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A-Level · Statistics

S9 Statistics: Hypothesis Testing Depth

EDEXCEL 9MA0 · Calculator allowed · about 110 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Hypothesis Testing Toolkit: Formulae and Standard Normal Critical Values

Original material written for Revision Library, styled on Edexcel A-level Mathematics (9MA0) Statistics content.

For a binomial test of H0: p = p0, use X ~ B(n, p0) and compute the relevant cumulative probability (a tail probability, or a p-value) directly, either from cumulative binomial tables or a calculator's binomial cdf function. For a normal test of H0: mu = mu0 with known sigma, standardise using z = (xbar - mu0)/(sigma/sqrt(n)) and compare z with a critical value, or convert to a critical region for xbar directly. For a correlation test of H0: rho = 0, compare the sample product moment correlation coefficient, r, with a critical value from a table of critical values for the sample size n (such a critical value will always be given where needed). Standard normal critical z-values used throughout this pack: one-tailed 10% (or two-tailed 20%) = 1.2816; one-tailed 5% (or two-tailed 10%, i.e. 5% in each tail) = 1.6449; one-tailed 1% (or two-tailed 2%) = 2.3263; two-tailed 5% (2.5% in each tail) = 1.9600; two-tailed 1% (0.5% in each tail) = 2.5758. When finding a binomial critical region, choose the region whose cumulative probability is as close as possible to the stated significance level, unless a question specifically requires the probability to remain below that level; the actual (achieved) significance level is almost never exactly equal to the nominal one, because the binomial distribution is discrete.

A garden centre in Yorkshire claims that 30% of the tulip bulbs it sells produce double blooms. A horticulturist, Priya, is investigating whether the true proportion of double-blooming bulbs is different from 30%. Let X be the number of bulbs producing double blooms in a random sample of n bulbs, and assume X ~ B(n, p) under the null hypothesis.

(a)Write down suitable null and alternative hypotheses for Priya's test.(2)

(b)State, with a reason, whether this is a one-tailed or two-tailed test.(1)

(Total for Question 1 is 3 marks)

A cafe in Leeds states that 20% of its customers choose a vegan pastry. The cafe manager, Kwame, believes this proportion has increased since a new vegan menu was introduced. In a random sample of 15 customers, 6 chose a vegan pastry. Let X be the number of customers, out of 15, choosing a vegan pastry, with X ~ B(15, p) under H0.

(a)State suitable null and alternative hypotheses for Kwame's test.(1)

(b)Test, at the 5% significance level, whether there is evidence that the proportion of customers choosing a vegan pastry has increased. State your conclusion in context.(4)

(Total for Question 2 is 5 marks)

A seed company claims that 65% of its sunflower seeds germinate. Following a wet spring, an allotment holder in Norwich, Siobhan, believes the germination rate has decreased. She plants a random sample of 20 seeds, of which 8 germinate. Let X be the number of seeds, out of 20, that germinate, with X ~ B(20, p) under H0: p = 0.65.

(a)Find the critical region for a test at the 5% significance level for whether the germination rate has decreased. The critical region should have probability as close as possible to 0.05.(4)

(b)State, with a reason, the conclusion of the test given that 8 of the 20 seeds germinated.(1)

(Total for Question 3 is 5 marks)

A machine at a bakery in Bristol is set to produce loaves such that 25% have a well-fired crust, according to specification. After maintenance work, the quality control manager, Tomasz, samples 18 loaves at random and finds that only 1 has a well-fired crust. Let X be the number of loaves, out of 18, with a well-fired crust, with X ~ B(18, p) under H0.

(a)State suitable hypotheses for a two-tailed test of whether the proportion of well-fired loaves has changed.(2)

(b)Assuming X ~ B(18, 0.25), find P(X ≤ 1).(3)

(c)Using a 10% significance level, test whether there is evidence that the proportion of well-fired loaves has changed. State your conclusion in context.(2)

(Total for Question 4 is 7 marks)

A call centre in Cardiff aims for 40% of calls to be resolved on first contact. The team leader, Fatima, wants to test whether this proportion has changed following a new staff training programme. She takes a random sample of 25 calls. Let X be the number of calls, out of 25, resolved on first contact, with X ~ B(25, p) under H0: p = 0.4.

(a)Find the critical region for a two-tailed test at the 10% significance level, such that the probability in each tail is as close as possible to 0.05.(6)

(b)Given that 5 of the 25 sampled calls were resolved on first contact, state the conclusion of the test.(1)

(Total for Question 5 is 7 marks)

The times taken by students at a sixth-form college in Manchester to complete an online numeracy assessment are normally distributed with mean 42 minutes and standard deviation 6 minutes, based on historical records. Following the introduction of a new revision app, the exams officer, Grace, takes a random sample of 40 students and finds a mean completion time of 40.1 minutes. Assume the standard deviation is unchanged.

(a)State suitable hypotheses for Grace's test of whether the mean completion time has decreased.(2)

(b)Show that the test statistic is z = -2.00, correct to 3 significant figures.(2)

(c)Using a critical value of -1.6449 for a one-tailed test at the 5% significance level, test whether there is evidence that the mean completion time has decreased. State your conclusion in context.(3)

(Total for Question 6 is 7 marks)

The mass of bags of flour produced at a factory in Preston is modelled as X ~ N(1000, 64) grams (mean 1000 g, standard deviation 8 g), as stated on the packaging. Following a change of flour supplier, quality inspector Olusegun wants to test whether the mean mass has changed, using a random sample of 25 bags.

(a)State suitable hypotheses, and find, in terms of xbar (the sample mean mass), the critical region for a two-tailed test at the 1% significance level.(5)

(b)Given that the mean mass of the sample of 25 bags is found to be 1003.2 g, comment on whether there is evidence that the mean mass has changed.(2)

(Total for Question 7 is 7 marks)

A pharmaceutical trial investigates a new insomnia treatment. Without treatment, the time taken by adults to fall asleep is normally distributed with mean 34 minutes and standard deviation 9 minutes, based on a large historical dataset. Researcher Amara records a mean time of 30.5 minutes to fall asleep in a random sample of 36 patients using the new treatment.

(a)Using a suitable hypothesis test at the 1% significance level, determine whether there is evidence that the treatment reduces the mean time taken to fall asleep. State your hypotheses clearly.(6)

(b)Interpret, in context, what is meant by the test in part (a) being 'significant at the 1% level'.(2)

(Total for Question 8 is 8 marks)

A researcher, Chen, investigates whether there is a linear relationship between the number of hours of daylight (x) and the number of visitors (y) to a National Trust garden in Kent, using data from a random sample of 12 days spread throughout the year. The product moment correlation coefficient for the sample is calculated as r = 0.62.

(a)State suitable hypotheses to test whether there is positive linear correlation between hours of daylight and number of visitors.(2)

(b)Given that the critical value for a one-tailed test at the 5% significance level, for a sample of size 12, is 0.4973, determine whether there is significant evidence of positive correlation.(3)

(c)Chen argues that the value of r shows that longer daylight hours directly cause more visitors. Comment on this claim.(2)

(Total for Question 9 is 7 marks)

A dental practice in Swansea states that 85% of patients attend their booked check-up appointments. The practice manager, Aisha, runs a reminder-text trial and wants to test, using a random sample of 30 patients, whether the attendance rate has increased. Let X be the number of patients, out of 30, who attend, with X ~ B(30, p) under H0.

(a)Write down suitable null and alternative hypotheses.(2)

(b)Using a 5% significance level, find the critical region for the test, such that the probability of incorrectly rejecting H0 is as close as possible to 5% while remaining less than 5%.(5)

(c)State the actual significance level of the test in part (b).(1)

(d)After the trial, 29 of the 30 sampled patients attended their appointment. State the conclusion of the test.(1)

(Total for Question 10 is 9 marks)

An orchard in Herefordshire has historically produced apples with weights normally distributed with mean 180 g and standard deviation 22 g. After introducing a new irrigation system, the orchard manager, Liam, samples 50 apples at random and records a sample mean weight of 186.4 g. Assume the standard deviation is unchanged.

(a)State suitable hypotheses to test whether the irrigation system has changed the mean weight of the apples.(2)

(b)Show that the test statistic is z = 2.06 (awrt), and hence carry out the test at the 5% significance level, stating your conclusion clearly.(5)

(c)Liam decides to repeat the test using a 1% significance level instead. State, with a reason, whether his conclusion would change.(2)

(d)Explain what is meant by a Type I error in this context, and state the probability of a Type I error for the test in part (c).(2)

(Total for Question 11 is 11 marks)

A students' union at a university in Nottingham believes that 55% of first-year students use the on-campus gym at least once a week. Following a new fitness marketing campaign, the sabbatical officer, Nadia, wants to investigate whether this proportion has changed. She takes a random sample of 22 first-year students. Let X be the number, out of the 22 sampled, who use the gym at least once a week, with X ~ B(22, p) under H0: p = 0.55.

(a)Nadia initially wants to test only whether the proportion has increased. State appropriate hypotheses for this one-tailed test.(2)

(b)Find the critical region for this one-tailed test at the 5% significance level, such that the probability of rejecting H0 is as close as possible to 0.05.(5)

(c)State the actual significance level of the test in part (b).(1)

(d)Nadia then reconsiders, and decides she should instead test whether the proportion has simply changed (in either direction), rather than only increased. Explain how this changes the hypotheses, and find the new critical region for a two-tailed test at the 10% significance level (5% in each tail), such that each tail's probability is as close as possible to 0.05. You are given that, under H0, P(X ≤ 7) = 0.0243 and P(X ≤ 8) = 0.0617.(3)

(e)A survey afterwards finds that 16 of the 22 students sampled use the gym at least once a week. Using your critical region from part (d), state the conclusion of the two-tailed test, and briefly explain, in context, one advantage of using a two-tailed test rather than a one-tailed test in this situation.(2)

(Total for Question 12 is 13 marks)

Mark scheme · S9 Statistics: Hypothesis Testing Depth

Question 1

(a) B1 H0: p = 0.3, where p is the population proportion of bulbs producing double blooms
(a) B1 H1: p ≠ 0.3 (oe)
(a) Answer: H0: p = 0.3, H1: p ≠ 0.3
(b) B1 two-tailed, since H1 tests for a difference in either direction (p ≠ 0.3) rather than a specific increase or decrease
(b) Answer: Two-tailed test.

Question 2

(a) B1 H0: p = 0.2, H1: p > 0.2, both stated correctly
(a) Answer: H0: p = 0.2, H1: p > 0.2
(b) M1 attempt P(X ≥ 6) using X ~ B(15, 0.2), e.g. via 1 - P(X ≤ 5)
(b) A1 P(X ≥ 6) = 0.0610 (awrt 3sf)
(b) M1 compare 0.0610 with 0.05
(b) A1ft 0.0610 > 0.05, so insufficient evidence, at the 5% level, that the proportion of customers choosing a vegan pastry has increased (do not reject H0)
(b) Answer: P(X ≥ 6) = 0.0610 (3sf); 0.0610 > 0.05, so insufficient evidence at the 5% level that the proportion has increased.

Question 3

(a) M1 attempt cumulative probabilities P(X ≤ c) for X ~ B(20, 0.65) for small c
(a) A1 P(X ≤ 8) = 0.0196 (awrt 3sf)
(a) A1 P(X ≤ 9) = 0.0532 (awrt 3sf)
(a) A1 critical region X ≤ 9 (since |0.0532-0.05| = 0.0032 is smaller than |0.0196-0.05| = 0.0304)
(a) Answer: Critical region: X ≤ 9 (i.e. {0,1,...,9}); actual significance level 0.0532.
(b) B1ft 8 lies within the critical region {0,...,9}, so reject H0; evidence, at the 5% level, that the germination rate has decreased
(b) Answer: 8 is in the critical region, so reject H0: evidence the germination rate has decreased.

Question 4

(a) B1 H0: p = 0.25, where p is the population proportion of well-fired loaves
(a) B1 H1: p ≠ 0.25 (oe)
(a) Answer: H0: p = 0.25, H1: p ≠ 0.25
(b) M1 attempt P(X=0) + P(X=1) using X ~ B(18, 0.25)
(b) A1 P(X = 0) = 0.75¹⁸ = 0.00564 (awrt)
(b) A1 P(X ≤ 1) = 0.0395 (awrt 3sf)
(b) Answer: P(X ≤ 1) = 0.0395 (3sf)
(c) M1 compare 0.0395 with 0.05 (half of the 10% two-tailed significance level)
(c) A1ft 0.0395 < 0.05, so reject H0; evidence, at the 10% level, that the proportion of well-fired loaves has changed
(c) Answer: 0.0395 < 0.05, so reject H0: evidence the proportion of well-fired loaves has changed.

Question 5

(a) M1 attempt cumulative probabilities for X ~ B(25, 0.4) in the lower tail
(a) A1 P(X ≤ 5) = 0.0294 and P(X ≤ 6) = 0.0736 (awrt 3sf)
(a) B1 lower critical region X ≤ 5 (0.0294 is closer to 0.05 than 0.0736: |0.05-0.0294| = 0.0206 < |0.0736-0.05| = 0.0236)
(a) M1 attempt cumulative probabilities for X ~ B(25, 0.4) in the upper tail
(a) A1 P(X ≥ 15) = 0.0344 and P(X ≥ 14) = 0.0778 (awrt 3sf)
(a) B1 upper critical region X ≥ 15 (0.0344 is closer to 0.05 than 0.0778: |0.05-0.0344| = 0.0156 < |0.0778-0.05| = 0.0278)
(a) Answer: Critical region: X ≤ 5 or X ≥ 15; combined actual significance level 0.0294 + 0.0344 = 0.0638.
(b) B1ft X = 5 lies in the critical region (X ≤ 5), so reject H0; evidence, at the 10% level, that the proportion of calls resolved on first contact has changed
(b) Answer: X = 5 is in the critical region, so reject H0: evidence the proportion has changed.

Question 6

(a) B1 H0: μ = 42, where μ is the population mean completion time in minutes
(a) B1 H1: μ < 42 (oe)
(a) Answer: H0: μ = 42, H1: μ < 42
(b) M1 use of z = (xbar - μ)/(σ/√n)
(b) A1 z = -2.00 cso
(b) Answer: z = -2.00 (-2.003 to 3dp)
(c) M1 compare -2.00 with -1.6449
(c) A1 -2.00 < -1.6449, so reject H0
(c) A1ft conclusion in context: evidence, at the 5% level, that the mean completion time has decreased
(c) Answer: -2.00 < -1.6449, so reject H0: evidence the mean completion time has decreased.

Question 7

(a) B1 H0: μ = 1000, H1: μ ≠ 1000
(a) M1 identify the two-tailed 1% critical value z = 2.5758
(a) M1 use xbar = μ ± z*σ/√n
(a) A1 lower boundary = 995.9 (awrt)
(a) A1 upper boundary = 1004.1 (awrt)
(a) Answer: Critical region: xbar < 995.9 or xbar > 1004.1 (g).
(b) B1 1003.2 does not lie in the critical region, since 1003.2 < 1004.1
(b) B1ft insufficient evidence, at the 1% level, that the mean mass of the bags has changed (do not reject H0)
(b) Answer: 1003.2 is not in the critical region, so insufficient evidence the mean mass has changed.

Question 8

(a) B1 H0: μ = 34, H1: μ < 34
(a) M1 standardise, z = (xbar - μ)/(σ/√n)
(a) A1 z = -2.33 (awrt, -2.333 to 3dp)
(a) B1 critical value -2.3263 (one-tailed, 1%) correctly stated for comparison
(a) M1 valid comparison (z < -2.3263, or equivalently a p-value less than 0.01)
(a) A1ft reject H0; evidence, at the 1% level, that the treatment reduces the mean time taken to fall asleep
(a) Answer: z = -2.33 (3sf); -2.33 < -2.3263, so reject H0: evidence the treatment reduces the mean time to fall asleep.
(b) B1 refers to the probability of obtaining a sample mean as low as (or lower than) 30.5 minutes, if the treatment genuinely had no effect (i.e. under H0, μ = 34)
(b) B1 this probability is less than 0.01 (1%), so a result this extreme is unlikely to have occurred by chance alone, giving evidence against H0
(b) Answer: It means that, if the treatment had no effect, a sample mean as low as 30.5 minutes (or lower) would occur with probability less than 0.01, so the result is unlikely to be due to chance.

Question 9

(a) B1 H0: ρ = 0, where ρ is the population product moment correlation coefficient
(a) B1 H1: ρ > 0 (oe)
(a) Answer: H0: ρ = 0, H1: ρ > 0
(b) B1 compares r = 0.62 with the critical value 0.4973
(b) B1 0.62 > 0.4973, so reject H0
(b) B1 conclusion in context: significant evidence, at the 5% level, of positive correlation between hours of daylight and the number of visitors
(b) Answer: 0.62 > 0.4973, so reject H0: significant evidence of positive correlation.
(c) B1 correlation does not imply causation
(c) B1 valid alternative explanation, e.g. a confounding/lurking variable such as warmer weather or school holidays could independently affect both daylight hours (via season) and visitor numbers, so a direct causal link cannot be concluded from correlation alone
(c) Answer: Correlation does not imply causation; a confounding variable (e.g. season or weather) may explain the association without daylight hours directly causing higher visitor numbers.

Question 10

(a) B1 H0: p = 0.85, where p is the population proportion of patients who attend
(a) B1 H1: p > 0.85 (oe)
(a) Answer: H0: p = 0.85, H1: p > 0.85
(b) M1 attempt to find upper-tail cumulative probabilities for X ~ B(30, 0.85), e.g. via Y = 30 - X ~ B(30, 0.15)
(b) A1 P(X ≥ 28) = 0.151 (awrt 3sf) - too large, so rejected
(b) A1 P(X ≥ 29) = 0.0480 (awrt 3sf) - valid, since less than 5%
(b) A1 correctly identifies P(X ≥ 29) = 0.0480 as the required boundary, since P(X ≥ 28) exceeds 5%
(b) A1 critical region X ≥ 29 (accept {29, 30})
(b) Answer: Critical region: X ≥ 29 (i.e. {29, 30}).
(c) B1 0.0480 (4.80%, awrt) (ft)
(c) Answer: 0.0480 (4.80%)
(d) B1ft 29 lies in the critical region (X ≥ 29), so reject H0; evidence, at the 5% level, that the attendance rate has increased
(d) Answer: 29 is in the critical region, so reject H0: evidence the attendance rate has increased.

Question 11

(a) B1 H0: μ = 180, where μ is the population mean weight in grams
(a) B1 H1: μ ≠ 180 (oe)
(a) Answer: H0: μ = 180, H1: μ ≠ 180
(b) M1 standardise, z = (xbar - μ)/(σ/√n)
(b) A1 z = 2.06 (2.057 to 3dp) cso
(b) B1 critical value ±1.96 (two-tailed, 5%) stated for comparison
(b) M1 valid comparison, |z| > 1.96 (2.057 > 1.96)
(b) A1ft reject H0; evidence, at the 5% level, that the mean weight of the apples has changed
(b) Answer: z = 2.06 (2.057, 3dp); 2.057 > 1.96, so reject H0: evidence the mean weight has changed.
(c) B1 new critical value ±2.5758 (two-tailed, 1%) identified, and 2.057 < 2.5758
(c) B1 conclusion changes: insufficient evidence, at the 1% level, that the mean weight of the apples has changed (do not reject H0)
(c) Answer: 2.057 < 2.5758, so at the 1% level there is insufficient evidence of a change (the conclusion changes).
(d) B1 a Type I error is rejecting H0 (concluding the mean weight has changed) when H0 is actually true (the mean weight of the apples is still 180 g)
(d) B1 P(Type I error) = 0.01
(d) Answer: A Type I error is rejecting H0 when it is true; P(Type I error) = 0.01 for the 1% level test.

Question 12

(a) B1 H0: p = 0.55
(a) B1 H1: p > 0.55 (oe)
(a) Answer: H0: p = 0.55, H1: p > 0.55
(b) M1 attempt upper-tail cumulative probabilities for X ~ B(22, 0.55)
(b) A1 P(X ≥ 17) = 0.0271 (awrt 3sf)
(b) A1 P(X ≥ 16) = 0.0705 (awrt 3sf)
(b) B1 correct comparison identifying X ≥ 16 as closer to 0.05 (|0.0705-0.05| = 0.0205 < |0.0271-0.05| = 0.0229)
(b) A1 critical region X ≥ 16 (accept {16,17,...,22})
(b) Answer: Critical region: X ≥ 16.
(c) B1 0.0705 (7.05%, awrt) (ft)
(c) Answer: 0.0705 (7.05%)
(d) B1 revised hypotheses H0: p = 0.55, H1: p ≠ 0.55 (two-tailed)
(d) B1 lower critical region X ≤ 8, since P(X ≤ 8) = 0.0617 is closer to 0.05 than P(X ≤ 7) = 0.0243 (|0.0617-0.05| = 0.0117 < |0.0243-0.05| = 0.0257)
(d) B1 overall critical region X ≤ 8 or X ≥ 16 (reusing the upper region found in part (b))
(d) Answer: H1: p ≠ 0.55; critical region: X ≤ 8 or X ≥ 16.
(e) B1ft X = 16 lies in the critical region (X ≥ 16), so reject H0; evidence, at the 10% level, that the proportion of first-year students using the gym has changed
(e) B1 valid advantage of the two-tailed test in context, e.g. it detects a change in either direction without assuming in advance that the campaign could only increase (rather than decrease) gym use, appropriate since Nadia was not certain of the direction of any change before collecting the data
(e) Answer: 16 is in the critical region, so reject H0: evidence the proportion has changed. Advantage: a two-tailed test does not require assuming the direction of change in advance.

Statistics: Hypothesis Testing Depth - Worksheets, Questions and Revision