The shape of the lesson
By the end of the lesson, learners can
- The language of a test: Every test follows the same script.
- Two-tailed tests and Poisson tests: In a two-tailed test, compare the tail probability with HALF the significance level.
- Using a normal approximation: When \(n\) is large (binomial with \(np, nq > 5\)) or \(\lambda > 15\) (Poisson), use a normal approximation with a continuity correction for the tail probability, exactly as in the approximation lessons.
- Hypotheses: \(H_0: p = p_0\) (or \(\lambda = \lambda_0\)); \(H_1\) with \(<\), \(>\) or \(\ne\) from the wording.
- Distribution under \(H_0\): \(B(n, p_0)\) or \(\text{Po}(\lambda_0)\), scaled to the observed period.
- Tail probability: observed value or more extreme, in the direction of \(H_1\).
- Compare with the significance level (half of it in each tail for a two-tailed test).
- Conclude in context, as evidence, never as proof.
- Large \(n\) or \(\lambda\): normal approximation with continuity correction.
One picture every idea returns to
Assume \(H_0\) is true. Find the probability of a result at least as extreme as the one observed (in the direction of \(H_1\)), or find the critical region. If that probability is below the significance level, the result lies in the critical region: reject \(H_0\). Always conclude in context and without certainty: "there is (in)sufficient evidence that ...".
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Is it chance, or a real change?. Paper 6. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The language of a test | Every test follows the same script. Model the worked example A one-tailed binomial test on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Two-tailed tests and Poisson tests | In a two-tailed test, compare the tail probability with HALF the significance level. Model the worked example Two-tailed on the board, one line at a time. Your turn: A machine breaks down at a mean rate of 3.2 per month. After a service it breaks down once in the next 2 months. Test at the 2% level whether the rate has fallen. | Pairs |
| 21 to 30 min | Teach and check: Using a normal approximation | When \(n\) is large (binomial with \(np, nq > 5\)) or \(\lambda > 15\) (Poisson), use a normal approximation with a continuity correction for the tail probability, exactly as in the approximation lessons. Model the worked example Poisson with a large mean on the board, one line at a time. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 24 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Using a normal approximation, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Compare the TAIL probability \(\text{P}(X \ge 9)\) with 5%, never the single probability \(\text{P}(X = 9)\). | And never write "\(H_0\) is true" or "this proves the die is biased": a test only gives evidence. |
| Write hypotheses about the PARAMETER (\(p\) or \(\lambda\)), not about the observed count, and scale \(\lambda\) to the period of the observation: 2 months at 3.2 per month means \(H_0: \lambda = 6.4\). | Return to Two-tailed tests and Poisson tests: ask a pair to state the correct version and show it on the lab before moving on. |
| The continuity correction points AWAY from the tail: \(X \ge 95\) becomes \(Y > 94.5\), and \(X \le 8\) becomes \(Y < 8.5\). | Return to Using a normal approximation: ask a pair to state the correct version and show it on the lab before moving on. |
Hypothesis tests with binomial and Poisson distributions: worksheet
6 questions, 24 marks. Written in the style of the papers; no past paper question is reproduced.
Explain what is meant by (a) the critical region (b) the significance level of a hypothesis test.
A coin is tossed 20 times and lands heads 15 times. Test at the 5% significance level whether the coin is biased towards heads.
Under \(H_0\), \(X \sim B(25, 0.2)\). Find the critical region for a test of \(H_1: p < 0.2\) at the 10% level, and state the actual significance level.
A machine breaks down at random at a mean rate of 3.2 times per month. After a service, it breaks down once in the following 2 months. Test at the 2% level whether the mean rate has decreased.
A shop believes that 40% of its customers pay in cash. In a random sample of 30 customers, 6 pay in cash. Test at the 10% level whether the proportion paying in cash is different from 40%.
A council claims that 60% of residents support a new park. In a random survey of 150 residents, 78 support it.
(a) Use a suitable approximation to test at the 5% level whether support is lower than the council claims.
(b) Explain why a normal approximation is valid here.
Answer key and marking notes
- (a) The set of values of the test statistic for which \(H_0\) is rejected.
- (b) The probability threshold for rejecting \(H_0\); equivalently, the probability of a result in the critical region when \(H_0\) is true.
- \(H_0: p = 0.5\); \(H_1: p > 0.5\). Under \(H_0\), \(X \sim B(20, 0.5)\).
- \(\text{P}(X \ge 15) = 0.0207\).
- \(0.0207 < 0.05\): reject \(H_0\).
- There is sufficient evidence at the 5% level that the coin is biased towards heads.
- \(\text{P}(X \le 2) = 0.0982 \le 0.10\); \(\text{P}(X \le 3) = 0.2340 > 0.10\).
- Critical region \(X \le 2\); actual significance level \(9.82\%\).
- \(H_0: \lambda = 6.4\) (2 months); \(H_1: \lambda < 6.4\). Under \(H_0\), \(X \sim \text{Po}(6.4)\).
- \(\text{P}(X \le 1) = e^{-6.4}(1 + 6.4) = 0.0123\).
- \(0.0123 < 0.02\): reject \(H_0\).
- There is sufficient evidence at the 2% level that the breakdown rate has decreased.
- \(H_0: p = 0.4\); \(H_1: p \ne 0.4\) (two-tailed: 5% in each tail).
- Under \(H_0\), \(X \sim B(30, 0.4)\) with mean 12; 6 is below, so use the lower tail.
- \(\text{P}(X \le 6) = 0.0172\).
- \(0.0172 < 0.05\): reject \(H_0\).
- There is sufficient evidence at the 10% level that the proportion paying in cash is not 40%.
- (a) \(H_0: p = 0.6\); \(H_1: p < 0.6\). \(X \sim B(150, 0.6) \approx N(90, 36)\).
- \(\text{P}(X \le 78) \approx \text{P}(Y < 78.5)\), \(z = \dfrac{78.5 - 90}{6} = -1.917\).
- \(\Phi(-1.917) = 1 - 0.9724 = 0.0276 < 0.05\): reject \(H_0\).
- There is sufficient evidence at the 5% level that support is lower than 60%.
- (b) \(np = 90\) and \(nq = 60\) are both greater than 5.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A manufacturer claims that 20% of its customers buy an extended warranty. A researcher believes that the true proportion is lower. She asks a random sample of 30 customers. Which are the correct hypotheses, where \(p\) is the proportion of all customers who buy a warranty?
- A\(H_0: p = 0.2\), \(H_1: p < 0.2\)
- B\(H_0: p < 0.2\), \(H_1: p = 0.2\)
- C\(H_0: p = 0.2\), \(H_1: p \ne 0.2\)
- D\(H_0: X = 6\), \(H_1: X < 6\)
A coin is thrown 12 times and shows 10 heads. Test at the 5% significance level whether the coin is biased towards heads. Which is correct?
- A\(\text{P}(X \ge 10) = 0.0193 < 0.05\), so reject \(H_0\)
- B\(\text{P}(X = 10) = 0.0161 < 0.05\), so reject \(H_0\)
- C\(\text{P}(X \ge 10) = 0.0193 < 0.05\), so do not reject \(H_0\)
- D\(\text{P}(X \le 10) = 0.997 > 0.05\), so do not reject \(H_0\)
\(X \sim B(20, p)\). A test of \(H_0: p = 0.25\) against \(H_1: p > 0.25\) is carried out at the 5% significance level. Find the critical region.
- A\(X \ge 9\)
- B\(X \ge 8\)
- C\(X \ge 7\)
- D\(X \le 1\)
Faults on a production line used to occur at random at a mean rate of 5 per week. After a new maintenance schedule, there was 1 fault in a randomly chosen week. Test at the 5% significance level whether the mean rate has decreased. Which is correct?
- A\(\text{P}(X \le 1) = 0.0404 < 0.05\): there is sufficient evidence that the mean rate has decreased
- B\(\text{P}(X = 1) = 0.0337 < 0.05\): there is sufficient evidence that the mean rate has decreased
- C\(\text{P}(X \le 1) = 0.0404 < 0.05\): this proves that the mean rate has decreased
- D\(\text{P}(X \ge 1) = 0.993 > 0.05\): there is insufficient evidence that the mean rate has decreased
In the past, 30% of applicants to a college were from overseas. In a random sample of 25 applicants this year, 12 were from overseas. Test at the 5% significance level whether the proportion has changed. Which is correct?
- A\(\text{P}(X \ge 12) = 0.0442 > 0.025\), so there is insufficient evidence that the proportion has changed
- B\(\text{P}(X \ge 12) = 0.0442 < 0.05\), so there is sufficient evidence that the proportion has changed
- C\(\text{P}(X = 12) = 0.0268 > 0.025\), so there is insufficient evidence that the proportion has changed
- D\(\text{P}(X \le 12) = 0.983 > 0.025\), so there is insufficient evidence that the proportion has changed
\(X \sim B(20, p)\). A test of \(H_0: p = 0.4\) against \(H_1: p \ne 0.4\) is carried out at the 10% significance level. Find the critical region.
- A\(X \le 3\) or \(X \ge 13\)
- B\(X \le 4\) or \(X \ge 12\)
- C\(X \ge 12\)
- D\(X \le 4\) or \(X \ge 13\)
Complaints to a council used to arrive at random at a mean rate of 2.5 per day. In a period of 10 days after a policy change there were 34 complaints. Using a suitable approximation, test at the 5% significance level whether the mean rate has increased. Which is correct?
- AProbability \(0.0446 < 0.05\): there is sufficient evidence that the mean rate has increased
- BProbability \(0.0359 < 0.05\): there is sufficient evidence that the mean rate has increased
- CProbability \(0.0287 < 0.05\): there is sufficient evidence that the mean rate has increased
- DProbability \(0.0446 < 0.05\): there is insufficient evidence that the mean rate has increased
Customers arrive at a help desk at random at a mean rate of 2.5 per hour. A manager will test whether the rate has decreased by counting the arrivals in a 3-hour period, using a 5% significance level. Find the critical region.
- A\(X \le 2\)
- B\(X \le 3\)
- C\(X \le 4\)
- D\(X \ge 13\)
Quiz key
\(H_0\) is the claim, \(p = 0.2\), and the researcher suspects the proportion is lower, so \(H_1: p < 0.2\) (one-tailed). The tempting \(p \ne 0.2\) is two-tailed, used only when no direction is suggested. Swapping the hypotheses puts the inequality in \(H_0\), and hypotheses must be about the parameter \(p\), not the observed count \(X\).
\(\text{P}(X \ge 10) = \dfrac{\binom{12}{10} + \binom{12}{11} + \binom{12}{12}}{2^{12}} = \dfrac{66 + 12 + 1}{4096} = 0.0193\). This is less than 0.05, so the result is in the critical region: reject \(H_0\); there is sufficient evidence at the 5% level that the coin is biased towards heads. The tempting \(\text{P}(X = 10)\) reaches the right decision by the wrong method: always compare a TAIL probability. Not rejecting when the probability is below the level reverses the rule, and \(\text{P}(X \le 10)\) is the wrong tail.
With \(p = 0.25\): \(\text{P}(X \ge 8) = 0.1018\), which is more than 0.05, and \(\text{P}(X \ge 9) = 0.0409\), which is less. So the critical region is \(X \ge 9\). The tempting \(X \ge 8\) has probability \(0.1018 > 0.05\), and \(X \ge 7\) (\(0.214\)) is larger still. \(X \le 1\) is in the wrong tail for \(H_1: p > 0.25\).
\(\text{P}(X \le 1) = e^{-5}(1 + 5) = 6e^{-5} = 0.0404\), which is less than 0.05, so reject \(H_0\): there is sufficient evidence at the 5% level that the mean rate has decreased. The tempting \(\text{P}(X = 1)\) uses a single probability, not the tail. A test gives evidence, never proof, and \(\text{P}(X \ge 1)\) is the wrong tail for a decrease.
\(X \sim B(25, 0.3)\) under \(H_0\), with mean 7.5. 12 is above the mean, so the tail is \(\text{P}(X \ge 12) = 1 - \text{P}(X \le 11) = 0.0442\). For a two-tailed test at 5%, compare with 0.025: \(0.0442 > 0.025\), so do not reject \(H_0\); there is insufficient evidence at the 5% level that the proportion has changed. The tempting second option compares with 0.05, as if the test were one-tailed. The third uses a single probability instead of the tail, and the fourth uses the wrong tail.
With \(p = 0.4\): \(\text{P}(X \le 3) = 0.0160\) but \(\text{P}(X \le 4) = 0.0510 > 0.05\), so the lower tail is \(X \le 3\). \(\text{P}(X \ge 13) = 0.0210\) but \(\text{P}(X \ge 12) = 0.0565 > 0.05\), so the upper tail is \(X \ge 13\). The tempting \(X \le 4\) or \(X \ge 12\) compares each tail with the full 10%. \(X \ge 12\) alone is the one-tailed 10% region, and \(X \le 4\) treats \(0.0510\) as if it were below 0.05.
\(H_0: \lambda = 25\), \(H_1: \lambda > 25\) for 10 days. \(\text{P}(X \ge 34) \approx \text{P}(Y > 33.5)\) with \(Y \sim N(25, 25)\): \(z = \dfrac{33.5 - 25}{5} = 1.7\), and \(1 - \Phi(1.7) = 1 - 0.9554 = 0.0446 < 0.05\). Reject \(H_0\): there is sufficient evidence at the 5% level that the mean rate has increased. The tempting \(0.0359\) uses 34 with no continuity correction, and \(0.0287\) uses 34.5, which leaves out 34 itself. The last option has the right probability but the wrong conclusion.
\(X \sim \text{Po}(7.5)\) under \(H_0\). \(\text{P}(X \le 2) = e^{-7.5}(1 + 7.5 + 28.125) = 0.0203\), and \(\text{P}(X \le 3) = 0.0203 + e^{-7.5}\dfrac{7.5^{3}}{6} = 0.0591 > 0.05\). So the critical region is \(X \le 2\). The tempting \(X \le 3\) has probability \(0.0591\), more than 5%; \(X \le 4\) (\(0.132\)) is larger still. \(X \ge 13\) is in the upper tail, which tests for an increase, not a decrease.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.