sample to sample, or a poorly chosen sample may be unrepresentative [1]. [4] Examiner insight: “A census takes too long” ignores the context. Tie the reason to the testing.

3. (a) Nine pairs, each probability 1/9 [1]. T = 2 for 22, 24, 26, 42, 62: P(T = 2) = 5/9 [1]. T = 4 for 44, 46, 64: 3/9. T = 6 for 66: 1/9 [1]. (b) E(T) = (2×5 + 4×3 + 6×1)/9 [1] = 28/9 ≈ 3.11 [1]. (c) It is calculated only from the sample values and contains no unknown parameter [1]. [6] Examiner insight: List all nine pairs first.

4. (a) H₀: p = 0.25, H₁: p < 0.25 [1]. (b) P(X ≤ 3) = 0.0551 > 0.05 [1]. P(X ≤ 2) = 0.0166 < 0.05 [1]. Critical region: X ≤ 2 [1]. (c) 3 is outside the critical region, so H₀ stands. At the 5% level the sample does not give enough evidence that the proportion signing up has fallen [1]. [5] Examiner insight: Show both boundary probabilities.

5. (a) H₀: p = 0.1, H₁: p > 0.1 [1]. (b) P(X ≥ 9) = 0.0155 > 0.01, so 9 is not in the region [1]. P(X ≥ 10) = 0.0051 < 0.01 [1]. Critical region X ≥ 10 [1]. Actual significance level 0.0051 (0.51%) [1]. (c) 9 is outside the critical region, so H₀ stands. At the 1% level the sample does not give enough evidence that the proportion defective has risen [1]. [6] Examiner insight: The same count can reject at 5% but not at 1%, so read the level carefully.

6. (a) H₀: p = 0.45, H₁: p ≠ 0.45 [1]. (b) Lower: P(X ≤ 4) = 0.0411 > 0.025 [1]; P(X ≤ 3) = 0.0120 < 0.025, region X ≤ 3 [1]. Upper: P(X ≥ 12) = 0.0537 > 0.025 [1]; P(X ≥ 13) = 0.0183 < 0.025, region X ≥ 13 [1]. Critical region: X ≤ 3 or X ≥ 13. (c) 0.0120 + 0.0183 = 0.0303 [1]. (d) 13 is in the critical region, so reject H₀. At the 5% level the sample gives enough evidence that the proportion cycling weekly has changed [1]. [7] Examiner insight: Compare each tail with 0.025, not 0.05, and state both regions.

7. (a) H₀: λ = 3.2 per day, H₁: λ > 3.2 [1]. Over 5 days X ~ Po(16) [1]. (b) P(X ≥ 23) = 0.0582 > 0.05 [1]; P(X ≥ 24) = 0.0367 < 0.05, so critical region X ≥ 24 [1]. (c) 25 is in the critical region, so reject H₀. At the 5% level the sample gives enough evidence that the daily rate of complaints has increased [1]. [5] Examiner insight: Scaling the mean to 16 is easy to skip.

8. (a) H₀: λ = 0.8 per minute, H₁: λ ≠ 0.8 [1]. X ~ Po(8) [1]. (b) Lower: P(X ≤ 4) = 0.0996 > 0.05 [1]; P(X ≤ 3) = 0.0424 < 0.05, region X ≤ 3 [1]. Upper: P(X ≥ 13) = 0.0638 > 0.05 [1]; P(X ≥ 14) = 0.0342 < 0.05, region X ≥ 14 [1]. Critical region X ≤ 3 or X ≥ 14. (c) 2 is in the critical region, so reject H₀. At the 10% level the sample gives enough evidence that the rate of failed login attempts has changed [1]. [7] Examiner insight: A two-tailed 10% test uses 5% in each tail.

9. (a) H₀: p = 0.15, H₁: p < 0.15 [1]. Mean = 200 × 0.15 = 30 [1]. Variance = 200 × 0.15 × 0.85 = 25.5, so X ≈ N(30, 25.5) [1]. (b) P(X ≤ 22) ≈ P(Y < 22.5) using the continuity correction [1]. z = (22.5 − 30)/√25.5 = −1.485 [1]. P(Z < −1.485) = 0.0687 [1]. (c) 0.0687 > 0.05, so H₀ stands. At the 5% level the sample does not give enough evidence that the proportion of cancelled bookings has fallen [1]. [7] Examiner insight: Using 22 instead of 22.5 changes the probability.

10. (a) H₀: p = 0.25, H₁: p > 0.25 [1]. (b) P(X ≥ 6) = 0.0544 > 0.05 [1]. P(X ≥ 7) = 0.0143 < 0.05 [1]. Critical region X ≥ 7 [1]. (c) 6 is outside the critical region: do not reject H₀ [1]. At the 5% level the sample does not give enough evidence that the proportion completing a streak has risen [1]. (d) 0.0143 (1.43%) [1]. (e) The sample is small, so a modest rise is unlikely to be detected [1]. Users must be independent with the same probability of a streak [1]. [9] Examiner insight: 0.0544 against 0.05 is a “do not reject”. Do not write that the app has had no effect.

11. (a) Sampling unit: one hourly period [1]. Sampling frame: the list of all 720 hourly periods in the month [1]. (b) H₀: λ = 2.5 per hour, H₁: λ < 2.5 [1]. Over 4 hours X ~ Po(10) [1]. (c) P(X ≤ 5) = 0.0671 > 0.05 [1]. P(X ≤ 4) = 0.0293 < 0.05 [1]. Critical region X ≤ 4 [1]. (d) 3 is in the critical region, so reject H₀. At the 5% level the sample gives enough evidence that the hourly rate of blockages has fallen [1]. [8] Examiner insight: The sampling units are hourly periods, not blockages.

Where marks are usually lost

• Writing hypotheses about X instead of p or λ.
• Leaving the Poisson mean unscaled for a longer interval.
• Setting a single tail against the whole level in a two-tailed test.
• A critical value that overshoots the level, or not checking the next value down.
• A missing or reversed continuity correction.
• A conclusion with no context, or “H₀ accepted”.

Next steps

• Revision notes and study guide
• Course hub and checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Unit S2: Statistics 2, section 4 (Hypothesis tests), topics 4.1 to 4.6. Published by Pearson Education Limited.