Statistics: Hypothesis Testing - Worksheets, Questions and Revision

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A-Level · Statistics

S5 Statistics: Hypothesis Testing

EDEXCEL 9MA0 · Calculator allowed · about 120 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A manufacturer states that, historically, the probability that a component produced by an automated lathe is defective is 0.1. After the lathe undergoes routine maintenance, the plant supervisor, Aisha, suspects that the probability of a component being defective has increased. She tests a random sample of 20 components and finds that 5 are defective. Let X be the number of defective components, out of 20, produced by the lathe after maintenance.

(a)State suitable null and alternative hypotheses for Aisha's test.(2)

(b)Assuming X ~ B(20, 0.1), find P(X ≥ 5).(3)

(c)Using a 5% significance level, test whether there is evidence that the probability a component is defective has increased. State your conclusion in context.(2)

(Total for Question 1 is 7 marks)

A fairground stallholder claims that a coin-toss game pays out a prize with probability 0.3 on each play. A regular customer, Ben, suspects that the true probability of winning differs from 0.3. He watches a random sample of 20 plays of the game and observes that a prize is won on only 2 occasions. Let X be the number of plays, out of 20, that result in a prize.

(a)State suitable hypotheses for a two-tailed test of Ben's suspicion.(2)

(b)Assuming X ~ B(20, 0.3), find P(X ≤ 2).(3)

(c)Using a 10% significance level, test whether there is evidence that the probability of winning a prize differs from 0.3. State your conclusion in context.(3)

(Total for Question 2 is 8 marks)

In a hypothesis test, a researcher uses a 5% significance level to test H0: p = 0.45 against H1: p ≠ 0.45, where p is a population proportion.

(a)Explain what is meant by the significance level of a hypothesis test.(1)

(b)State what is meant by a Type I error in the context of this test.(1)

(c)The researcher instead decides to carry out the test using a 1% significance level. State, with a reason, the effect this would have on the width of the critical region.(2)

(d)A second researcher carries out the same test on the same sample data and calculates a p-value of 0.023. Explain why this p-value would lead to a different conclusion from a test at the 1% significance level, but the same conclusion as a test at the 5% significance level.(2)

(Total for Question 3 is 6 marks)

Historically, 20% of seeds of a certain plant fail to germinate. A gardener, Rhys, treats a random sample of 25 seeds with a new fertiliser, hoping it decreases the failure rate. Let X be the number of seeds, out of 25, that fail to germinate.

(a)State suitable hypotheses for Rhys's test.(2)

(b)Assuming X ~ B(25, 0.2), find the critical region for the test, using a 5% significance level. You should show the probabilities used to justify the critical region.(4)

(c)State the actual (achieved) significance level of the test.(1)

(d)In the sample, 1 of the 25 seeds failed to germinate. State, with a reason, the conclusion of the test in context.(2)

(Total for Question 4 is 9 marks)

The number of hours of rainfall (x) and the crop yield (y, in tonnes per hectare) are recorded for 9 randomly selected farms in a region. Priya calculates the product moment correlation coefficient for the data to be r = 0.658.

(a)State suitable hypotheses to test, at the 5% significance level, whether there is positive correlation between rainfall and crop yield.(2)

(b)Given that the critical value for a one-tail test at the 5% significance level, with n = 9, is 0.5822, determine whether there is significant evidence of positive correlation between rainfall and crop yield. Justify your answer.(2)

(c)State one assumption that is necessary for this hypothesis test to be valid.(1)

(d)Priya's colleague, Marcus, claims that this result proves that increased rainfall causes an increase in crop yield. Comment on whether Marcus's claim is justified.(2)

(Total for Question 5 is 7 marks)

A telephone sales company finds that, historically, 8% of calls made result in a completed sale. Following a change to the sales script, the sales manager, Kwame, wants to test whether the probability of a call resulting in a sale has changed (either increased or decreased). A random sample of 40 calls is monitored. Let X be the number of calls, out of 40, that result in a sale.

(a)State suitable hypotheses for a two-tailed test of Kwame's question.(2)

(b)Assuming X ~ B(40, 0.08), find the critical region for a two-tailed test at the 10% significance level. You must show sufficient working to justify both parts of the critical region.(5)

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

(d)In the sample, 8 of the 40 calls resulted in a completed sale. State, with a reason, the conclusion of the test.(2)

(Total for Question 6 is 10 marks)

The mass, in grams, of flour in bags produced by a filling machine is modelled by a normal distribution with mean 1000 and standard deviation 8. Following a service of the machine, a random sample of 25 bags is taken and found to have a sample mean mass of 1004.2 g. The standard deviation is assumed to be unchanged. Sanjay wants to test, at the 5% significance level, whether the mean mass of flour in a bag has increased.

(a)State suitable hypotheses for Sanjay's test.(2)

(b)Show that the value of the test statistic is z = 2.625.(3)

(c)Using a critical value of 1.6449, test at the 5% significance level whether there is evidence that the mean mass has increased. State your conclusion in context.(3)

(Total for Question 7 is 8 marks)

A machine fills bottles with liquid such that the volume, in ml, is normally distributed with standard deviation 5. The machine is set so that the mean volume is 500 ml. During a routine quality check, a random sample of 36 bottles is taken and found to have a sample mean volume of 497.9 ml. Test, at the 1% significance level, whether there is evidence that the mean volume differs from 500 ml.

(a)State suitable hypotheses for a two-tailed test.(2)

(b)Calculate the value of the test statistic, giving your answer to 3 significant figures.(3)

(c)Given that the critical value is ±2.5758, determine whether there is significant evidence, at the 1% level, that the mean volume differs from 500 ml. State your conclusion in context.(3)

(Total for Question 8 is 8 marks)

A large company's technical support team finds that, historically, 25% of support tickets are resolved on first contact. After introducing a new troubleshooting script, the team leader, Fatima, wants to test whether the proportion of tickets resolved on first contact has increased. She takes a random sample of 30 tickets handled using the new script. Let X be the number of tickets, out of 30, resolved on first contact.

(a)State suitable hypotheses for Fatima's test.(2)

(b)Assuming X ~ B(30, 0.25), find the critical region for the test in the form X ≥ k, using a 5% significance level. You must show sufficient working to justify your answer.(4)

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

(d)Given that 13 of the 30 tickets sampled were resolved on first contact, state the conclusion of the test, giving a reason.(2)

(e)State one condition needed for X to be well modelled by a binomial distribution in this context.(1)

(Total for Question 9 is 10 marks)

A sports scientist records the number of hours of sleep, x, and the reaction time, y (in milliseconds), for 10 randomly selected athletes. The summary statistics are Sxx = 45.6, Syy = 812.4, Sxy = -165.3.

(a)Calculate the value of the product moment correlation coefficient, r, giving your answer to 3 significant figures.(3)

(b)State suitable hypotheses to test, at the 1% significance level, whether there is evidence of negative correlation between hours of sleep and reaction time.(2)

(c)Given that the critical value for a one-tail test at the 1% significance level, with n = 10, is -0.7155, test whether there is evidence of negative correlation. State your conclusion in context.(3)

(Total for Question 10 is 8 marks)

It is known that, under a standard treatment, the probability that a patient recovers within a week is 0.4. A new drug is trialled on a random sample of 20 patients, and X, the number who recover within a week, is recorded. A test of H0: p = 0.4 against H1: p > 0.4 is to be carried out.

(a)Given that X ~ B(20, 0.4), show that P(X ≥ 13) = 0.0210, correct to 3 significant figures.(3)

(b)In the trial, 13 patients recovered within a week. State, giving a reason, the conclusion of the test at (i) the 5% significance level, (ii) the 1% significance level.(3)

(c)Find the set of values of the significance level, α%, for which the observation X = 13 would lead to the conclusion that the new drug increases the probability of recovery within a week.(2)

(Total for Question 11 is 8 marks)

The time, in minutes, taken by staff to complete a task is modelled by a normal distribution with standard deviation 4.5. It is believed that the mean completion time is 30 minutes. Following a new training programme, a random sample of n staff members' completion times is recorded and the sample mean, xbar, is calculated. A test of H0: μ = 30 against H1: μ < 30 is to be carried out at the 1% significance level, assuming the standard deviation remains 4.5.

(a)Find, in terms of n, the critical region for the sample mean, xbar, giving your answer in the form xbar < c.(3)

(b)Given that a random sample of n = 50 staff members has a sample mean completion time of 28.7 minutes, test at the 1% significance level whether there is evidence that the training has decreased the mean completion time.(4)

(c)Find the minimum sample size, n, for which a sample mean of 28.7 minutes would indicate a significant decrease in the mean completion time at the 1% significance level.(4)

(Total for Question 12 is 11 marks)

Mark scheme · S5 Statistics: Hypothesis Testing

Question 1

(a) B1 H0: p = 0.1, where p is the probability a component is defective
(a) B1 H1: p > 0.1 (oe, one-tail test)
(a) Answer: H0: p = 0.1, H1: p > 0.1
(b) M1 using X ~ B(20,0.1) and P(X≥5) = 1 - P(X≤4)
(b) A1 P(X≤4) = 0.9568 (awrt)
(b) A1 P(X≥5) = 0.0432 (awrt cao)
(b) Answer: P(X ≥ 5) = 0.0432 (awrt 3sf)
(c) M1 compare 0.0432 with 0.05 and reject H0 since 0.0432 < 0.05
(c) A1 conclusion in context (ft): sufficient evidence, at the 5% level, that the probability a component is defective has increased
(c) Answer: 0.0432 < 0.05, so reject H0; there is significant evidence that the proportion of defective components has increased.

Question 2

(a) B1 H0: p = 0.3, where p is the probability of winning a prize
(a) B1 H1: p ≠ 0.3 (oe, two-tail test)
(a) Answer: H0: p = 0.3, H1: p ≠ 0.3
(b) M1 using X ~ B(20,0.3) and summing P(X=0)+P(X=1)+P(X=2)
(b) A1 two of P(X=0)=0.000798, P(X=1)=0.00684, P(X=2)=0.02785 correct (awrt)
(b) A1 P(X≤2) = 0.0355 (awrt cao)
(b) Answer: P(X ≤ 2) = 0.0355 (awrt 3sf)
(c) B1 compares 0.0355 with 0.05 (half of the 10% two-tail level)
(c) M1 0.0355 < 0.05 so reject H0
(c) A1 conclusion in context (ft): significant evidence, at the 10% level, that the probability of winning a prize differs from 0.3 (appears lower than claimed)
(c) Answer: 0.0355 < 0.05, so reject H0; there is significant evidence that the true probability of winning differs from 0.3.

Question 3

(a) B1 the probability of incorrectly rejecting H0 when H0 is actually true (oe)
(a) Answer: The probability of rejecting H0 when H0 is true.
(b) B1 rejecting H0 (concluding p ≠ 0.45) when in fact p = 0.45 (oe)
(b) Answer: Concluding that p ≠ 0.45 when in fact p = 0.45.
(c) B1 the critical region becomes smaller/narrower
(c) B1 reason: a lower significance level requires stronger evidence against H0 before it is rejected (oe)
(c) Answer: The critical region becomes narrower, because a lower significance level requires more extreme evidence before H0 is rejected.
(d) M1 0.023 < 0.05 so H0 would be rejected at the 5% level
(d) A1 0.023 > 0.01 so H0 would not be rejected at the 1% level, hence the differing conclusions are correctly explained
(d) Answer: 0.023 < 0.05 (reject at 5%) but 0.023 > 0.01 (do not reject at 1%), so the two significance levels give different conclusions.

Question 4

(a) B1 H0: p = 0.2, where p is the probability a seed fails to germinate
(a) B1 H1: p < 0.2 (oe, one-tail test)
(a) Answer: H0: p = 0.2, H1: p < 0.2
(b) M1 attempt to find P(X≤1) or P(X≤2) using B(25,0.2)
(b) A1 P(X≤1) = 0.0274 (awrt)
(b) A1 P(X≤2) = 0.0982 (awrt), showing 0.0274 ≤ 0.05 < 0.0982
(b) A1 correct critical region: X ≤ 1 (cao)
(b) Answer: Critical region: X ≤ 1
(c) B1 actual significance level = 0.0274 (2.74%) (ft)
(c) Answer: 2.74%
(d) M1 1 lies in the critical region {0,1} (ft)
(d) A1 reject H0; evidence, at the 5% level, that the fertiliser has decreased the probability a seed fails to germinate
(d) Answer: Since X = 1 is in the critical region, reject H0; there is evidence the fertiliser has decreased the failure probability.

Question 5

(a) B1 H0: ρ = 0, where ρ is the population product moment correlation coefficient
(a) B1 H1: ρ > 0 (oe, one-tail test)
(a) Answer: H0: ρ = 0, H1: ρ > 0
(b) M1 compare r = 0.658 with the critical value 0.5822
(b) A1 0.658 > 0.5822 so reject H0; significant evidence of positive correlation between rainfall and crop yield
(b) Answer: 0.658 > 0.5822, so reject H0; significant evidence of positive correlation.
(c) B1 the pairs (rainfall, crop yield) come from a population that follows a bivariate normal distribution (oe)
(c) Answer: The population of (rainfall, crop yield) pairs is bivariate normally distributed.
(d) B1 correlation does not imply causation
(d) B1 there may be other factors (e.g. soil quality, temperature, farming practice) that affect crop yield, so causation cannot be concluded from correlation alone (oe)
(d) Answer: No, Marcus's claim is not justified; correlation does not prove causation, as other factors could explain the relationship.

Question 6

(a) B1 H0: p = 0.08, where p is the probability a call results in a sale
(a) B1 H1: p ≠ 0.08 (oe, two-tail test)
(a) Answer: H0: p = 0.08, H1: p ≠ 0.08
(b) M1 attempt lower tail: find P(X=0) and P(X≤1) using B(40,0.08)
(b) A1 P(X=0) = 0.0356 (awrt) and P(X≤1) = 0.159 (awrt), showing the lower critical region is X = 0
(b) M1 attempt upper tail: find P(X≥6) and P(X≥7)
(b) A1 P(X≥7) = 0.0377 (awrt) and P(X≥6) = 0.0967 (awrt), showing the upper critical region is X ≥ 7
(b) A1 correct critical region stated: X = 0 or X ≥ 7 (cao)
(b) Answer: Critical region: X = 0 or X ≥ 7
(c) B1 actual significance level = 0.0356 + 0.0377 = 0.0733 (7.33%) (ft)
(c) Answer: 7.33%
(d) M1 X = 8 lies in the critical region {7,8,...,40} (ft)
(d) A1 reject H0; evidence, at the 10% level, that the probability of a call resulting in a sale has changed
(d) Answer: X = 8 is in the critical region, so reject H0; evidence the probability has changed.

Question 7

(a) B1 H0: μ = 1000, where μ is the population mean mass of flour
(a) B1 H1: μ > 1000 (oe, one-tail test)
(a) Answer: H0: μ = 1000, H1: μ > 1000
(b) M1 use of z = (xbar - μ)/(σ/√n)
(b) A1 correct substitution: z = (1004.2 - 1000)/(8/√25)
(b) A1 z = 2.625 (cso)
(b) Answer: z = 2.625
(c) M1 compare 2.625 with 1.6449
(c) A1 2.625 > 1.6449 so reject H0
(c) A1 conclusion in context: significant evidence, at the 5% level, that the mean mass of flour in a bag has increased
(c) Answer: 2.625 > 1.6449, so reject H0; evidence the mean mass has increased.

Question 8

(a) B1 H0: μ = 500, where μ is the population mean volume
(a) B1 H1: μ ≠ 500 (oe, two-tail test)
(a) Answer: H0: μ = 500, H1: μ ≠ 500
(b) M1 use of z = (xbar - μ)/(σ/√n)
(b) A1 correct substitution: z = (497.9 - 500)/(5/√36)
(b) A1 z = -2.52 (awrt 3sf)
(b) Answer: z = -2.52
(c) M1 compare |z| = 2.52 with 2.5758
(c) A1 2.52 < 2.5758 so do not reject H0
(c) A1 conclusion in context: insufficient evidence, at the 1% level, that the mean volume differs from 500 ml
(c) Answer: 2.52 < 2.5758, so do not reject H0; no significant evidence the mean volume has changed.

Question 9

(a) B1 H0: p = 0.25, where p is the probability a ticket is resolved on first contact
(a) B1 H1: p > 0.25 (oe, one-tail test)
(a) Answer: H0: p = 0.25, H1: p > 0.25
(b) M1 attempt cumulative binomial probability, e.g. P(X≥12) or P(X≤11), using B(30,0.25)
(b) A1 P(X≥12) = 0.0506 (awrt) and P(X≥13) = 0.0215 (awrt) (or equivalent cumulative values)
(b) A1 correct comparison: 0.0506 > 0.05 (fails) but 0.0215 ≤ 0.05 (satisfies)
(b) A1 correct critical region: X ≥ 13 (cao)
(b) Answer: Critical region: X ≥ 13
(c) B1 actual significance level = 0.0215 (2.15%) (ft)
(c) Answer: 2.15%
(d) M1 X = 13 lies in the critical region {13,14,...,30} (ft)
(d) A1 reject H0; evidence, at the 5% level, that the proportion of tickets resolved on first contact has increased
(d) Answer: X = 13 is in the critical region, so reject H0; evidence the proportion resolved on first contact has increased.
(e) B1 each ticket is resolved independently of the others / the probability a ticket is resolved on first contact is constant for every ticket (oe)
(e) Answer: Each ticket is resolved independently, with a constant probability of resolution on first contact.

Question 10

(a) M1 use of r = Sxy/√Sxx*Syy
(a) A1 √45.6*812.4 = √37045.44 = 192 (awrt, unsimplified value acceptable)
(a) A1 r = -0.859 (awrt 3sf)
(a) Answer: r = -0.859
(b) B1 H0: ρ = 0, where ρ is the population product moment correlation coefficient
(b) B1 H1: ρ < 0 (oe, one-tail test)
(b) Answer: H0: ρ = 0, H1: ρ < 0
(c) M1 compare r = -0.859 with the critical value -0.7155
(c) A1 -0.859 < -0.7155 (more extreme) so reject H0
(c) A1 conclusion in context: significant evidence, at the 1% level, of negative correlation between hours of sleep and reaction time (more sleep is associated with faster reaction times)
(c) Answer: -0.859 < -0.7155, so reject H0; significant evidence of negative correlation between sleep and reaction time.

Question 11

(a) M1 attempt P(X≤12) or an equivalent complement approach using B(20,0.4)
(a) A1 P(X≤12) = 0.9790 (awrt)
(a) A1 P(X≥13) = 1 - 0.9790 = 0.0210 (cso)
(a) Answer: P(X ≥ 13) = 0.0210
(b) B1 (i) 0.0210 < 0.05 so reject H0 at the 5% level
(b) B1 (ii) 0.0210 > 0.01 so do not reject H0 at the 1% level
(b) B1 both comparisons stated correctly with correct reasoning
(b) Answer: (i) Reject H0 at 5% (0.0210 < 0.05). (ii) Do not reject H0 at 1% (0.0210 > 0.01).
(c) M1 recognise H0 is rejected whenever α ≥ p-value
(c) A1 α ≥ 2.11% (accept awrt 2.1%)
(c) Answer: α ≥ 2.11% (awrt 2.1%)