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Estimation & Confidence Intervals

Practice exam-style IB Math AI questions for Estimation & Confidence Intervals, aligned with the syllabus and grouped by topic.

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Question 1
HL • Paper 1
Easy
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HL • Paper 1
Easy
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A town council wants to survey residents about its new bus service. One proposed questionnaire item is:

“Do you agree that the excellent new buses should run more often?”

A

two problems with this questionnaire item.

[2]
B

Write a precise, unbiased and structured replacement for this item.

[2]
C

The council also asks residents to their annual household income. why responses to this item should be collected anonymously.

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

A fitness application records the weekly exercise time of its users. A researcher intends to use these data to the weekly exercise time of all adults in a country.

A

one reason why these data may not be appropriate for the intended population.

[1]
B

two additional variables that could help the researcher whether the application users are representative of the population.

[2]
C

an improved method of collecting data for the intended .

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

For a chi-square goodness-of-fit test, journey times are placed into five adjacent categories. The expected frequencies are 3.23.2, 4.84.8, 11.011.0, 18.018.0 and 13.013.0, in increasing order of journey time. One parameter of the proposed distribution was estimated from the sample.

A

why the categories cannot be used in their present form.

[1]
B

a suitable change to the categorisation and your answer.

[2]
C

the number of degrees of freedom for the test after making this change.

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

A psychologist gives the same stress questionnaire to 4040 employees on two occasions, three weeks apart. The correlation coefficient between the two sets of scores is r=0.910r=0.910.

A

the name of this method of testing reliability.

[1]
B

, with a reason, what the value of rr suggests about the reliability of the questionnaire.

[2]
C

why this result does not establish that the questionnaire is valid.

[1]
D

Outline how criterion-related validity could be investigated for this questionnaire.

[2]

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Question 5
HL • Paper 1
Medium
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HL • Paper 1
Medium
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A school creates two equivalent forms of a mathematics test. The correlation between students' scores on the two forms is 0.9400.940. However, all questions on both forms assess algebraic manipulation, although the course also includes functions, geometry and statistics.

A

the reliability method used.

[1]
B

what the correlation suggests about the reliability of the test.

[1]
C

Evaluate the content validity of the test.

[2]
D

one modification that would improve the content validity.

[1]
Question 6
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The mass, in grams, of a packet produced by a machine is normally distributed with mean 4242 and standard deviation 88. A random sample of 1616 packets is selected. Let Xˉ\bar X denote the sample mean mass.

A

the distribution of Xˉ\bar X.

[2]
B

P(40<Xˉ<45)P(40<\bar X<45).

[2]
C

why the normal distribution used in part (a) is exact rather than an approximation.

[1]
Question 7
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A non-normal population has mean 3030 and standard deviation 1414. Two simulations repeatedly select independent random samples and record the sample mean. Simulation A uses samples of size 55, while simulation B uses samples of size 5050.

A

the expected centres of the two simulated sampling distributions.

[1]
B

the standard deviation of the sample means in each simulation.

[2]
C

which simulation should produce a sampling distribution that is closer to normal.

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

The time taken to complete a task is normally distributed with known population standard deviation 4.64.6 minutes. A random sample of 4040 workers has a mean completion time of 31.831.8 minutes.

A

a 92.5%92.5\% confidence interval for the population mean completion time.

[3]
B

this interval in context.

[2]

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Question 9
HL • Paper 1
Medium
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HL • Paper 1
Medium
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The dissolved oxygen concentration in a lake is assumed to be normally distributed. For a random sample of 1818 water samples, the mean concentration is 12.4 mg L−112.4\text{ mg L}^{-1} and the sample standard deviation is 2.1 mg L−12.1\text{ mg L}^{-1}.

A

why a tt-distribution should be used to a confidence interval for the population mean.

[1]
B

a 95.5%95.5\% confidence interval for the population mean dissolved oxygen concentration.

[3]
C

A biologist claims that the population mean concentration is 14.0 mg L−114.0\text{ mg L}^{-1}. what the interval suggests about this claim.

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

A researcher repeatedly takes random samples of the same size from a population and calculates a 95.5%95.5\% confidence interval for the population mean from each sample.

A

the repeated-sampling meaning of the 95.5%95.5\% confidence level.

[2]
B

The researcher constructs 200200 such intervals. the expected number that contain the true population mean.

[1]
C

how the width of the intervals would change if the confidence level were increased while the sample size and observed standard deviation remained fixed.

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

The masses, in kilograms, of six independently selected food containers are denoted by X1,X2,…,X6X_1, X_2, \dots, X_6. Each mass is normally distributed with mean 2.42.4 and standard deviation 0.30.3. The total shipping mass is

Y=X1+X2+⋯+X6+0.8Y=X_1+X_2+\dots+X_6+0.8

where 0.80.8 kg is the mass of the shipping box.

A

the mean and variance of YY.

[3]
B

the probability that the total shipping mass exceeds 1616 kg.

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

The daily amount spent by a visitor at a market is strongly right-skewed, with mean €1818 and standard deviation €1212. A random sample of 4949 visitors is selected and Xˉ\bar X is their mean daily expenditure.

A

why a normal model may be used for the distribution of Xˉ\bar X.

[2]
B

an approximate distribution for Xˉ\bar X.

[1]
C

P(Xˉ<20)P(\bar X<20).

[2]

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Question 13
HL • Paper 1
Medium
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HL • Paper 1
Medium
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The lifetime of a type of battery is normally distributed with mean 7575 hours and standard deviation 1212 hours. A random sample of nn batteries is selected.

A

why the sample mean lifetime is normally distributed for every positive integer value of nn.

[1]
B

the minimum value of nn such that the probability that the sample mean is within 33 hours of 7575 is at least 0.950.95.

[4]
Question 14
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The standard deviation of the diameter of components produced by a factory is known to be 7.27.2 mm. A random sample will be used to a 99.5%99.5\% confidence interval for the population mean diameter.

A

the minimum sample size required for the width of the confidence interval to be less than 33 mm.

[3]
B

, with a reason, how the required sample size would change if a 95.5%95.5\% confidence interval of the same maximum width were used.

[2]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A random sample of 3636 observations is taken from a normal population with known standard deviation σ\sigma. A 90.5%90.5\% confidence interval for the population mean is

52.1<μ<57.952.1<\mu<57.9
A

the sample mean.

[1]
B

the margin of error.

[1]
C

σ\sigma.

[2]
D

Using the same sample data, a 95.5%95.5\% confidence interval for the population mean.

[2]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Two independent random samples of customer ratings are taken for brands A and B. The ratings for each brand are assumed to come from a normal population.

For brand A, n=25n=25, xˉ=82\bar x=82 and s=10s=10.

For brand B, n=30n=30, xˉ=78\bar x=78 and s=9s=9.

A

a 90%90\% confidence interval for the population mean rating of brand A.

[2]
B

a 90%90\% confidence interval for the population mean rating of brand B.

[2]
C

Using these intervals, on the claim that brand A has a higher population mean rating than brand B.

[2]

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Question 17
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A university investigates student satisfaction with its online library. The proposed questionnaire includes the item: “How much do you agree that the convenient and excellent online library has improved your studies?” Responses are recorded on a scale from 1 to 5. A random sample of 36 students gives a mean response of 3.42 and a sample standard deviation of 0.88.

A

Consider the design of the questionnaire.

I.

two problems with the proposed item.

[2]
II.

Write a precise, unbiased and structured replacement item that measures frequency of use.

[2]
B

The researchers assume that satisfaction scores in the population are normally distributed.

I.

the reliability method in which the same questionnaire is given to the same students on two occasions, and why a high correlation would not establish validity.

[2]
II.

a 95%95\% confidence interval for the population mean satisfaction score and it in context.

[4]
Question 18
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A researcher models the number of hours of screen use per week. For a chi-square goodness-of-fit test, the expected frequencies in six adjacent categories, from lowest to highest screen use, are 2.42.4, 4.64.6, 8.28.2, 14.814.8, 20.020.0 and 10.010.0. Two parameters of the model were estimated from the sample. In a separate study, weekly screen use has mean 1212 hours and standard deviation 99 hours, and its distribution is strongly right-skewed. An independent random sample of 64 people is selected from this population.

Screen use category [h/week]

Expected frequency

0≤x<50\le x<5

2.4

5≤x<105\le x<10

4.6

10≤x<1510\le x<15

8.2

15≤x<2015\le x<20

14.8

20≤x<2520\le x<25

20.0

25≤x<3025\le x<30

10.0

A

Consider the chi-square goodness-of-fit test.

I.

why the original categorisation is unsuitable and an appropriate revision.

[2]
II.

the number of degrees of freedom for the revised test.

[2]
B

An independent random sample of 64 people is selected from the separate study.

I.

an approximate distribution for the sample mean weekly screen use, giving its parameters, and the approximation.

[2]
II.

the probability that the sample mean is between 10.510.5 and 13.513.5 hours.

[3]
Question 19
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A gift hamper contains three independently selected bags of coffee and two independently selected boxes of tea. The mass XX, in grams, of a coffee bag is normally distributed with mean 480480 and standard deviation 1212. The mass YY, in grams, of a tea box is normally distributed with mean 320320 and standard deviation 88. The empty hamper has mass 5050 grams. Let HH be the total mass of a completed hamper.

A

Consider the distribution of HH.

I.

the mean and variance of HH.

[2]
II.

why HH is normally distributed and P(H>2170)P(H>2170).

[2]
B

A random sample of 25 completed hampers, whose masses are independent, is selected, and Hˉ\bar H is the mean of their masses.

I.

the distribution of the sample mean mass Hˉ\bar H.

[2]
II.

the value of dd such that P(2130−d<Hˉ<2130+d)=0.95P(2130-d<\bar H<2130+d)=0.95.

[3]
Question 20
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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The amount paid for an individual home-insurance claim is strongly right-skewed, with population mean €240 and population standard deviation €300. Claim amounts are independent. An insurer studies the mean amount paid for a random sample of 100 claims.

A

A random sample of 100 claims is selected.

I.

an approximate distribution for the sample mean claim amount and your answer.

[2]
II.

the probability that the sample mean exceeds €300.

[2]
B

The insurer wants the sample mean to be close to the population mean.

I.

the condition P(∣Xˉ−240∣<40)≥0.95P(|\bar X-240|<40)\geq0.95 leads to n≥216.1n\geq216.1.

[2]
II.

the minimum sample size and on whether use of the central limit theorem is reasonable.

[3]

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Question 21
HL • Paper 2
Hard
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HL • Paper 2
Hard
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The temperature at which a type of sensor activates is assumed to be normally distributed. For a random sample of 15 sensors, the mean activation temperature is 6.84∘C6.84^\circ\text{C} and the sample standard deviation is 1.12∘C1.12^\circ\text{C}. The population standard deviation is unknown.

A

Consider a confidence interval for the population mean activation temperature.

I.

why a tt-distribution with 14 degrees of freedom should be used.

[2]
II.

a 90%90\% confidence interval for the population mean.

[3]
B

Temperatures in degrees Fahrenheit are related to temperatures in degrees Celsius by F=1.8C+32F=1.8C+32.

I.

write the 90%90\% confidence interval for the population mean activation temperature in degrees Fahrenheit.

[2]
II.

a 99%99\% confidence interval in degrees Celsius and why it is wider than the interval from part (a)(ii).

[3]
Question 22
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A streaming platform asks 400 active subscribers to the number of hours of television watched per week. The sample mean is 7.87.8 hours. For active subscribers, the population standard deviation is known to be 2.42.4 hours. A researcher wants to use the results to mean weekly television viewing for all adults in the country.

A

Consider the suitability of the collected data.

I.

Give two reasons why the data may not be valid for the researcher's intended population.

[2]
II.

two relevant variables that could be collected to assess how representative the sample is.

[2]
B

For this part, treat the 400 active subscribers as a random, independent sample from the population of active subscribers.

I.

a 95%95\% confidence interval for the population mean viewing time of active subscribers.

[3]
II.

why the narrow interval does not applying the to all adults.

[2]
Question 23
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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A computer repeatedly selects independent random samples of the same size from a population with fixed mean μ\mu. For each sample it constructs an 80%80\% confidence interval for μ\mu. A total of 250 intervals are constructed, of which 207 contain μ\mu.

A

Consider the number of intervals that contain μ\mu.

I.

the expected number of intervals that contain μ\mu and the expected number that do not.

[2]
II.

Model the number containing μ\mu by X∼B(250,0.8)X\sim B(250,0.8). P(X≥207)P(X\geq207).

[2]
B

For a new set of 250 intervals, the simulation is repeated using 99%99\% confidence intervals constructed by the same method as before.

I.

the expected number, out of 250, that will fail to contain μ\mu, and why the actual number need not equal this value.

[2]
II.

how the widths of the new intervals with those of the original intervals, assuming the same sample data are used.

[2]
Question 24
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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The time spent on three independent stages of a delivery process is represented by AA, BB and CC, measured in minutes. The distributions are A∼N(18,16)A\sim N(18,16), B∼N(12,9)B\sim N(12,9) and C∼N(8,4)C\sim N(8,4). A fixed administration time of 5 minutes is added. The total delivery time is T=A+B+C+5T=A+B+C+5.

A

Consider the total time for one delivery.

I.

the distribution of TT.

[2]
II.

the probability that a delivery takes less than 50 minutes.

[3]
B

A random sample of 16 independent deliveries is selected. Let Tˉ\bar T denote the mean total delivery time for these deliveries.

I.

the distribution of Tˉ\bar T.

[2]
II.

P(41<Tˉ<45)P(41<\bar T<45).

[2]

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Question 25
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A national energy agency estimates monthly household electricity consumption. A simple random sample of households was contacted. Of these, 100 households completed the survey, with mean consumption 126 kWh126\ \text{kWh}. These completed responses may be subject to non-response bias. From previous national records, the population standard deviation is known to be 40 kWh40\ \text{kWh}. Only 42%42\% of households contacted completed the survey, and urban households had a higher response rate than rural households.

A

Use the responding households to the population mean.

I.

a 90%90\% confidence interval for the population mean monthly consumption.

[2]
II.

two reasons why the interval may not give a valid for all households, despite being based on a random initial sample.

[3]
B

The agency plans a new survey using the same known population standard deviation.

I.

the minimum number of completed responses required for a 95%95\% confidence interval to have margin of error at most 4 kWh4\ \text{kWh}.

[3]
II.

a sampling modification that would address the unequal urban and rural response rates, and its benefit.

[2]
Question 26
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A filling machine dispenses coffee into jars. The mass dispensed is normally distributed with unknown mean μ\mu and known standard deviation 12.8 g12.8\ \text{g}. A random sample of 6464 jars has mean mass 503.2 g503.2\ \text{g}.

A
I.

the distribution of the sample mean in terms of μ\mu.

[2]
II.

a 90%90\% confidence interval for μ\mu.

[2]
B

the interval from part (a)(ii) in context.

[1]
C

The manufacturer wants a 99%99\% confidence interval with margin of error at most 2 g2\ \text{g}. the minimum required sample size.

[3]
Question 27
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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An archaeologist measures the depths, in centimetres, of a particular layer at 1414 randomly selected locations. The depths are assumed to come from a normal population. The sample has mean 127.4 cm127.4\ \text{cm} and standard deviation 8.6 cm8.6\ \text{cm}.

A
I.

why a tt-distribution with 1313 degrees of freedom is appropriate.

[2]
II.

a 95%95\% confidence interval for the population mean depth.

[2]
B

a 99%99\% confidence interval for the population mean depth.

[2]
C

A previous study reported a population mean depth of 130 cm130\ \text{cm}. the two intervals and on this reported value.

[2]
Question 28
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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A school wants to the mean number of hours of sleep obtained by its senior students on a school night. It sends a questionnaire only to students attending an optional early-morning study club. One item asks, “How many hours of healthy sleep do you normally get?” A total of 120120 students respond, with mean 6.426.42 hours and sample standard deviation 1.351.35 hours. Assume the sleep times in the intended population are normally distributed.

A
I.

one sampling problem and how it may affect validity.

[2]
II.

Write a more precise and unbiased questionnaire item for measuring sleep duration.

[2]
B

a 90%90\% confidence interval for the population mean sleep time, treating the responses as a random sample.

[2]
C

why the narrow interval in part (b) does not resolve the validity problem identified in part (a)(i).

[2]

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Question 29
HL • Paper 3
Hard
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HL • Paper 3
Hard
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The time between successive calls to a help centre has a strongly right-skewed distribution with mean 1010 minutes and standard deviation 1010 minutes. A simulation repeatedly selects independent samples and records their means. Three simulations use sample sizes 44, 3636 and 100100. Their sampling distributions are labelled AA, BB and CC, but the labels do not the sample sizes.

A
I.

the expected centre of each sampling distribution.

[1]
II.

the standard error for each sample size.

[2]
B

Match the three sample sizes to histograms AA, BB and CC. your answer using both shape and spread.

[3]
C

why the simulation provides evidence for the central limit theorem but cannot it.

[2]
Question 30
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A company develops a wrist device to measure blood glucose concentration. For 1616 volunteers, the difference

D=device reading−laboratory readingD=\text{device reading}-\text{laboratory reading}

is assumed normally distributed, and the 1616 observed differences are assumed to form an independent random sample from the target population. The sample mean difference is 1.8 mmol L−11.8\ \text{mmol L}^{-1} and the sample standard deviation is 2.4 mmol L−12.4\ \text{mmol L}^{-1}. In a separate test, repeated readings from the wrist device have correlation 0.960.96.

A
I.

what the correlation of 0.960.96 suggests about the device.

[1]
II.

why this correlation does not establish criterion-related validity.

[2]
B

a 95%95\% confidence interval for the population mean difference.

[3]
C

evaluate the criterion-related validity of the device.

[2]
Question 31
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A music-streaming company uses responses posted through its application to the daily listening time of all people aged 1616 to 2525 in a country. Its questionnaire asks, “How many minutes do you spend enjoying our excellent music service each day?” For 100100 respondents, the mean is 7474 minutes. The population standard deviation of reported listening time is taken to be 3030 minutes.

A
I.

two threats to the validity of the .

[2]
II.

Write a precise and unbiased replacement question.

[2]
B

a 99%99\% confidence interval for the population mean reported listening time, assuming the respondents form a random sample.

[2]
C

Assess whether the interval supports an for all people aged 1616 to 2525 in the country.

[2]
Question 32
HL • Paper 3
Hard
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HL • Paper 3
Hard
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Two independent greenhouses use different lighting systems. Plant heights are assumed normally distributed in each greenhouse. For system AA, n=20n=20, xˉ=34.2 cm\bar x=34.2\ \text{cm} and s=3.5 cms=3.5\ \text{cm}. For system BB, n=22n=22, xˉ=36.0 cm\bar x=36.0\ \text{cm} and s=4.1 cms=4.1\ \text{cm}.

A
I.

a 95%95\% confidence interval for the population mean height under system AA.

[2]
II.

a 95%95\% confidence interval for the population mean height under system BB.

[2]
B

Using the intervals, on the claim that system BB produces taller plants on average.

[2]
C

The researcher wants to halve the approximate margin of error for each system without changing the confidence level or anticipated variability. how each sample size should change.

[2]

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Question 33
HL • Paper 3
Hard
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HL • Paper 3
Hard
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The duration of a sea turtle's nesting visit is strongly right-skewed. Historical research indicates that the population standard deviation is 2020 minutes, but the current population mean is unknown. A random sample of 6464 current visits has mean duration 5656 minutes.

A
I.

why an approximately normal model may be used for the sample mean.

[1]
II.

the standard error of the sample mean.

[2]
B

a 95%95\% confidence interval for the current population mean nesting duration.

[2]
C

the minimum sample size required for a 90%90\% confidence interval to have total width at most 88 minutes.

[3]
Question 34
HL • Paper 2
Hard
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HL • Paper 2
Hard
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The concentration of a chemical in bottles of cleaning solution has known population standard deviation 3.6 g L−13.6\ \text{g L}^{-1}. A random sample of 64 bottles has mean concentration 18.7 g L−118.7\ \text{g L}^{-1}.

A

Use the sample to the population mean concentration.

I.

a 98%98\% confidence interval for the population mean concentration.

[3]
II.

the interval in context.

[2]
B

The manufacturer plans a new random sample, using the same known population standard deviation.

I.

the minimum sample size required for a 95%95\% confidence interval to have margin of error at most 0.5 g L−10.5\ \text{g L}^{-1}.

[3]
II.

Another interval based on the original sample is (17.8,19.6)(17.8,19.6). its confidence level.

[3]
Question 35
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A language school designs two forms, A and B, of a vocabulary test. Both forms are completed by the same 10 students. The correlation between the two sets of scores is 0.930.93. Let DD be a student's score on form A minus the student's score on form B. The differences are assumed to come from a normal population and have sample mean 4.84.8 and sample standard deviation 3.13.1.

A

Consider the reliability and validity of the tests.

I.

Name the reliability method and the value 0.930.93.

[2]
II.

how content validity of the forms could be investigated.

[2]
B

The school investigates whether the forms have the same population mean score.

I.

a 95%95\% confidence interval for the population mean difference μD\mu_D.

[3]
II.

Using the correlation and the confidence interval, evaluate the claim that the forms are equivalent.

[3]
Question 36
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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The breaking strength of a type of climbing cord is assumed to be normally distributed. A random sample of 20 cords has mean breaking strength 52.4 kN52.4\ \text{kN} and sample standard deviation 6.5 kN6.5\ \text{kN}. An engineer constructs the confidence interval (49.6,55.2) kN(49.6,55.2)\ \text{kN} for the population mean.

A

the confidence level of the engineer's interval.

I.

the margin of error and the implied positive tt critical value.

[2]
II.

the confidence level.

[3]
B

The engineer instead requires a 95%95\% confidence interval based on the same sample.

I.

the 95%95\% confidence interval.

[2]
II.

why this interval is wider than the engineer's original interval.

[2]

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Question 37
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A finite population contains 120 measurements from a non-normal distribution with mean 70 and standard deviation 16. A simulation selects independent random samples with replacement and records the sample mean. Simulation A uses samples of size 16, while simulation B uses samples of size 64.

A

the two sampling distributions.

I.

the expected centre and the standard error for each simulation.

[2]
II.

For Simulation B, the approximate probability that the sample mean is between 68 and 73.

[3]
B

Separately from the simulations, a researcher considers all unordered samples of size 8 drawn without replacement from the 120 distinguishable population elements.

I.

the number of distinct samples of size 8.

[2]
II.

why running a large but finite number of simulations cannot the central limit theorem.

[2]
Question 38
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Two independent farms grow the same variety of tomato using different fertilizers. Tomato masses at each farm are assumed to be normally distributed. Farm A samples 16 tomatoes and obtains xˉA=102 g\bar x_A=102\ \text{g} and sA=12 gs_A=12\ \text{g}. Farm B samples 20 tomatoes and obtains xˉB=96 g\bar x_B=96\ \text{g} and sB=10 gs_B=10\ \text{g}.

A

separate confidence intervals for the two population means.

I.

a 95%95\% confidence interval for the population mean mass at Farm A.

[3]
II.

a 95%95\% confidence interval for the population mean mass at Farm B.

[3]
B

The fertilizer supplier claims that Farm A's fertilizer produces tomatoes with a higher population mean mass.

I.

Using the intervals, on the supplier's claim that the fertilizer used at Farm A produces tomatoes with a higher population mean mass than the fertilizer used at Farm B.

[2]
II.

one limitation of using overlap of the two intervals as a formal comparison of the population means.

[2]
Question 39
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A physiotherapist records the time, in seconds, taken by each of 12 patients to complete a mobility task before and after a treatment. For each patient, the reduction in completion time is calculated as the time before treatment minus the time after treatment. The reductions are assumed to come from a normal population. Assume that the 12 patient reductions form an independent random sample from this population. They have sample mean 1.851.85 seconds and sample standard deviation 2.402.40 seconds.

A

the population mean reduction.

I.

why the analysis should use the paired reductions rather than treating the before and after measurements as independent samples.

[2]
II.

a 95%95\% confidence interval for the population mean reduction.

[3]
B

The physiotherapist also considers a 99%99\% confidence interval.

I.

a 99%99\% confidence interval for the population mean reduction.

[3]
II.

the conclusions suggested by the 95%95\% and 99%99\% intervals about whether the treatment reduces mean completion time.

[3]
Question 40
HL • Paper 3
Hard
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HL • Paper 3
Hard
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The amount paid for an individual household insurance claim is strongly right-skewed, with mean 24002400 euros and standard deviation 18001800 euros. Independent claims are selected at random. Let Xˉ\bar X be the mean amount paid for a sample of 8181 claims.

A
I.

why a normal model may be used for Xˉ\bar X.

[1]
II.

an approximate distribution for Xˉ\bar X.

[2]
B

the probability that the sample mean exceeds 27502750 euros.

[2]
C

the minimum sample size required so that P(∣Xˉ−2400∣<300)≥0.95P(|\bar X-2400|<300)\geq0.95.

[3]

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Question 41
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A gift package contains three independently selected packets of tea and two independently selected packets of biscuits. The mass XX, in grams, of a tea packet is normally distributed with mean 1818 and standard deviation 22. The mass YY, in grams, of a biscuit packet is normally distributed with mean 1212 and standard deviation 1.51.5. The box has a fixed mass of 55 grams. The total mass is
T=X1+X2+X3+Y1+Y2+5T=X_1+X_2+X_3+Y_1+Y_2+5

A
I.

E(T)E(T).

[1]
II.

Var⁡(T)\operatorname{Var}(T) and the distribution of TT.

[3]
B

For a random sample of 2525 complete packages, the probability that their mean total mass is less than 81.581.5 grams.

[2]
C

In practice, packets filled during the same production period may be positively correlated. how this would affect the probability calculated in part (b).

[2]
Question 42
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A biologist tests whether seed-germination times follow a proposed distribution. Six adjacent time categories have observed frequencies 44, 55, 1111, 1212, 2020 and 1818, and expected frequencies 2.62.6, 4.94.9, 9.59.5, 14.014.0, 22.022.0 and 17.017.0, respectively. Two parameters of the proposed distribution were estimated from the sample.

Adjacent category

Observed frequency

Expected frequency

1

4

2.6

2

5

4.9

3

11

9.5

4

12

14.0

5

20

22.0

6

18

17.0

A
I.

why the categories cannot be used in their present form and a suitable modification.

[2]
II.

the number of degrees of freedom after this modification.

[1]
B

Using the revised categories, the chi-square test statistic and the corresponding pp-value.

[3]
C

At the 5%5\% significance level, the result and one disadvantage of combining categories.

[2]
Question 43
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A random sample of 2525 measurements is taken from a normal population with unknown mean and standard deviation. A 95%95\% confidence interval for the population mean is
41.7<μ<48.341.7<\mu<48.3

A
I.

the sample mean and margin of error.

[2]
II.

the sample standard deviation.

[2]
B

Using the same sample statistics, a 90%90\% confidence interval for the population mean.

[2]
C

Using s=8s=8 as a planning value, the minimum sample size required for a 95%95\% tt-interval to have margin of error at most 22.

[2]
Question 44
HL • Paper 3
Hard
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HL • Paper 3
Hard
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An island has 88 marked ecological plots. A researcher studies samples of 33 distinct plots selected without replacement. Each sample is chosen by simple random sampling, so each of the (83)=56\binom{8}{3}=56 unordered samples is equally likely. She proposes listing every possible unordered sample to obtain the exact sampling distribution of the sample mean. A colleague instead proposes generating 50005000 random samples by computer.

A
I.

the number of distinct unordered samples of 33 plots.

[2]
II.

one advantage of listing all 5656 samples rather than simulating 50005000 samples.

[1]
B

The population mean of the eight measurements is 18.518.5. the expected value of the sample mean and why it has this value.

[2]
C

The simulated distribution appears approximately symmetric. Evaluate the claim that this verifies the central limit theorem for all populations.

[3]

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Question 45
HL • Paper 3
Hard
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HL • Paper 3
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The output, in kilograms, from a machine during one production cycle is normally distributed. Under setting XX, the output has mean 5050 and standard deviation 66. Under setting YY, it has mean 5353 and standard deviation 88. Independent samples of 99 cycles under XX and 1616 cycles under YY are taken. Let
D=Yˉ−XˉD=\bar Y-\bar X

A
I.

E(D)E(D) and Var⁡(D)\operatorname{Var}(D).

[2]
II.

the distribution of DD.

[2]
B

the probability that the observed sample mean under setting YY exceeds that under setting XX.

[2]
C

Equal sample sizes nn are now to be used for both settings. the minimum nn such that P(D>0)≥0.95P(D>0)\geq0.95.

[2]
Question 46
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A university calculates an admissions index from an aptitude score AA and an interview score BB using
S=0.4A+0.6BS=0.4A+0.6B
For the applicant population, A∼N(70,82)A\sim N(70,8^2) and B∼N(60,102)B\sim N(60,10^2). Initially, assume that AA and BB are independent.

A
I.

E(S)E(S).

[1]
II.

Var⁡(S)\operatorname{Var}(S) and the distribution of SS.

[3]
B

A random sample of 4949 applicants from a later cohort has mean admissions index 65.565.5. Assuming the population standard deviation remains 6.86.8, a 98%98\% confidence interval for the later cohort's population mean index.

[2]
C

Suppose that AA and BB are actually positively correlated. how this affects the interval in part (b), assuming the same sample mean.

[2]
Question 47
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A researcher repeatedly selects independent random samples of size 3636 from a normal population with mean μ\mu and known standard deviation 1515. For each sample, a 95%95\% confidence interval for μ\mu is calculated. A set of 4040 intervals is produced.

Sample

Lower

Upper

1

-5.6

4.2

2

-3.8

6.0

3

-7.2

2.6

4

-2.1

7.7

5

-8.5

1.3

6

-4.5

5.3

7

-3.0

6.8

8

-6.4

3.4

9

-1.5

8.3

10

-6.9

2.9

11

-4.0

5.8

12

-2.7

7.1

13

-9.0

0.8

14

-1.9

7.9

15

-5.0

4.8

16

-3.3

6.5

17

-5.9

3.9

18

-2.3

7.5

19

-7.9

1.9

20

-4.1

5.7

21

0.2

10.0

22

-9.4

0.4

23

-1.1

8.7

24

-7.7

2.1

25

-3.6

6.2

26

-5.2

4.6

27

-2.0

7.8

28

-6.7

3.1

29

-4.4

5.4

30

-10.1

-0.3

31

-9.7

0.1

32

-1.8

8.0

33

-7.0

2.8

34

-3.9

5.9

35

-5.5

4.3

36

-2.5

7.3

37

-8.3

1.5

38

-4.7

5.1

39

-5.0

4.8

40

0.4

10.2

A
I.

the expected number of the 4040 intervals that contain μ\mu.

[1]
II.

the probability that at least 3939 of the intervals contain μ\mu, assuming the intervals are generated independently.

[2]
B

the margin of error of each interval.

[2]
C

The researcher wants to halve the margin of error. the required sample size and why it is incorrect to say that a particular calculated interval has a 95%95\% probability of containing μ\mu.

[3]
Question 48
HL • Paper 2
Hard
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HL • Paper 2
Hard
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An ecologist studies the number of insects caught in traps. For a chi-square goodness-of-fit test, the expected frequencies in six adjacent count categories are 3.83.8, 7.27.2, 14.014.0, 19.019.0, 11.011.0 and 5.05.0. One parameter of the proposed model was estimated from the observed data. The ecologist also measures insect mass. A random sample of 24 insects from a normal population has mean mass 7.4 mg7.4\ \text{mg} and sample standard deviation 1.8 mg1.8\ \text{mg}.

Section

Item

Value

Chi-square GOF

Category 1 expected frequency

3.8

Chi-square GOF

Category 2 expected frequency

7.2

Chi-square GOF

Category 3 expected frequency

14.0

Chi-square GOF

Category 4 expected frequency

19.0

Chi-square GOF

Category 5 expected frequency

11.0

Chi-square GOF

Category 6 expected frequency

5.0

Chi-square GOF

Estimated parameters, m

1

Mass sample

Sample size, n

24

Mass sample

Sample mean mass [mg]

7.4

Mass sample

Sample standard deviation [mg]

1.8

Mass sample

Population model

Normal

A

Consider the categorisation for the chi-square goodness-of-fit test.

I.

a suitable revision to the categories and it.

[2]
II.

the number of degrees of freedom for the revised test.

[2]
B

The ecologist estimates the population mean insect mass.

I.

a 96%96\% confidence interval for the population mean mass.

[3]
II.

A pesticide claims that the population mean insect mass in the area is less than 8.0 mg8.0\ \text{mg}. Evaluate this claim using the interval, and one limitation of the conclusion.

[4]

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Distributions

Hypothesis Testing