Hypothesis tests

• Tail, not point. Halve the level for two tails. Conclude with “evidence”, in context, at a stated percentage.
• Tail, not point. Halve the level for two tails. P(Type I) is the actual region probability for a discrete test. Conclude with evidence, in context, at a stated level.

• 6.5.1 I understand the nature of a hypothesis test, the difference between one-tailed and two-tailed tests, and the terms null hypothesis, alternative hypothesis, significance level, rejection region (or critical region), acceptance region and test statistic
• 6.5.2 I can formulate hypotheses and carry out a hypothesis test in the context of a single observation from a population which has a binomial or Poisson distribution, using (a) direct evaluation of probabilities (b) a normal approximation to the binomial or the Poisson distribution, where appropriate
• 6.5.3 I can formulate hypotheses and carry out a hypothesis test concerning the population mean in cases where the population is normally distributed with known variance or where a large sample is used
• 6.5.4 I understand the terms Type I error and Type II error in relation to hypothesis tests
• 6.5.5 I can calculate the probabilities of making Type I and Type II errors in specific situations involving tests based on a normal distribution or direct evaluation of binomial or Poisson probabilities

Accuracy for this chapter. Give a tail probability to 3 significant figures, but compare it with the level to as many figures as the decision needs: 0.0127 against 0.01, 0.0044 against 0.005. Keep \(\sigma/\sqrt{n}\), \(s/\sqrt{n}\), \(s^2\) and a critical \(\bar{x}\) unrounded (51.028125, not 51.03) until the end. Give \(z\) to 3 decimal places and read \(\Phi\) with the ADD column for the third decimal as written: \(\Phi(2.237) = 0.9871 + 0.0002 = 0.9873\). Quote critical values exactly as printed: 1.960, not 1.96. The table stops at \(z = 2.99\); beyond it, write \(\Phi(z) > 0.9986\) rather than inventing a value.

• “\(\text{P}(X = 2) = 0.0278 < 0.05\), so reject H\(_0\).” Correct Tail, not point. The test asks how likely a result at least as extreme as the one observed is, so it uses \(\text{P}(X \le 2) = 0.0355\) for H\(_1\): \(p < p_0\), or \(\text{P}(X \ge x)\) for H\(_1\): \(p > p_0\). Here the decision happens to be the same, but the method is not, and elsewhere the two disagree: for \(\text{B}(10, 0.5)\), H\(_1\): \(p > 0.5\) and \(x = 8\), the point probability \(\text{P}(X = 8) = 0.0439\) is below 0.05 while the tail \(\text{P}(X \ge 8) = 0.0547\) is not.
• “Reject H\(_0\). The mean has increased.” Correct Conclude with evidence, in context, at a stated level. “There is evidence at the 5% level that the mean mass of the bags has increased.” Or, when not rejecting: “There is insufficient evidence at the 5% level that the mean has changed.” Never “H\(_0\) is true”: a test never proves the null hypothesis.
• “The test is at 5%, so P(Type I error) = 0.05.” (for a binomial test) Correct For a discrete test, P(Type I) is the actual probability of the rejection region. With region \(X \ge 7\) for B(15, 0.2) it is \(0.0181\). Only a test on a continuous (normal) statistic has P(Type I) exactly equal to the level.
• “Two-tailed at 5%: the tail is 0.03, less than 0.05, so reject.” Correct A two-tailed test splits the level between the tails: compare the one-tail probability with 2.5%. \(0.03 > 0.025\): do not reject.
• “H\(_0\): \(\bar{x} = 51.2\)” or “H\(_0\): \(p < 0.3\)” Correct Hypotheses are about the population parameter (\(p\), \(\lambda\), \(\mu\)), and H\(_0\) carries the equals sign: H\(_0\): \(\mu = 50\), H\(_1\): \(\mu > 50\).
• “\(z = \dfrac{51.2 - 50}{2.5}\)” Correct A test for a mean is about \(\bar{X}\), whose standard deviation is \(\sigma/\sqrt{n}\): \(z = \dfrac{51.2 - 50}{2.5/\sqrt{16}} = 1.92\).
• “H\(_0\): \(\bar{x} = 51.2\).” Repair Hypotheses are about the population parameter, not the sample: H\(_0\): \(\mu = 50\).
• “H\(_0\): \(p < 0.3\), H\(_1\): \(p = 0.3\).” Repair H\(_0\) carries the equals sign, because the calculation is done assuming it; the inequality belongs to H\(_1\): H\(_0\): \(p = 0.3\), H\(_1\): \(p < 0.3\).
• “The question says ‘has changed’, so H\(_1\): \(\mu > 50\).” Repair “Changed” has no direction, so H\(_1\): \(\mu \neq 50\), two-tailed. Only “increased” (or a word like it) gives \(>\).
• “\(\text{P}(X = 2) = 0.0278 < 0.05\), so reject H\(_0\).” Repair Use the tail \(\text{P}(X \le 2) = 0.0355\), the probability of a result at least as extreme as the one observed.
• “Two-tailed at 5%: the tail is 0.03 < 0.05, so reject.” Repair Compare the one-tail probability with 2.5%: \(0.03 > 0.025\), do not reject.
• “Reject H\(_0\), so the mean has increased.” Repair “Reject H\(_0\); there is evidence at the 5% level that the mean has increased.” The evidence, the level and the context all belong in the sentence.
• “Do not reject H\(_0\), so H\(_0\) is true.” Repair “There is insufficient evidence at the 5% level to reject H\(_0\).” A test never proves H\(_0\).
• “Poisson test, rate 0.5 per week, 6 observed in 4 weeks: H\(_0\): \(\lambda = 0.5\), \(\text{P}(X \ge 6)\) under Po(0.5).” Repair Scale the mean to the interval observed: under H\(_0\) the count in 4 weeks is Po(2).
• “\(\text{P}(X \ge 96) \approx \text{P}(Y > 96)\).” Repair Apply the continuity correction so that 96 is inside the tail: \(\text{P}(Y > 95.5)\).
• “\(z = \dfrac{51.2 - 50}{2.5}\).” Repair The test is about \(\bar{X}\), so divide by \(\sigma/\sqrt{n} = 2.5/\sqrt{16} = 0.625\): \(z = 1.92\).
• “Large-sample test: \(s^2 = \dfrac{\Sigma(x - \bar{x})^2}{n}\).” Repair Use the unbiased estimate with divisor \(n - 1\), as MF19 prints it.
• “P(Type I error) = 0.05” for the binomial test with rejection region \(X \ge 7\). Repair For a discrete test it is the actual probability of the region: \(\text{P}(X \ge 7 \mid p = 0.2) = 0.0181\).
• “P(Type II) \(= \text{P}(\bar{X} \le 51.03 \mid \mu = 50)\).” Repair A Type II error happens when H\(_0\) is false, so compute under the given true value: \(\text{P}(\bar{X} \le 51.028125 \mid \mu = 52) = 0.0600\).
• “A Type I error is accepting H\(_0\) when it is false.” Repair That is a Type II error. A Type I error is rejecting H\(_0\) when it is true.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. This is the last section of Paper 6, and it uses all the others: the binomial and Poisson distributions (chapters 26 and 28), the normal approximations (chapters 27 and 28) and the sample mean (chapter 31). When you revise any of those chapters, finish by turning one of its questions into a test: add hypotheses, a level and a conclusion. Recalling a method inside a new problem is worth more than another pass over this chapter on its own.

• Chapter 32 · Probability & Statistics 2 · How it is assessed
• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Probability & Statistics 2 content, examined in Paper 6. Paper 6 (Probability & Statistics 2) is offered only as part of the A Level, where it is 20%. It assumes the whole of the Paper 5 content and the calculus of Paper 3. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A hypothesis-test question on Paper 6 is set in a context: a claimed proportion, an average rate, a target mean. A question may ask for a test at a stated level, and may go on to ask for the rejection region, a Type I or Type II error stated in context, or the probability of one. The hypotheses, the distribution used and the conclusion are each part of the answer, not decoration around the number.
• MF19 prints the binomial and Poisson formulae, the formula for \(s^2\), the Central Limit Theorem line, the normal table \(\Phi(z)\) and the critical values. It prints no test procedure, no definition of the terms and no Type I or Type II formula: all of those must be known. See the MF19 card.
• Write the tail probability to 3 significant figures, but compare it with the level to as many figures as the decision needs (0.0127 against 0.01). Keep \(\sigma/\sqrt{n}\) and \(s/\sqrt{n}\) unrounded, give \(z\) to 3 decimal places, read \(\Phi\) to 4 decimal places with the ADD column. A conclusion without the hypotheses and the comparison line is an unsupported answer.