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A-Level · Further Statistics 1 (Hypothesis Testing for a Poisson or Binomial Parameter)

FP.FS4 Further Statistics: Hypothesis Testing, Type I and Type II Errors

AQA 7367 · Calculator allowed · about 120 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Hypothesis Testing for a Poisson or Binomial Parameter

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For a hypothesis test on a Poisson mean lambda or a binomial probability p, a null hypothesis H0 states a fixed value of the parameter, and an alternative hypothesis H1 states that the parameter has increased, has decreased (a one-tailed test), or is simply different (a two-tailed test). A test statistic X, assumed to follow the distribution stated in H0, is compared against a critical region: the set of values of X for which H0 is rejected in favour of H1.

Because X is discrete, the probability of X falling in a proposed critical region can rarely be made to equal a chosen significance level exactly. The usual convention is to choose the largest critical region whose probability does not exceed the significance level requested. This probability is called the actual significance level of the test, and it is also equal to P(Type I error), the probability of wrongly rejecting a true H0.

A Type II error occurs when H0 is not rejected even though H1 is actually true. Its probability depends on the true value of the parameter, and is found by calculating the probability that X falls outside the critical region, using the true (alternative) value of the parameter in place of the value stated in H0. The power of a test is defined as 1 - P(Type II error): the probability of correctly rejecting a false H0. Power increases as the true parameter value moves further from the value stated in H0, and is also affected by the significance level chosen and the sample size used.

A car insurance company models the number of claims made per month by a certain class of driver using a Poisson distribution, and regularly carries out hypothesis tests on the mean claim rate.

(a)State what is meant by the critical region (rejection region) of a hypothesis test.(1)

(b)Explain why, for a test statistic with a discrete distribution such as Poisson or binomial, the actual significance level of a test is not usually exactly equal to the significance level originally requested.(2)

(c)A test is carried out, at a chosen significance level, of H0: the mean monthly claim rate is unchanged, against H1: the mean monthly claim rate has increased. Define, in the context of this test, a Type I error and a Type II error.(2)

(d)State an equation linking the power of a hypothesis test to P(Type II error).(1)

(Total for Question 1 is 6 marks)

A call centre in Preston models the number of customer complaints received per day, under normal conditions, by X ~ Po(4). Following a change to the automated phone system, the manager wants to test, at the 5% significance level, whether the mean number of complaints per day has decreased.

(a)State suitable null and alternative hypotheses, then find the critical region for the test and state the actual significance level.(3)

(b)On a particular day following the change, the number of complaints received is 0. State, with a reason, the conclusion of the test.(2)

(c)Suppose that the change has in fact reduced the true mean number of complaints per day to 2. Find the probability that this test results in a Type II error.(3)

(d)Hence state the power of the test in this case.(1)

(Total for Question 2 is 9 marks)

A garden centre in Exeter finds that, historically, 30% of tomato seedlings supplied by a particular grower are affected by blight in damp weather. A random sample of 20 seedlings is taken, and X is the number affected. A new fungicide is applied to the grower's stock, and the centre wants to test, at the 5% significance level, whether the fungicide has reduced the proportion affected below 0.3.

(a)State suitable hypotheses, then find the critical region for the test and state the actual significance level.(4)

(b)After treatment, 1 of the 20 sampled seedlings is found to be affected by blight. State, with a reason, the conclusion of the test.(2)

(c)Suppose that the fungicide has in fact reduced the true proportion of seedlings affected to 0.15. Find the probability that the test results in a Type II error.(3)

(d)Hence state the power of the test in this case.(1)

(e)State, without further calculation, what would happen to the probability of a Type II error found in part (c) if the significance level of the test were increased from 5% to 10%. Justify your answer.(2)

(Total for Question 3 is 12 marks)

The number of minor faults found on newly manufactured circuit boards from a particular production line is modelled, under normal conditions, by X ~ Po(6). Quality control engineers want to test, at the 10% significance level, whether the mean number of faults per board has changed from 6 (a two-tailed test).

(a)State suitable hypotheses, then find the critical region for this two-tailed test and state the actual significance level.(6)

(b)A randomly chosen board is found to have 10 faults. State, with a reason, the conclusion of the test.(2)

(c)Suppose that a fault in the manufacturing process has in fact increased the true mean number of faults per board to 9. Find the probability that the test results in a Type II error.(3)

(d)Hence state the power of the test in this case.(1)

(e)Comment on the suitability of this test for detecting the increase in mean described in part (c).(1)

(Total for Question 4 is 13 marks)

The number of typing errors per page of a large legal document is modelled, under normal conditions, by X ~ Po(10). A proofreading test uses the critical region X ≤ 3 or X ≥ 18 to test whether the mean number of errors per page differs from 10.

(a)Find the probability of a Type I error for this test.(5)

(Total for Question 5 is 5 marks)

A machine at a plastics factory in Telford produces components, 20% of which have historically been classed as premium grade. After the machine is recalibrated, an engineer wants to test, at the 5% significance level, whether the proportion of premium-grade components has increased. A random sample of 25 components is taken, and X is the number classed as premium grade.

(a)State suitable hypotheses, then find the critical region for the test and state the actual significance level.(4)

(b)In the sample, 10 components are classed as premium grade. State, with a reason, the conclusion of the test.(2)

(c)Suppose that the recalibration has in fact increased the true proportion of premium-grade components to 0.35. Find the probability that the test results in a Type II error.(4)

(d)Hence state the power of the test in this case.(1)

(e)State, without further calculation, what would happen to the power of the test found in part (d) if the sample size were increased from 25 to 50, with the significance level kept at 5%. Justify your answer.(2)

(Total for Question 6 is 13 marks)

A fairground game at a summer fete in Durham is designed so that a player wins on any go with probability 0.4, independently of other goes. The organiser wants to test, at the 10% significance level, whether the true win probability differs from 0.4, based on X, the number of wins in 15 goes.

(a)State suitable hypotheses, then find the critical region for this two-tailed test and state the actual significance level.(6)

(b)In one trial of 15 goes, there are 3 wins. State, with a reason, the conclusion of the test.(2)

(c)State one advantage of using a 10% significance level rather than a 5% significance level for this test, together with a corresponding disadvantage.(1)

(Total for Question 7 is 9 marks)

Referring again to the garden centre's fungicide trial in Question 3, where X ~ B(20, 0.3) under H0, suppose instead that the garden centre wants the actual significance level of the one-tailed (decrease) test to not exceed 1%.

(a)Find the resulting critical region, and state the actual significance level.(3)

(b)Explain briefly why this stricter test is less likely to detect a genuine reduction in the proportion of affected seedlings than the 5% test used in Question 3(a).(2)

(Total for Question 8 is 5 marks)

A hospital microbiology lab models the number of bacterial colonies per agar plate, under safe conditions, as a Poisson random variable with a known mean. Two proposed testing procedures are available to check whether the mean number of colonies per plate has risen above the safe threshold: Test A uses a 1% significance level, and Test B uses a 10% significance level. Failing to detect a genuine rise in bacterial colonies (a Type II error) could pose a public health risk, while wrongly signalling a rise when none has occurred (a Type I error) would trigger a costly, unnecessary shutdown of the affected ward for deep cleaning.

Discuss, with reference to Type I and Type II errors, which of the two significance levels the lab should adopt, and justify a recommendation.

(Total for Question 9 is 6 marks)

A textile mill in Bradford inspects rolls of fabric for flaws. Under normal production, the number of flaws found on a 20m roll is modelled by X ~ Po(8). Following a change of yarn supplier, the mill wants to test, at the 5% significance level, whether the mean number of flaws per roll has decreased.

(a)State suitable hypotheses, then find the critical region for the test and state the actual significance level.(4)

(b)Suppose that, in fact, the new yarn has reduced the true mean number of flaws per roll to 5. Find the probability of a Type II error.(3)

(c)Hence state the power of the test in this case.(1)

(d)The mill's quality manager wants the power of the test to be at least 0.5 when the true mean is 5, without inspecting more than one roll. Suggest one change to the test that could achieve this, and state one disadvantage of your suggestion.(2)

(Total for Question 10 is 10 marks)

A market-research company uses a 12-sided spinner which should show each of the numbers 1 to 12 with equal probability. To check whether the spinner is biased towards even numbers, an assistant spins it 12 times and records X, the number of times an even number occurs. If the spinner is fair, X ~ B(12, 0.5). In one trial, an even number occurs on 9 of the 12 spins.

(a)Calculate the two-tailed p-value associated with this result.(4)

(b)State the conclusion of the test at the 5% significance level, with a reason.(1)

(Total for Question 11 is 5 marks)

Referring again to the textile mill's test in Question 10, where X ~ Po(8) under H0 and the critical region for the 5% one-tailed (decrease) test was found to be X ≤ 3, suppose instead that the new yarn has reduced the true mean number of flaws per roll to 3 (rather than 5, as in Question 10).

(a)Using the same critical region as in Question 10, find the probability of a Type II error, and hence the power of the test, in this case.(5)

(b)Compare the power found in part (a) with the power found in Question 10(c), and explain, with reference to the two Poisson distributions involved, why the powers differ.(3)

(Total for Question 12 is 8 marks)

Mark scheme · FP.FS4 Further Statistics: Hypothesis Testing, Type I and Type II Errors

Question 1

(a) B1 the set (range) of values of the test statistic that would lead to the rejection of H0 in favour of H1 (oe)
(a) Answer: The critical region is the set of values of the test statistic for which H0 is rejected in favour of H1.
(b) B1 because the test statistic is discrete, its cumulative probabilities take only a limited set of values, so a critical region with probability exactly equal to the requested level usually cannot be found (oe)
(b) B1 the convention is to choose the largest critical region whose probability does not exceed the requested level, so the actual significance level found is usually slightly less than (and never more than) the level requested (oe)
(b) Answer: Cumulative probabilities for a discrete distribution take only a limited set of values, so it is usually impossible to match the requested level exactly; the largest critical region not exceeding the requested level is chosen instead, giving an actual significance level that is usually a little below what was requested.
(c) B1 Type I error: concluding that the mean claim rate has increased when it has not actually changed (i.e. rejecting a true H0) (oe)
(c) B1 Type II error: concluding that there is not enough evidence that the mean claim rate has increased when it actually has increased (i.e. not rejecting a false H0) (oe)
(c) Answer: Type I error: wrongly concluding the mean claim rate has increased when it has not. Type II error: wrongly concluding there is insufficient evidence of an increase when the mean claim rate has in fact increased.
(d) B1 power = 1 - P(Type II error), cao
(d) Answer: power = 1 - P(Type II error)

Question 2

(a) M1 states H0: λ = 4, H1: λ < 4, and attempts to compare P(X≤0) and P(X≤1) with 0.05
(a) A1 P(X≤0) = e^-4 = awrt 0.0183 (< 0.05) but P(X≤1) = awrt 0.0916 (> 0.05), so critical region is X = 0, cao
(a) A1 actual significance level = awrt 0.0183
(a) Answer: Critical region: X = 0. Actual significance level = awrt 0.0183.
(b) B1 0 lies in the critical region found in part (a) (ft)
(b) B1 reject H0; there is evidence, at the 5% level, that the mean number of complaints per day has decreased (ft, in context)
(b) Answer: 0 is in the critical region {0}, so H0 is rejected; there is evidence at the 5% level that the mean daily complaint rate has decreased.
(c) M1 recognises that a Type II error occurs here when X does not fall in the critical region {0}, i.e. when X ≥ 1, using the true distribution Po(2)
(c) M1 uses P(X≥1 | λ=2) = 1 - P(X=0 | λ=2) = 1 - e^-2
(c) A1 awrt 0.865
(c) Answer: P(Type II error) = 1 - e^-2 = awrt 0.865
(d) B1 power = 1 - 0.865 = awrt 0.135 (ft from part (c))
(d) Answer: power = awrt 0.135

Question 3

(a) M1 states H0: p = 0.3, H1: p < 0.3, with X ~ B(20, 0.3) under H0, and attempts a lower-tail search comparing cumulative probabilities with 0.05
(a) M1 compares P(X≤2) and P(X≤3) with 0.05
(a) A1 critical region is X ≤ 2 (i.e. {0, 1, 2}), cao
(a) A1 actual significance level = awrt 0.0355
(a) Answer: Critical region: X ≤ 2 (i.e. {0, 1, 2}). Actual significance level = awrt 0.0355.
(b) B1 1 lies in the critical region {0, 1, 2} found in part (a) (ft)
(b) B1 reject H0; there is evidence, at the 5% level, that the fungicide has reduced the proportion of seedlings affected by blight (ft, in context)
(b) Answer: 1 is in the critical region {0, 1, 2}, so H0 is rejected; there is evidence at the 5% level that the fungicide has reduced the proportion affected.
(c) M1 recognises the Type II error region is the complement of the critical region, i.e. X ≥ 3, using the true distribution B(20, 0.15)
(c) M1 uses P(X≥3 | p=0.15) = 1 - P(X≤2 | p=0.15)
(c) A1 awrt 0.595
(c) Answer: P(Type II error) = awrt 0.595
(d) B1 power = 1 - 0.595 = awrt 0.405 (ft from part (c))
(d) Answer: power = awrt 0.405
(e) B1 the probability of a Type II error would decrease (oe)
(e) B1 a larger significance level allows a larger critical region to be used; more of the alternative distribution's probability then falls inside this larger critical region (i.e. leads to rejecting H0), leaving less probability in the acceptance region, so P(Type II error) falls (oe)
(e) Answer: The probability of a Type II error would decrease, because a 10% test uses a larger critical region, so a greater proportion of the B(20, 0.15) distribution falls inside it, correctly leading to rejection of H0.

Question 4

(a) M1 states H0: λ = 6, H1: λ ≠ 6, and searches the lower tail comparing P(X≤1) and P(X≤2) with 0.05
(a) A1 lower critical region X ≤ 1 (i.e. {0, 1}), actual lower-tail level = awrt 0.0174
(a) M1 searches the upper tail comparing P(X≥10) and P(X≥11) with 0.05
(a) A1 upper critical region X ≥ 11 (i.e. {11, 12, 13, ...}), actual upper-tail level = awrt 0.0426
(a) A1 combined critical region stated as {0, 1} union {11, 12, 13, ...}
(a) A1 total actual significance level = 0.0174 + 0.0426 = awrt 0.0600
(a) Answer: Critical region: {0, 1} union {11, 12, 13, ...}. Total actual significance level = awrt 0.0600.
(b) B1 10 does not lie in the critical region {0, 1} union {11, 12, ...} found in part (a) (ft)
(b) B1 do not reject H0; there is insufficient evidence, at the 10% level, that the mean number of faults per board has changed (ft, in context)
(b) Answer: 10 is not in the critical region, so H0 is not rejected; there is insufficient evidence at the 10% level that the mean has changed.
(c) M1 identifies the acceptance (non-rejection) region as 2 ≤ X ≤ 10, using the true distribution Po(9)
(c) M1 uses P(2≤X≤10 | λ=9) = P(X≤10 | λ=9) - P(X≤1 | λ=9)
(c) A1 awrt 0.705
(c) Answer: P(Type II error) = awrt 0.705
(d) B1 power = 1 - 0.705 = awrt 0.295 (ft from part (c))
(d) Answer: power = awrt 0.295
(e) B1 valid comment referencing the low power value, e.g. the power is only awrt 0.295, so the test would fail to detect this particular increase in mean on roughly 70% of occasions; the test is not very sensitive to a change of this size (oe)
(e) Answer: The power is only about 0.295, so the test would fail to detect an increase in mean to 9 on around 70% of occasions; it is not a very sensitive test for a change of this size.

Question 5

(a) M1 attempts P(X≤3) with λ = 10
(a) A1 P(X≤3) = awrt 0.0103
(a) M1 attempts P(X≥18) = 1 - P(X≤17) with λ = 10
(a) A1 P(X≥18) = awrt 0.0143
(a) A1 P(Type I error) = 0.0103 + 0.0143 = awrt 0.0246
(a) Answer: P(Type I error) = awrt 0.0246

Question 6

(a) M1 states H0: p = 0.2, H1: p > 0.2, with X ~ B(25, 0.2) under H0, and attempts an upper-tail search comparing P(X≥8) and P(X≥9) with 0.05
(a) A1 critical region is X ≥ 9 (i.e. {9, 10, ..., 25}), cao
(a) A1 actual significance level = awrt 0.0468
(a) B1 correctly notes P(X≥8) = awrt 0.109 exceeds 0.05, confirming X=8 cannot be included
(a) Answer: Critical region: X ≥ 9 (i.e. {9, 10, ..., 25}). Actual significance level = awrt 0.0468.
(b) B1 10 lies in the critical region {9, 10, ..., 25} found in part (a) (ft)
(b) B1 reject H0; there is evidence, at the 5% level, that the proportion of premium-grade components has increased (ft, in context)
(b) Answer: 10 is in the critical region {9, ..., 25}, so H0 is rejected; there is evidence at the 5% level that the proportion of premium-grade components has increased.
(c) M1 identifies the Type II error region as the acceptance region X ≤ 8, using the true distribution B(25, 0.35)
(c) M1 attempts to sum P(X=0) to P(X=8) with p = 0.35
(c) A1 awrt 0.467
(c) B1 correctly identifies this cumulative value itself as P(Type II error), distinguishing it from the power
(c) Answer: P(Type II error) = awrt 0.467
(d) B1 power = 1 - 0.467 = awrt 0.533 (ft from part (c))
(d) Answer: power = awrt 0.533
(e) B1 the power of the test would increase (oe)
(e) B1 a larger sample size reduces the variability (standard error) of the sample proportion, so the sampling distributions under H0 (p=0.2) and under the true value (p=0.35) overlap less; a greater proportion of the alternative distribution then falls in the (relatively better-targeted) critical region, so P(Type II error) falls and power rises (oe)
(e) Answer: The power would increase, because with a larger sample the distributions of X under H0 and under the true p = 0.35 overlap less, so the critical region captures a larger proportion of the true distribution, reducing P(Type II error).

Question 7

(a) M1 states H0: p = 0.4, H1: p ≠ 0.4, with X ~ B(15, 0.4) under H0, and searches the lower tail comparing P(X≤2) and P(X≤3) with 0.05
(a) A1 lower critical region X ≤ 2 (i.e. {0, 1, 2}), actual lower-tail level = awrt 0.0271
(a) M1 searches the upper tail comparing P(X≥9) and P(X≥10) with 0.05
(a) A1 upper critical region X ≥ 10 (i.e. {10, 11, ..., 15}), actual upper-tail level = awrt 0.0338
(a) A1 combined critical region stated as {0, 1, 2} union {10, 11, ..., 15}
(a) A1 total actual significance level = 0.0271 + 0.0338 = awrt 0.0609
(a) Answer: Critical region: {0, 1, 2} union {10, 11, ..., 15}. Total actual significance level = awrt 0.0609.
(b) B1 3 does not lie in the critical region {0, 1, 2} union {10, ..., 15} found in part (a) (ft)
(b) B1 do not reject H0; there is insufficient evidence, at the 10% level, that the true win probability differs from 0.4 (ft, in context)
(b) Answer: 3 is not in the critical region, so H0 is not rejected; there is insufficient evidence at the 10% level that the win probability differs from 0.4.
(c) B1 any one correctly matched advantage and disadvantage, e.g. advantage: a larger critical region gives higher power, so a genuine change in win probability is more likely to be detected; disadvantage: this also gives a higher probability of a Type I error, wrongly concluding the win probability has changed when it has not (oe)
(c) Answer: Advantage: a 10% test has a larger critical region and higher power, so it is more likely to detect a genuine change in win probability. Disadvantage: it also has a higher probability of a Type I error (wrongly rejecting a true H0) than a 5% test would.

Question 8

(a) M1 compares P(X≤0), P(X≤1) and P(X≤2) with 0.01
(a) A1 critical region is X ≤ 1 (i.e. {0, 1}), cao
(a) A1 actual significance level = awrt 0.00764
(a) Answer: Critical region: X ≤ 1 (i.e. {0, 1}). Actual significance level = awrt 0.00764.
(b) B1 the critical region {0, 1} used here is smaller than the critical region {0, 1, 2} used in Question 3(a), so the acceptance region is correspondingly larger (oe)
(b) B1 for the same true alternative value of p, a larger proportion of the alternative distribution now falls in the acceptance region, so P(Type II error) is larger (power is lower) than in Question 3 (oe)
(b) Answer: The critical region {0, 1} is smaller than {0, 1, 2}, so more of the alternative (p = 0.15) distribution falls in the acceptance region here than in Question 3; P(Type II error) is therefore larger, and the power lower, than for the 5% test.

Question 9

Level 1 (1-2): Makes a simple, largely one-sided statement, e.g. states that a smaller significance level reduces the probability of a Type I error (or that a larger significance level reduces the probability of a Type II error), with little development or link to the context given.
Level 2 (3-4): Gives some balanced reasoning referring to both Type I and Type II errors and links this to at least one consequence described in the context (health risk or cost of shutdown), but the discussion is not fully developed or does not reach a clearly justified recommendation.
Level 3 (5-6): Gives a fully balanced, well-reasoned discussion that correctly explains the Type I/Type II error trade-off as the significance level changes, links this clearly to both consequences described in the context, and reaches a justified recommendation supported by that reasoning.
Indicative content:
- Reducing the significance level (e.g. from 10% to 1%) makes the critical region smaller, so P(Type I error) decreases.
- A smaller critical region means a smaller proportion of the alternative (raised-mean) distribution falls inside it, so P(Type II error) increases (power decreases) for a fixed true mean above the threshold.
- In this context, a Type II error means failing to detect a genuine rise in bacterial colonies, risking patient and staff health.
- A Type I error means wrongly concluding contamination has risen when it has not, triggering an unnecessary and costly ward shutdown.
- Since the consequence of a Type II error (health risk) is generally judged more serious than a Type I error (cost and disruption), the larger significance level (Test B, 10%), which gives higher power and a lower P(Type II error), may be preferred despite its higher false-alarm rate.
- A well-justified alternative conclusion favouring Test A could still be credited, provided it correctly links significance level to the Type I/Type II error probabilities and gives sound context-based reasoning (e.g. if shutdowns are judged to carry very severe cost or disruption).

Question 10

(a) M1 states H0: λ = 8, H1: λ < 8, and attempts to compare P(X≤3) and P(X≤4) with 0.05
(a) M1 correctly evaluates both cumulative probabilities
(a) A1 critical region is X ≤ 3 (i.e. {0, 1, 2, 3}), cao
(a) A1 actual significance level = awrt 0.0424
(a) Answer: Critical region: X ≤ 3 (i.e. {0, 1, 2, 3}). Actual significance level = awrt 0.0424.
(b) M1 identifies the acceptance region as X ≥ 4, using the true distribution Po(5)
(b) M1 uses P(X≥4 | λ=5) = 1 - P(X≤3 | λ=5)
(b) A1 awrt 0.735
(b) Answer: P(Type II error) = awrt 0.735
(c) B1 power = 1 - 0.735 = awrt 0.265 (ft from part (b))
(c) Answer: power = awrt 0.265
(d) B1 valid suggestion, e.g. increase the significance level (widen the critical region, for example to a 10% test), so that more of the Po(5) distribution falls inside the (larger) critical region
(d) B1 valid corresponding disadvantage, e.g. this increases the probability of a Type I error (wrongly concluding the mean has decreased when it has not) (oe)
(d) Answer: Suggestion: increase the significance level (e.g. to 10%), widening the critical region so more of the true Po(5) distribution falls inside it, raising the power above 0.5. Disadvantage: this also increases the probability of a Type I error.

Question 11

(a) M1 attempts P(X≥9) using X ~ B(12, 0.5)
(a) A1 P(X≥9) = 299/4096 = awrt 0.0730
(a) M1 doubles this probability for a two-tailed test
(a) A1 p-value = awrt 0.146
(a) Answer: p-value = awrt 0.146
(b) B1 0.146 > 0.05, so do not reject H0; there is insufficient evidence, at the 5% level, that the spinner is biased towards even numbers (ft, in context)
(b) Answer: Since 0.146 > 0.05, H0 is not rejected; there is insufficient evidence that the spinner is biased towards even numbers.

Question 12

(a) M1 recognises that, since the critical region is X ≤ 3, the power here equals P(X≤3 | λ=3), and the Type II error probability is its complement
(a) M1 attempts to sum P(X=0) to P(X=3) with λ = 3
(a) A1 power = awrt 0.647
(a) A1 P(Type II error) = 1 - 0.647 = awrt 0.353
(a) B1 clearly labels which value is the power and which is P(Type II error), avoiding confusing the two
(a) Answer: power = awrt 0.647; P(Type II error) = awrt 0.353
(b) B1 correctly states the power is higher when the true mean is 3 (awrt 0.647) than when the true mean is 5 (awrt 0.265, from Question 10)
(b) B1 gives a valid reason referencing distance from H0 / overlap of distributions, e.g. a true mean of 3 is further from the hypothesised mean of 8 than a true mean of 5 is, so the null (λ=8) and alternative distributions overlap less, and a greater proportion of the alternative distribution falls inside the critical region X≤3 (oe)
(b) B1 states the general principle: for a fixed significance level and sample size, the power of a test increases as the true parameter value moves further from the value stated in H0 (oe)
(b) Answer: The power is higher when the true mean is 3 (0.647) than when it is 5 (0.265, from Question 10), because a true mean of 3 is further from the hypothesised value of 8, so the null and alternative distributions overlap less and more of the alternative distribution falls in the critical region X≤3; in general, power increases as the true parameter moves further from the value stated in H0.

Further Statistics: Hypothesis Testing, Type I and Type II Errors - Worksheets, Questions and Revision

FP.FS4 Further Statistics: Hypothesis Testing, Type I and Type II Errors

Hypothesis Testing for a Poisson or Binomial Parameter

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12