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Hypothesis Testing

Practice exam-style IB Math AI questions for Hypothesis Testing, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

A researcher investigates whether a new fertilizer increases the mean height of a species of plant. Under the null hypothesis, 10 00010\,000 simulated random samples are generated. Of these, 327327 produce an increase at least as large as the increase observed in the researcher's sample.

A

the pp-value from the simulation.

[1]
B

the decision at the 5%5\% significance level and it in context.

[2]
C

what the simulations represent.

[1]
Question 2
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

A company compares the population mean battery lives of two brands. A two-tailed test gives a pp-value of 0.0630.063.

A

the hypotheses, using μA\mu_A and μB\mu_B for the two population means.

[1]
B

the decision at the 5%5\% significance level.

[1]
C

the decision at the 10%10\% significance level.

[1]
D

why the pp-value is not the probability that the null hypothesis is true.

[1]
Question 3
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A random sample of commuters is classified by their usual method of travel and whether they live in an urban or rural area. The results are shown in the table. A χ2\chi^2 test is used to whether usual method of travel is independent of area of residence.

Area of residence

Car

Public transport

Bicycle

Total

Urban

20

25

25

70

Rural

30

15

15

60

Total

50

40

40

130

A

the null and alternative hypotheses.

[2]
B

the expected frequency of urban commuters who travel by car.

[2]
C

The test gives χ2=6.27\chi^2=6.27 and p=0.0436p=0.0436. whether usual method of travel is independent of area of residence at the 5%5\% significance level. your answer.

[2]
Question 4
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A manufacturer claims that packets of fruit sweets contain red, orange, yellow and green sweets in the proportions 0.300.30, 0.300.30, 0.250.25 and 0.150.15, respectively. A random sample of 120120 sweets is classified by colour. The observed frequencies are shown in the table.

Colour

Observed frequency

Red

30

Orange

32

Yellow

34

Green

24

Total

120

A

the expected frequency of green sweets under the manufacturer's claim.

[1]
B

the value of the χ2\chi^2 test statistic.

[2]
C

the pp-value and whether the data are consistent with the manufacturer's claim at the 5%5\% significance level.

[3]

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Question 5
SL • Paper 1
Medium
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SL • Paper 1
Medium
Calculator Permitted

Visitors to a nature reserve are classified by age group and by the activity they choose. The observed frequencies are shown in a 3×33\times3 contingency table. A test for independence gives χ2=10.2\chi^2=10.2. At the 5%5\% significance level, the relevant critical value is 9.4889.488.

Age group

Hiking

Cycling

Wildlife watching

Total

18-30

32

20

20

72

31-50

17

15

16

48

51+

11

25

24

60

Total

60

60

60

180

A

the alternative hypothesis.

[1]
B

the expected frequency for the specified cell with row total 7272, column total 6060 and grand total 180180.

[2]
C

the number of degrees of freedom and the conclusion of the test.

[2]
Question 6
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A coach compares the recovery times of athletes using treatment A with those using treatment B. The coach claims that the population mean recovery time for treatment A is greater than that for treatment B. A pooled two-sample tt-test, calculated as t=xˉA−xˉBSEt=\dfrac{\bar{x}_A-\bar{x}_B}{SE}, gives t=1.83t=1.83 with 2828 degrees of freedom. The calculator displays a two-tailed pp-value of 0.07790.0779.

A

the null and alternative hypotheses.

[2]
B

the pp-value appropriate for the coach's claim.

[1]
C

the conclusion at the 5%5\% significance level.

[2]
Question 7
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A streaming service predicts that viewers choose four film genres in the proportions 0.400.40, 0.300.30, 0.200.20 and 0.100.10. In a random sample of 200200 viewers, the observed frequencies are 6060, 7070, 4747 and 2323, respectively. A χ2\chi^2 goodness-of-fit test is performed at the 10%10\% significance level. The critical value is 6.2516.251.

A

the expected frequency for the first genre.

[1]
B

the contribution of the first genre to the χ2\chi^2 statistic.

[2]
C

The complete test statistic is χ2=8.34\chi^2=8.34. whether the predicted genre distribution is a satisfactory fit.

[2]
Question 8
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The mass of cereal in a box is normally distributed with known population standard deviation 6 g6\ \text{g}. The labelled population mean is 50 g50\ \text{g}. A random sample of 3636 boxes has mean mass 52.1 g52.1\ \text{g}. A consumer organization tests whether the population mean mass is greater than the labelled value.

A

the hypotheses.

[1]
B

the value of the test statistic.

[2]
C

the pp-value and the conclusion at the 5%5\% significance level.

[3]

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Question 9
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A manufacturer claims that the population mean operating time of a sensor is 7272 hours. The population standard deviation is unknown. A random sample of 1212 sensors has mean operating time 74.574.5 hours and sample standard deviation 4.204.20 hours. Operating times are assumed to be normally distributed. An engineer tests whether the population mean exceeds 7272 hours.

A

why a tt-test, rather than a zz-test, is appropriate.

[1]
B

the test statistic.

[2]
C

the pp-value and the conclusion at the 5%5\% significance level.

[3]
Question 10
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A basketball player claims that the probability of scoring a free throw is greater than 0.400.40. During 2020 independent free throws, the player scores 1313 times. The probability of scoring is assumed to remain constant.

A

the hypotheses and the distribution used under the null hypothesis.

[2]
B

the pp-value.

[2]
C

the conclusion at the 5%5\% significance level.

[1]
Question 11
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

Two independent groups of students use different revision methods. Their scores on the same test are summarized below. Scores in both populations are assumed to be normally distributed with equal unknown variances.

Method A: nA=14n_A=14, xˉA=18.2\bar{x}_A=18.2, sA=2.40s_A=2.40

Method B: nB=12n_B=12, xˉB=20.1\bar{x}_B=20.1, sB=2.10s_B=2.10

A teacher wishes to test whether the population mean scores for the two methods are different.

A

the hypotheses for the test.

[1]
B

Carry out an appropriate test and the pp-value.

[3]
C

the conclusion of the test at the 5%5\% significance level.

[2]
Question 12
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Eight participants complete a task before and after a training session. For each participant, the difference is defined as

d=score after training−score before trainingd=\text{score after training}-\text{score before training}

The observed differences are 2,1,3,−1,4,2,0,32,1,3,-1,4,2,0,3. The differences are assumed to come from a normal population.

A

suitable hypotheses to test whether the training increases the population mean score.

[1]
B

why the data should be analysed using a one-sample test rather than an independent two-sample test.

[1]
C

Carry out an appropriate test and the conclusion at the 5%5\% significance level.

[4]

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Question 13
HL • Paper 1
Medium
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HL • Paper 1
Medium
Calculator Permitted

The number of faults recorded by a monitoring system in one hour is modelled by a Poisson distribution. The usual population mean is 2.52.5 faults per hour. During eight independent hours after a software update, a total of 2929 faults is recorded. An engineer tests whether the update has increased the mean fault rate.

A

the hypotheses and the distribution of the total number of faults under the null hypothesis.

[2]
B

the pp-value.

[2]
C

the conclusion at the 5%5\% significance level.

[1]
D

a Type I error in this context.

[1]
Question 14
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

For a random sample of 1010 cities, a researcher records annual public transport expenditure per resident and the percentage of journeys made by public transport. The sample product moment correlation coefficient is r=0.762r=0.762. The paired variables are assumed to follow a bivariate normal distribution.

A

hypotheses to test whether there is a non-zero linear correlation in the population.

[2]
B

Use an appropriate test to the pp-value.

[2]
C

the conclusion at the 5%5\% significance level.

[1]
D

why this conclusion does not establish that greater expenditure causes a higher percentage of public transport journeys.

[1]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A company tests

H0:π=0.20againstH1:π>0.20H_0:\pi=0.20\qquad\text{against}\qquad H_1:\pi>0.20

using a random sample of 2525 independent items. Let XX be the number of defective items. The significance level is 5%5\%.

A

the critical region, choosing the region that maximizes the probability of a Type I error while keeping it below 0.050.05.

[2]
B

the actual probability of a Type I error.

[1]
C

Given that the true proportion of defective items is 0.350.35, the probability of a Type II error.

[2]
D

the power of the test when the true proportion is 0.350.35.

[1]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The lifetime, in hours, of a component is normally distributed with known population standard deviation 1212 hours. A test of

H0:μ=100againstH1:μ>100H_0:\mu=100\qquad\text{against}\qquad H_1:\mu>100

uses the mean lifetime Xˉ\bar X of a random sample of 3636 components. The significance level is 5%5\%.

A

the critical value of Xˉ\bar X and the critical region.

[3]
B

When the true population mean is 105105 hours, the probability of a Type II error.

[2]
C

the power of the test when the true population mean is 105105 hours.

[1]
D

the effect on the probability of a Type II error if the significance level is reduced while the sample size remains unchanged.

[1]

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Question 17
SL • Paper 2
Medium
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SL • Paper 2
Medium
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A museum records the payment method and age group of a random sample of 200 visitors. The observed frequencies are shown in the table. A χ2\chi^2 test for independence is performed at the 5%5\% significance level.

Payment method

Under 25

25–49

50 or over

Row total

Cash

18

24

30

72

Card

32

28

20

80

App

25

15

8

48

Column total

75

67

58

200

A
I.

the null and alternative hypotheses.

[2]
II.

the expected frequency of visitors aged under 25 who paid with cash.

[2]
B

The observed rows are (18,24,30)(18,24,30), (32,28,20)(32,28,20) and (25,15,8)(25,15,8). Carry out the test and the conclusion in context.

[4]
Question 18
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A seed supplier claims that packets contain seeds from five varieties in the proportions 0.100.10, 0.200.20, 0.400.40, 0.200.20 and 0.100.10. A random sample of 200 seeds gives observed frequencies 1414, 4848, 7272, 4343 and 2323.

Seed variety

Claimed proportion

Observed frequency

Variety 1

0.10

14

Variety 2

0.20

48

Variety 3

0.40

72

Variety 4

0.20

43

Variety 5

0.10

23

A
I.

suitable hypotheses for a goodness-of-fit test.

[2]
II.

the expected frequencies.

[2]
B

Carry out the test at the 5%5\% significance level and the result.

[4]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A delivery company claims that unsuccessful deliveries occur equally often on each of the five working days. During 200 unsuccessful deliveries, the frequencies from Monday to Friday are 3232, 4141, 2929, 5050 and 4848.

Weekday

Unsuccessful deliveries

Monday

32

Tuesday

41

Wednesday

29

Thursday

50

Friday

48

A
I.

the expected frequency for each day under the company's claim.

[2]
II.

the χ2\chi^2 statistic and its degrees of freedom.

[2]
B

the pp-value and why conclusions at the 5%5\% and 10%10\% significance levels are different.

[4]
Question 20
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A conservationist compares the mean time taken to a bird species using two field guides. Independent random samples give the following summaries.

Guide A: nA=10n_A=10, xˉA=42.3\bar{x}_A=42.3 seconds, sA=5.10s_A=5.10 seconds.

Guide B: nB=12n_B=12, xˉB=38.1\bar{x}_B=38.1 seconds, sB=4.60s_B=4.60 seconds.

Times in both populations are assumed normal with equal unknown variances.

A
I.

hypotheses to test whether the population mean time using guide A is greater than that using guide B.

[2]
II.

why a pooled two-sample tt-test is appropriate.

[2]
B

Carry out the test at the 5%5\% significance level and the result.

[4]

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Question 21
SL • Paper 2
Hard
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SL • Paper 2
Hard
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Two independent laboratories measure the concentration of a mineral in samples from two water sources. The measurements are assumed to come from normal populations with equal unknown variances.

Source P: nP=15n_P=15, xˉP=6.42\bar{x}_P=6.42, sP=0.51s_P=0.51.

Source Q: nQ=13n_Q=13, xˉQ=6.05\bar{x}_Q=6.05, sQ=0.46s_Q=0.46.

A
I.

hypotheses to test whether the two population means are different.

[2]
II.

the test statistic and the pp-value.

[2]
B

and the conclusions at the 5%5\% and 10%10\% significance levels.

[4]
Question 22
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

The mass MM, in grams, of a product is claimed to follow N(50,82)N(50,8^2). A sample of 300 products is grouped into the intervals M<42M<42, 42≤M<5042\leq M<50, 50≤M<5850\leq M<58 and M≥58M\geq58. The observed frequencies are 3737, 112112, 111111 and 4040.

Mass interval [g]

Observed frequency

M < 42

37

42 ≤ M < 50

112

50 ≤ M < 58

111

M ≥ 58

40

A
I.

P(M<42)P(M<42) and P(42≤M<50)P(42\leq M<50).

[2]
II.

the four expected frequencies, correct to one decimal place.

[2]
B
I.

hypotheses for a goodness-of-fit test.

[2]
II.

Carry out the test at the 5%5\% significance level and the result. If you did not obtain the expected frequencies in part (a), use 47.6,102.4,102.4,47.647.6,102.4,102.4,47.6.

[4]
Question 23
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A school compares the mean waiting times at two canteens. The observed difference in sample means is 2.82.8 minutes. Under the null hypothesis of equal population means, 5000 random reallocations of the recorded waiting times are performed. In 118 reallocations, the difference is at least 2.82.8 minutes in the specified direction.

Simulation outcome

Count (out of 5000)

Difference < 2.8 min

4882

Difference ≥ 2.8 min

118

Total

5000

A
I.

the one-tailed pp-value.

[2]
II.

the decisions at the 5%5\% and 1%1\% significance levels.

[2]
B

what the simulation represents and why the estimated pp-value is not the probability that the null hypothesis is true.

[4]
Question 24
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

An online course provider records the device used and whether each of 190 randomly selected learners completed a course. The observed frequencies are shown in the table.

Device

Completed

Did not complete

Total

Laptop

50

20

70

Tablet

35

25

60

Phone

25

35

60

Total

110

80

190

A
I.

Given that the laptop row is (50,20)(50,20), the expected frequencies for this row.

[2]
II.

the degrees of freedom and the alternative hypothesis.

[2]
B

The tablet row is (35,25)(35,25) and the phone row is (25,35)(25,35). Carry out a χ2\chi^2 test at the 1%1\% significance level.

[4]

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Question 25
HL • Paper 2
Hard
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HL • Paper 2
Hard
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The breaking force of a type of cable is assumed to be normally distributed. The population standard deviation is unknown. A sample of 16 cables has mean breaking force 102.8 kN102.8\ \text{kN} and sample standard deviation 5.60 kN5.60\ \text{kN}. A manufacturer claims that the population mean is 100 kN100\ \text{kN}.

A
I.

hypotheses to test the manufacturer's claim against a two-sided alternative.

[2]
II.

why a one-sample tt-test is appropriate.

[2]
B

Carry out the test at the 5%5\% significance level and the result.

[4]
Question 26
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For a random sample of 12 farms, a researcher records annual rainfall and crop yield. The sample product moment correlation coefficient is r=0.492r=0.492. The researcher claims that greater rainfall is associated with greater crop yield.

A
I.

hypotheses for an appropriate test of the researcher's claim.

[2]
II.

the assumption about the joint population distribution that is required, and define the parameter being tested.

[2]
B

Carry out the test at the 5%5\% significance level and one limitation of the conclusion.

[4]
Question 27
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A bottling company fills containers with juice. Fill volume is normally distributed with population mean μ\mu and known population standard deviation 8 ml8\ \text{ml}. A random sample of 4949 containers is used to test

H0:μ=120againstH1:μ>120H_0:\mu=120\qquad\text{against}\qquad H_1:\mu>120

The significance level is 5%5\%.

A
I.

what a Type I error would mean in this context.

[1]
II.

Given that the sample mean is 122.4 ml122.4\ \text{ml}, the test statistic and the conclusion.

[2]
B

the critical value of Xˉ\bar X and the critical region.

[2]
C

When the true population mean is 123 ml123\ \text{ml}, the probability of a Type II error and the power of the test.

[3]
Question 28
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Ten musicians test a new breathing exercise. For each musician, the increase in the time for which a note can be sustained is calculated as

d=time after exercise−time before exercised=\text{time after exercise}-\text{time before exercise}

The sample of differences has mean dˉ=1.80 s\bar d=1.80\ \text{s} and sample standard deviation sd=1.40 ss_d=1.40\ \text{s}. The differences are assumed to come from a normal population.

Musician

Before / s

After / s

dd / s

1

18.0

21.5

3.5

2

19.4

22.7

3.3

3

20.1

23.2

3.1

4

17.8

20.6

2.8

5

21.0

23.8

2.8

6

24.5

24.6

0.1

7

23.1

23.4

0.3

8

22.2

22.8

0.6

9

18.7

19.4

0.7

10

26.0

26.8

0.8

A
I.

why a one-sample tt-test of the differences is appropriate.

[1]
II.

the test statistic for testing whether the exercise increases the population mean sustaining time.

[2]
B

The one-tailed pp-value is 0.001410.00141. the conclusion at the 5%5\% significance level.

[2]
C

The conductor considers an increase practically useful only if the population mean increase exceeds 1.00 s1.00\ \text{s}. Test this claim at the 5%5\% significance level and on the difference between statistical and practical significance.

[3]

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Question 29
HL • Paper 3
Hard
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HL • Paper 3
Hard
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Two independent cooling systems are tested using separate electronic devices. Cooling times are assumed to be normally distributed with equal but unknown population variances. Summary statistics are shown for the two systems.

System

Sample size n

Mean cooling time [min]

Sample standard deviation [min]

A

12

52.4

4.1

B

10

48.6

3.7

A
I.

hypotheses to test whether system A has a greater population mean cooling time than system B.

[1]
II.

the pooled of the common population standard deviation.

[2]
B

Carry out the test at the 5%5\% significance level.

[3]
C

A difference is considered operationally important only if μA−μB>2\mu_A-\mu_B>2 minutes. Test this claim and the result.

[2]
Question 30
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A marine scientist investigates the relationship between water temperature and the growth rate of an algae species at 1414 locations. The paired variables are assumed to follow a bivariate normal distribution. For all 1414 locations, the sample product moment correlation coefficient is r=0.667r=0.667.

A
I.

hypotheses for testing whether there is a positive linear correlation in the population.

[1]
II.

the test statistic.

[2]
B

The one-tailed pp-value is 0.004610.00461. the conclusion at the 1%1\% significance level.

[2]
C

After the potentially influential location is removed, the remaining data give r=0.363r=0.363 and a two-tailed pp-value of 0.220.22. Discuss what this suggests about the original claim.

[3]
Question 31
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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A national park classifies 200200 volunteers by age group and preferred conservation activity. The observed frequencies are shown in the table. A chi-squared test is used to whether age group and preferred activity are independent.

Age group

Trail maintenance

Habitat restoration

Wildlife monitoring

Total

Junior

24

10

26

60

Senior

19

25

26

70

Adult

27

35

8

70

Total

70

70

60

200

A
I.

the null and alternative hypotheses.

[1]
II.

the expected frequency for junior volunteers who prefer trail maintenance.

[2]
B

A GDC gives χ2=24.98\chi^2=24.98. the degrees of freedom and the conclusion at the 5%5\% significance level.

[3]
C

The largest contribution to χ2\chi^2 is 8.058.05, from adult volunteers choosing wildlife monitoring. this contribution and Cramér's measure V=χ2/(nmin⁡(r−1,c−1))V=\sqrt{\chi^2/(n\min(r-1,c-1))}.

[2]
Question 32
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A renewable-energy company predicts that maintenance requests occur in five categories with probabilities 0.100.10, 0.200.20, 0.300.30, 0.250.25 and 0.150.15. A random sample of 200200 requests gives observed frequencies 1212, 4848, 7171, 4343 and 2626, respectively.

Category

Predicted probability

Observed frequency

1

0.10

12

2

0.20

48

3

0.30

71

4

0.25

43

5

0.15

26

A
I.

the expected frequencies under the company's model.

[2]
II.

the number of degrees of freedom.

[1]
B

the chi-squared statistic and its pp-value.

[3]
C

the conclusions at the 5%5\% and 10%10\% significance levels and why the significance level must be selected before examining the data.

[2]

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Question 33
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A language-learning application claims that a learner answers each vocabulary question correctly with probability 0.500.50. A researcher tests whether an adaptive feature increases this probability. Each learner answers 2020 independent questions. One learner answers 1515 correctly. A simulation under the null hypothesis produces 10 00010\,000 samples, of which 218218 contain at least 1515 correct answers.

A
I.

the pp-value using the simulation.

[1]
II.

the exact binomial pp-value.

[2]
B

the critical region for a 5%5\% significance level and the conclusion for the learner.

[2]
C

When the true success probability is 0.700.70, the probability of a Type II error and the power of the test, and what the simulation represents.

[3]
Question 34
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A digital randomizer is intended to produce each of six symbols with equal probability. In 120120 trials, the observed frequencies are 1111, 1717, 2222, 1919, 2121 and 3030.

Symbol

Observed frequency

Symbol 1

11

Symbol 2

17

Symbol 3

22

Symbol 4

19

Symbol 5

21

Symbol 6

30

Total

120

A
I.

suitable hypotheses for a chi-squared goodness-of-fit test.

[1]
II.

the expected frequency for each symbol and the number of degrees of freedom.

[2]
B

Carry out the goodness-of-fit test at the 5%5\% significance level.

[3]
C

After inspecting the data, an analyst proposes a separate upper-tailed binomial test only for the sixth symbol. why using this post hoc test without adjustment may be misleading.

[2]
Question 35
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A call centre models the number of complaints received in a 3030-minute interval by a Poisson distribution with mean 22. During 100100 intervals, the observed numbers of complaints are grouped as 00, 11, 22, 33, and at least 44, with observed frequencies 1010, 2525, 3131, 1717 and 1717.

Complaints in a 30-minute interval

Observed frequency

0

10

1

25

2

31

3

17

At least 4

17

Total

100

A
I.

the expected frequency for intervals with exactly two complaints.

[2]
II.

the expected frequency for intervals with at least four complaints.

[1]
B

The complete expected frequencies are 13.5313.53, 27.0727.07, 27.0727.07, 18.0418.04 and 14.2914.29. Carry out a chi-squared goodness-of-fit test at the 5%5\% significance level.

[3]
C

why the conclusion does not that complaints follow a Poisson distribution, and one contextual assumption needed for the model.

[2]
Question 36
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The thickness of a glass panel is normally distributed with known population standard deviation 0.24 mm0.24\ \text{mm}. A sample of 25 panels is used to test
H0:μ=8.00againstH1:μ<8.00H_0:\mu=8.00\qquad\text{against}\qquad H_1:\mu<8.00
at the 1%1\% significance level.

A
I.

the critical value of Xˉ\bar X and the critical region.

[2]
II.

the probability of a Type I error.

[2]
B

When the true population mean is 7.85 mm7.85\ \text{mm}, the probability of a Type II error and the power of the test. If you did not obtain the critical value, use 7.8883… mm7.8883\ldots\ \text{mm}. Use the unrounded critical value in subsequent calculations; answers using 7.888 mm7.888\ \text{mm} are also accepted.

[4]

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Question 37
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A germination test uses 30 independent seeds. A supplier tests
H0:π=0.30againstH1:π>0.30H_0:\pi=0.30\qquad\text{against}\qquad H_1:\pi>0.30
at the 5%5\% significance level. Let XX be the number of seeds that germinate.

A
I.

the distribution of XX under H0H_0.

[1]
II.

the critical region that maximizes the probability of a Type I error while keeping it below 0.050.05; the actual probability.

[4]
B

When the true germination probability is 0.500.50, the probability of a Type II error and the power of the test.

[4]
Question 38
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The number of emergency calls received each day is modelled by a Poisson distribution. The usual population mean is 4 calls per day. Assume that the numbers of calls on different days are independent. Calls over five days are used to test
H0:λ=4againstH1:λ<4H_0:\lambda=4\qquad\text{against}\qquad H_1:\lambda<4
at the 5%5\% significance level. Let YY be the total number of calls over five days.

A
I.

the distribution of YY under H0H_0.

[1]
II.

the critical region and the actual probability of a Type I error.

[4]
B

Suppose the true population mean is 3 calls per day. the probability of a Type II error and the power of the test.

[4]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Ten patients have their systolic blood pressure measured before and after a relaxation programme. Define
d=blood pressure after−blood pressure befored=\text{blood pressure after}-\text{blood pressure before}
The differences, in mmHg\text{mmHg}, are −4,−2,−5,1,−3,−6,−1,−4,0,−5-4,-2,-5,1,-3,-6,-1,-4,0,-5. The population of differences is assumed normal.

A
I.

hypotheses to test whether the programme reduces the population mean blood pressure.

[2]
II.

why the data should be analysed using a one-sample test rather than an independent two-sample test.

[3]
B

Carry out the test at the 1%1\% significance level and the result.

[4]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A wildlife charity claims that more than 60%60\% of released turtles reach a protected feeding area. Of 40 independently tracked turtles, 30 reach the area. The probability of success is assumed constant.

A
I.

the hypotheses and the distribution used under the null hypothesis.

[2]
II.

why the test is upper-tailed.

[2]
B

the exact pp-value and the conclusion at the 5%5\% significance level. a Type I error in context.

[4]

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Question 41
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A conservation programme normally finds that 30%30\% of tagged turtle nests survive a severe storm. After a new protective barrier is installed, a researcher tests

H0:π=0.30againstH1:π<0.30H_0:\pi=0.30\qquad\text{against}\qquad H_1:\pi<0.30

using 4040 independently selected nests. Let XX be the number of nests that survive. The significance level is 5%5\%.

A
I.

the distribution of XX under H0H_0.

[1]
II.

the critical region.

[2]
B

the actual probability of a Type I error and why it is not exactly 0.050.05.

[2]
C

When the true survival probability is 0.180.18, the probability of a Type II error and the power of the test.

[3]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The number of emergency calls received by a rural clinic each night is modelled by a Poisson distribution with mean 22 calls per night. A new routing system is intended to reduce the mean number of calls received by the clinic. Data are collected over eight nights. Let XX be the total number of calls received.

X

P(X≤x)

5

0.0014

6

0.0040

7

0.0100

8

0.0220

9

0.0433

10

0.0774

11

0.1270

12

0.1931

A
I.

suitable hypotheses and the distribution of XX under the null hypothesis.

[2]
II.

the critical region for a 5%5\% significance level.

[1]
B

During the eight nights, 88 calls are received. the decision and it in context.

[2]
C

When the true mean is 1.51.5 calls per night, the probability of a Type II error and the power of the test.

[3]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A laboratory claims that the population mean acidity index of a solution is 8.008.00. A random sample of 1515 batches has mean index 7.847.84 and sample standard deviation 0.6200.620. Acidity indices are assumed to be normally distributed.

A
I.

why a tt-test is appropriate for testing whether the population mean is less than 8.008.00.

[1]
II.

the test statistic.

[2]
B

The one-tailed pp-value is 0.1670.167. the conclusion at the 5%5\% significance level.

[2]
C

For planning a future study, assume the population standard deviation is known to be 0.6200.620. the minimum sample size needed for a 5%5\% lower-tailed test to have at least 80%80\% power when the true mean is 7.607.60.

[3]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The breaking force of a safety clip is normally distributed with known population standard deviation 4 N4\ \text{N}. A manufacturer tests

H0:μ=50againstH1:μ≠50H_0:\mu=50\qquad\text{against}\qquad H_1:\mu\ne50

using the mean of a random sample of 2525 clips. The significance level is 1%1\%.

A
I.

why the significance level is divided equally between two tails.

[1]
II.

the two critical values of Xˉ\bar X.

[2]
B

the critical region and the probability of a Type I error.

[2]
C

the probability of a Type II error and the power when the true population mean is 53 N53\ \text{N}.

[3]

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Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A medical screening device normally gives a false alarm on 20%20\% of harmless samples. After recalibration, 2020 independent harmless samples are tested. Let XX be the number producing an alarm. The manufacturer tests

H0:π=0.20againstH1:π>0.20H_0:\pi=0.20\qquad\text{against}\qquad H_1:\pi>0.20

at the 5%5\% significance level.

Expression

Probability

P(X≥7∣X∼B⁡(20,0.20))P(X\geq 7\mid X\sim\operatorname{B}(20,0.20))

0.0867

P(X≥8∣X∼B⁡(20,0.20))P(X\geq 8\mid X\sim\operatorname{B}(20,0.20))

0.0321

P(X≤7∣X∼B⁡(20,0.40))P(X\leq 7\mid X\sim\operatorname{B}(20,0.40))

0.4159

A
I.

the critical region.

[2]
II.

the actual probability of a Type I error.

[1]
B

When the true alarm probability is 0.400.40, the probability of a Type II error and the power.

[3]
C

The manufacturer states that a Type I error is more serious than a Type II error. the practical consequence of each error in this context and one change that gives greater protection against a Type I error.

[2]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A telescope normally records an average of 22 transient signals per hour. After a detector upgrade, signals are counted over five hours. Let XX be the total number recorded. A one-tailed test is used to whether the upgrade has increased the mean signal rate.

A
I.

the hypotheses and the distribution of XX under H0H_0.

[2]
II.

the critical region at the 5%5\% significance level.

[1]
B

A total of 1717 signals is recorded. the pp-value and the conclusion.

[2]
C

When the true mean rate is 33 signals per hour, the probability of a Type II error and the power.

[3]
Question 47
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The mass of medicine in a capsule is normally distributed with known population standard deviation 1.8 mg1.8\ \text{mg}. A random sample of 36 capsules is used to test
H0:μ=50againstH1:μ≠50H_0:\mu=50\qquad\text{against}\qquad H_1:\mu\ne50
at the 5%5\% significance level.

A
I.

the two critical values of Xˉ\bar X and the critical region.

[4]
II.

why the probability of a Type I error is 0.050.05 even though the critical region has two parts.

[2]
B

Suppose the true population mean is 51.0 mg51.0\ \text{mg}. the probability of a Type II error and the power of the test. If you did not obtain the critical boundaries, use 49.41249.412 and 50.58850.588.

[4]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A health researcher investigates three independent outcome variables after introducing a workplace programme. Under their respective null hypotheses, the tests produce pp-values 0.0180.018, 0.0410.041 and 0.2200.220. The researcher initially tests each outcome at the 5%5\% significance level.

A
I.

which outcomes are statistically significant at the 5%5\% level.

[1]
II.

Assuming the three tests are independent and all three null hypotheses are true, the probability of at least one Type I error.

[2]
B

the common significance level aa for each test such that the probability of at least one Type I error is 0.050.05, assuming independence.

[3]
C

Using the adjusted level from part (b), evaluate the researcher's evidence.

[2]

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Estimation & Confidence Intervals

Probability