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Topic 4.12 · SL and HL

Chi GOF, Independence, T-test (Null / Alternative hypothesis, S.L. / p-values): notes and practice questions

Summary
  • Hypothesis Test: Uses sample data to test a statement about a population.
  • Null Hypothesis (H0H_0): Assumes no difference, no change, no association, or data follows a specific distribution.
  • Alternative Hypothesis (H1H_1): Opposes H0H_0, claims a difference, association, or data does not follow a distribution.
  • Significance Level (α\alpha): Probability threshold for rejecting H0H_0.
  • p-value: Probability of obtaining observed (or more extreme) results assuming H0H_0 is true.
  • Test Statistic: Calculated numerical value from sample data (e.g., χ2\chi^2, tt, zz).
  • Critical Value: Threshold for the test statistic, defining the critical region.
  • Degrees of Freedom (ν\nu): Parameter for χ2\chi^2 and tt-tests.
  • Type II Error (HL only): Failing to reject the null hypothesis when it is actually false.
  • Decision Rule (p-value):
  • If p-value<αp\text{-value} < \alpha: Reject H0H_0.
  • If p-value>αp\text{-value} > \alpha: Accept H0H_0.
  • Decision Rule (Test Statistic, HL only):
  • If Test Statistic > Critical Value: Reject H0H_0.
  • If Test Statistic < Critical Value: Accept H0H_0.
  • Chi-Squared Test for Independence: Tests if two categorical variables are independent.
  • H0H_0: Variable X is independent of Variable Y.
  • H1H_1: Variable X is not independent of Variable Y.
  • Degrees of Freedom: ν=(rows−1)×(columns−1) \nu = (\text{rows} - 1) \times (\text{columns} - 1)
  • Compares observed frequencies to expected frequencies (if independent).
  • Chi-Squared Goodness of Fit (GOF) Test: Tests if data fits a specific theoretical distribution.
  • H0H_0: The data can be modelled by the specified distribution.
  • H1H_1: The data cannot be modelled by the specified distribution.
  • Uniform Distribution: Expected frequency is the same for all categories.
  • Binomial/Normal: Calculate probabilities for intervals, then multiply by total sample size for expected frequencies.
  • T-Tests and Z-Tests for the Mean: Investigate claims about the population mean (μ\mu).
  • One-Sample Z-Test: Used when population variance (σ2\sigma^2) is known.
  • One-Sample T-Test: Used when population variance is unknown; calculate unbiased sample variance:

sn−12=nn−1sn2 s_{n-1}^2 = \frac{n}{n-1} s_n^2

  • Two-Sample T-Test: Compares means of two different, independent normally distributed populations (μ1,μ2\mu_1, \mu_2).
  • H0H_0: μ1=μ2\mu_1 = \mu_2
  • H1H_1: μ1≠μ2\mu_1 \neq \mu_2 (or <,><, >).
  • Paired T-Test: Compares two sets of data from the same sample by testing the mean difference (μD\mu_D).
  • H0H_0: μD=0\mu_D = 0
  • H1H_1: μD≠0\mu_D \neq 0 (or <,><, >).
  • GDC Tips:
  • Independence Tests: Enter observed frequencies into a matrix.
  • GOF Tests: Enter observed frequencies into one list, calculated expected frequencies into a second list.
  • T-Tests: Enter raw data into a list, or summary statistics directly.
  • SL vs. HL Distinctions:
  • SL: Basic χ2\chi^2 (independence, GOF), t-test.
  • HL: Type I/II errors, hypothesis testing for correlation, Poisson distributions, critical regions.
  • Exam Tips:
  • Conclusion Wording: State "There is insufficient evidence to reject the null hypothesis," not "proving it true." Conclude in context.
  • Paired vs. Two-Sample: Use paired t-test if data points correspond to the same individuals under two conditions.
  • IA Warning: χ2\chi^2 outcome may be inaccurate if any expected values are less than 5, or if there is only 1 degree of freedom.
  • One-Tailed vs. Two-Tailed: H1H_1 suggesting a specific direction (>,<>, <) is one-tailed; H1H_1 stating a difference (≠\neq) is two-tailed.

How it is examined

The longest question in the SL statistics section, usually five to eight marks across several parts: state the hypotheses, give the degrees of freedom, produce the pp-value, compare it to the significance level, then state the conclusion in context. The conclusion mark is only earned by referring back to the situation, not by writing "reject H0H_0". Every constraint above is an easy one to break when writing a question: four rows maximum, expected frequencies above 5, upper tail only, significance level from {1%, 5%, 10%}, unpaired samples, pooled two-sample tt-test.

Given in the booklet

The χ2\chi^2 statistic, χcalc2=∑(fo−fe)2fe\chi^2_{calc} = \sum \dfrac{(f_o - f_e)^2}{f_e}.

Key ideas
  • The formulation of null and alternative hypotheses, H0H_0 and H1H_1.
  • Significance levels.
  • pp-values.
  • Expected and observed frequencies.
Not assessed

Enrichment only, so not examinable: Yates' continuity correction.

Linking questions

  • Other contexts: in psychology, the Mann-Whitney U test is common. When and why is it thought to be a more reliable test there?
  • Links to other subjects: fieldwork (biology, psychology, environmental systems and societies, sports exercise and health science, geography).
  • TOK: why have some research journals banned pp-values from their articles because they deem them too misleading? In practical terms, is saying that a result is significant the same as saying it is true? How is the term "significant" used differently in different areas of knowledge?
  • Use of technology: use of simulations to generate data.

Practice questions

52 questions · 43 medium · 9 hard
Showing 20 of 20

Question 1

MediumPaper 1 · calculator5 marks
(a)

A lighting company technician is testing two new brands of energy-efficient light bulbs, Brand X and Brand Y. He claims that Brand X light bulbs have a shorter mean lifespan than Brand Y light bulbs.

He records the lifespans, in hours, from a random selection of light bulbs from each brand. The data is shown in the table.

Lifespan of Brand X (hours)125012801230126012401270
Lifespan of Brand Y (hours)126512751255127012601280

In order to test his claim, the technician performs a t-test at a 10% level of significance. It is assumed that the lifespans of light bulbs are normally distributed and the samples have equal variances.

State, in words, the null hypothesis (H0H_0).

[1]
(b)

Calculate the p-value for this test.

[2]
(c)

State whether the result of the test supports the technician's claim. Justify your reasoning.

[2]

Question 2

HardPaper 1 · calculator11 marks
(a)

(a) A botanist is studying a rare species of flower. They record the number of petals (pp) on a sample of these flowers. The data is presented in the frequency table below:

pp

5

6

7

8

9

10

Frequency

3

8

20

15

7

2

Find an unbiased estimate of the population mean number of petals for this rare flower species.

[2]
(b)

(b) Find an unbiased estimate of the population variance of the number of petals for this rare flower species.

[3]
(c)(i)

(c) A botanist suspects that the average number of petals for this rare species is different from the average of 7.0 petals observed in a more common, related species. She sets up a hypothesis test with the null hypothesis H0:μ=7.0H_0: \mu = 7.0.

(i) State the alternative hypothesis.

[1]
(c)(ii)

(ii) Given that all assumptions for this test are satisfied, carry out an appropriate hypothesis test. State and justify your conclusion, using a 10% significance level.

[5]

Question 3

MediumPaper 1 · calculator8 marks
(a)

A bakery owner, "The Daily Loaf", wants to determine if the sales of their signature "Morning Glory Muffin" are consistent throughout the work week. They believe that the number of muffins sold is the same each day.

To test this, they record the number of muffins sold each weekday during a particular week. This data is shown in the table.

Day | Monday | Tuesday | Wednesday | Thursday | Friday

Number of muffins sold | 120 | 135 | 115 | 140 | 150

A goodness of fit test at the 5% significance level is used on this data to determine whether the owner's belief is suitable.

The critical value for the test is 9.49 and the hypotheses are:

H0H_0: The number of Morning Glory Muffins sold is consistent across weekdays.

H1H_1: The number of Morning Glory Muffins sold is not consistent across weekdays.

Find an estimate for how many Morning Glory Muffins the owner expects to sell each day.

[1]
(b)(i)

Write down the degrees of freedom for this test.

[1]
(b)(ii)

Calculate the χ2\chi^2 statistic for this data.

[3]
(b)(iii)

Using the critical value of 9.49, state the conclusion to the test. Give a reason for your answer.

[3]

Question 4

HardPaper 2 · calculator14 marks
(a)

A manufacturer claims that the lifespan of a new batch of LED light bulbs follows a normal distribution with a mean of 50005000 hours and a standard deviation of 10001000 hours. To test this claim, a random sample of 250250 light bulbs was selected, and their lifespans were recorded. The observed frequencies are shown in the table below.

Lifespan (hours)Observed Frequency
x<3500x < 35001010
3500≤x<45003500 \le x < 45005050
4500≤x<55004500 \le x < 55009595
5500≤x<65005500 \le x < 65007070
x≥6500x \ge 65002525

(a) Copy and complete the following table of expected frequencies, assuming the manufacturer's claim is true. Give your answers to two decimal places.

[4]
(b)

Using a χ2\chi^2 distribution at the 5%5\% level of significance, test the hypothesis that the lifespan of the light bulbs follows a normal distribution with mean 50005000 hours and standard deviation 10001000 hours.

You should state the null and alternative hypotheses, clearly show your working for the χ2\chi^2 statistic, and justify your conclusion.

The correct critical value may be selected from the following table, where X(5%)2X^2_{(5\%)} is the value such that P(X>X(5%)2)=0.05P(X > X^2_{(5\%)}) = 0.05.

Degrees of freedomX(5%)2X^2_{(5\%)}
113.843.84
225.995.99
337.827.82
449.499.49
5511.0711.07
[10]

Question 5

MediumPaper 1 · calculator6 marks
(a)

A botanist is investigating the effectiveness of two new fertilizers, 'GrowthBoost' (G) and 'VitaCrop' (V), on the height increase of a specific plant species. They treat 10 plants with GrowthBoost and another 10 plants with VitaCrop, recording the height increase (in cm) over a month. The aim is to determine if GrowthBoost leads to a significantly greater height increase than VitaCrop.

The results obtained are summarized in the following table:

Height increase with GrowthBoost (cm) | 15.0 | 15.5 | 14.8 | 15.3 | 15.6 | 15.1 | 15.4 | 14.9 | 15.7 | 15.2

Height increase with VitaCrop (cm) | 14.9 | 15.4 | 14.7 | 15.2 | 15.1 | 14.8 | 15.0 | 14.6 | 15.3 | 14.7

A t-test is to be performed at the 5% significance level.

(a) Write down the null and alternative hypotheses.

[2]
(b)

(b) Find the p-value for this test.

[2]
(c)

(c) Write down the conclusion to the test. Give a reason for your answer.

[2]

Question 6

HardPaper 1 · calculator9 marks
(a)

(a) A company produces a new board game that includes a four-sided spinner. The spinner is designed to be fair, meaning each side (labelled 1, 2, 3, 4) should have an equal probability of being spun. During a quality control test, the spinner is spun 120120 times. The observed frequencies are:

Number on spinner11223344
Frequency2828323225253535

Find the expected frequencies for each number if the spinner is fair.

[2]
(b)

(b) Write down the number of degrees of freedom for this test.

[1]
(c)

(c) The critical value for a goodness of fit test at the 1%1\% significance level with the appropriate degrees of freedom is 11.34511.345.

Determine the results of a goodness of fit test to find out whether the observed data fits a uniform distribution. Remember to write down the null and alternative hypotheses.

[6]

Question 7

MediumPaper 1 · calculator7 marks
(a)

A confectionary company claims that its bags of "Rainbow Bites" candies contain an equal proportion of six different colors: Red, Orange, Yellow, Green, Blue, and Purple. A consumer group suspects this claim is false. They open a large bag containing 120 candies and count the number of each color. The observed frequencies are shown in the table below:

ColorRedOrangeYellowGreenBluePurple
Frequency182315251722

The consumer group carries out a χ2\chi^2 goodness of fit test at a 5% significance level.

(a) Write down the null and alternative hypotheses.

[1]
(b)

(b) Write down the degrees of freedom.

[1]
(c)

(c) Write down the expected frequency of any color.

[1]
(d)

(d) Find the p-value for the test.

[2]
(e)

(e) State the conclusion of the test. Give a reason for your answer.

[2]

Question 8

HardPaper 2 · calculator16 marks
(a)

A factory produces electronic components. A quality control inspector randomly selects a batch of 33 components and checks for defects. This process is repeated for 100100 batches. The number of defective components in each batch is recorded as shown in the table below.

Number of defects (xx)00112233
Frequency50503535121233

The manager claims that the number of defective components in a batch of 33 follows a binomial distribution B(3,0.2)B(3, 0.2).

Show that the expected frequency for a batch having 00 defective components is 51.251.2.

[3]
(b)

Find the table of expected frequencies for the number of defective components in 100100 batches, assuming the binomial distribution B(3,0.2)B(3, 0.2).

[3]
(c)

State the null and alternative hypotheses for a goodness-of-fit test. Justify any necessary adjustments to the categories for the test and state the degrees of freedom.

[4]
(d)

Calculate the Chi-squared test statistic for this data, using the adjusted categories and assuming a 5%5\% significance level. The critical value for this test is 5.9915.991.

[4]
(e)

State the conclusion for the test, justifying your answer.

[2]

Question 9

MediumPaper 1 · calculator6 marks
(a)

A team of agricultural scientists is investigating the effectiveness of two new fertilizers, BioGrow and NutriBloom, on the growth of a specific plant species. They applied each fertilizer to a separate group of 10 plants and measured the height, in cm, of each plant after a month. The aim is to determine if BioGrow leads to significantly taller plants compared to NutriBloom.

The results obtained are summarized in the following table:

FertilizerSample Size (n)Sample Mean Height (cm)Sample Standard Deviation (cm)
BioGrow1046.95.56
NutriBloom1043.14.82

A t-test is to be performed at the 5% significance level, assuming equal population variances.

(a) Write down the null and alternative hypotheses.

[2]
(b)

(b) Find the p-value for this test.

[2]
(c)

(c) Write down the conclusion to the test. Give a reason for your answer.

[2]

Question 10

HardPaper 2 · calculator21 marks
(a)(i)

The lifespans, tt, of 250 LED light bulbs are recorded in the following table.

Lifespan (hours)Frequency
0≤t<10000 \le t < 100020
1000≤t<15001000 \le t < 150060
1500≤t<20001500 \le t < 200090
2000≤t<25002000 \le t < 250055
2500≤t<30002500 \le t < 300025

This table is used to create a cumulative frequency graph.

Write down the mid-interval value of the class 0≤t<10000 \le t < 1000.

[1]
(a)(ii)

Calculate an estimate of the mean lifespan of the 250 light bulbs.

[3]
(b)

Use the cumulative frequency curve (which would be provided in an exam) to estimate the interquartile range. Assume the lower quartile (Q1Q_1) is 13001300 hours and the upper quartile (Q3Q_3) is 21502150 hours.

[3]
(c)

A light bulb from the data set had a lifespan of 34003400 hours.

Use your answer to part (b) to estimate whether this light bulb's lifespan is an outlier for this data. Justify your answer.

[3]
(d)

It is believed that the lifespans of these LED light bulbs follow a normal distribution with mean 17401740 hours and standard deviation 450450 hours.

It is decided to perform a χ2\chi^2 goodness of fit test on the data to determine whether this sample of 250 light bulbs could have plausibly been drawn from an underlying distribution N(1740,4502)N(1740, 450^2).

Write down the null and the alternative hypotheses for the test.

[2]
(e)(i)

As part of the test, the following table is created.

Lifespan of light bulb (hours)Observed frequencyExpected frequency
t<1000t < 10002014.0
1000≤t<15001000 \le t < 15006060.1
1500≤t<20001500 \le t < 200090a
2000≤t<25002000 \le t < 25005560.1
t≥2500t \ge 250025b

Find the value of aa and the value of bb. Give your answers to one decimal place.

[5]
(e)(ii)

Hence, perform the test to a 5% significance level, clearly stating the conclusion in context.

[4]

Question 11

MediumPaper 1 · calculator8 marks
(a)

A new trendy cafe is trying to understand customer behaviour regarding their signature 'Aurora Brew' drink. They hypothesize that the number of 'Aurora Brew' orders is evenly distributed throughout the day's main periods.

To test this model, they record the number of 'Aurora Brew' orders during four distinct time slots over a typical day. This data is shown in the table below.

Time SlotNumber of Orders
Morning (8 AM - 11 AM)45
Lunch (11 AM - 2 PM)62
Afternoon (2 PM - 5 PM)38
Evening (5 PM - 8 PM)55

(a) Find an estimate for how many 'Aurora Brew' orders the cafe expects to receive during each time slot, according to their model.

[1]
(b)(i)

(b) A goodness of fit test at the 5% significance level is used on this data to determine whether the cafe's model is suitable. The critical value for the test is 7.815.

(i) State the null and alternative hypotheses for this test.

[2]
(b)(ii)

(ii) Write down the degrees of freedom for this test.

[1]
(b)(iii)

(iii) Calculate the χ2\chi^2 test statistic and write down the conclusion to the test. Give a reason for your answer.

[4]

Question 12

HardPaper 2 · calculator25 marks
(a)(i)

The lifespan of a new type of LED bulb, in hours, can be modelled by a normal distribution with a mean of 1200012000 hours and a standard deviation of 800800 hours.

A randomly selected LED bulb is chosen.

(a) Calculate the probability that its lifespan is

(i) less than 1100011000 hours.

[4]
(a)(ii)

(ii) between 1150011500 hours and 1250012500 hours.

[4]
(b)

(b) 15%15\% of LED bulbs have a lifespan of more than hh hours.

Calculate the value of hh.

[2]
(c)(i)

A manufacturer wants to determine if a sample of 250250 LED bulbs from a new production batch could have been chosen from a normally distributed population with a mean of 1200012000 hours and a standard deviation of 800800 hours.

They perform a χ2\chi^2 goodness of fit test at the 5%5\% significance level. They begin by creating the following frequency table:

Lifespan, hh (hours)Observed frequencyExpected frequency
h≤11000h \le 11000151526.41226.412
11000<h≤1200011000 < h \le 12000100100a
12000<h≤1300012000 < h \le 13000110110b
h>13000h > 13000252526.41226.412

(c) Calculate, correct to four significant figures, the value of

(i) a.

[2]
(c)(ii)

(ii) b.

[2]
(d)

The hypotheses for the manufacturer's test are:

H0H_0: The lifespans of the LED bulbs are drawn from a normally distributed population with mean 1200012000 hours and standard deviation 800800 hours.

H1H_1: The lifespans of the LED bulbs are not drawn from a normally distributed population with mean 1200012000 hours and standard deviation 800800 hours.

(d) Write down the degrees of freedom for this test.

The critical value for this test is 7.8157.815.

[1]
(e)

(e) Perform the χ2\chi^2 goodness of fit test and state your conclusion, justifying your reasoning.

[4]
(f)

A competitor claims that their new 'Brand A' LED bulbs last longer on average than the manufacturer's 'Brand B' LED bulbs.

Random samples of 1212 Brand A bulbs and 1010 Brand B bulbs are chosen, and their lifespans (in hours) are measured:

Lifespans of Brand A bulbs (hours):

125001250011800118001320013200121001210012900129001190011900127001270012300123001310013100120001200012600126001220012200

Lifespans of Brand B bulbs (hours):

1210012100115001150012800128001190011900124001240011700117001200012000122001220011600116001230012300

The competitor performs a t-test at the 5%5\% significance level. It is assumed that the populations are normally distributed and have equal variances.

(f) Write down the null and alternative hypotheses for this test.

[2]
(g)

(g) Perform the t-test and state the conclusion, justifying your reasoning.

[4]

Question 13

MediumPaper 1 · calculator6 marks
(a)

A team of agricultural scientists is investigating the effectiveness of two new organic fertilizers, 'TerraGrow' and 'VitaBloom', on the growth of a specific type of leafy green vegetable. They hypothesize that one fertilizer might lead to significantly taller plants.

They conducted an experiment where 14 identical plant seedlings were randomly divided into two independent groups. One group was treated with TerraGrow, and the other with VitaBloom. After four weeks, the height of each plant, in cm, was measured.

The data collected is shown in the following table:

TerraGrow (cm)VitaBloom (cm)
12.511.8
14.112.0
13.013.5
15.212.2
13.811.5
14.513.0
12.912.8

At the 5% level of significance, a t-test was used to compare the mean plant heights produced by the two fertilizers. Each data set is assumed to be normally distributed, and the population variances are assumed to be the same.

Let μT\mu_T be the population mean height for plants treated with TerraGrow and μV\mu_V be the population mean height for plants treated with VitaBloom. The null hypothesis for this test is H0:μT−μV=0H_0: \mu_T - \mu_V = 0.

State the alternative hypothesis.

[1]
(b)

Calculate the p-value for this test.

[2]
(c)(i)

State the conclusion of the test. Justify your answer.

[2]
(c)(ii)

State what your conclusion means in context.

[1]

Question 14

HardPaper 2 · calculator21 marks
(a)

A company manufactures specialized medical devices. Each device consists of a main circuit board and a protective casing. The weight of the circuit board, CC, is normally distributed with a mean of 120120 g and a standard deviation of 55 g. The weight of the protective casing, PP, is normally distributed with a mean of 3030 g and a standard deviation of 22 g. The weights of the circuit board and the casing are independent.

Find the probability that a randomly chosen complete device has a total weight of less than 145145 g.

[5]
(b)

A batch of 1010 such devices is to be packed into a container. The container has a maximum weight capacity of 15101510 g. The weight of each device is independent.

Find the probability that the total weight of the 1010 devices is greater than the capacity of the container.

[4]
(c)(i)

The company sources a critical microchip component from two different suppliers, Supplier X and Supplier Y. An engineer claims that microchips from Supplier Y have a lower response time than those from Supplier X. To test this claim, a random sample is taken from each supplier.

The eight microchips in the sample from Supplier X have response times, in milliseconds (ms), of:

19.65,19.65,19.79,20.75,20.97,21.15,22.28,22.3719.65, 19.65, 19.79, 20.75, 20.97, 21.15, 22.28, 22.37.

Find the

(i) mean response time for the sample from Supplier X.

[3]
(c)(ii)

Find the

(ii) unbiased estimate of the population variance for the sample from Supplier X.

[3]
(d)

The seven microchips in the sample from Supplier Y have a mean response time of 19.519.5 ms and an unbiased estimate of the population standard deviation (sn−1s_{n-1}) of 1.051.05 ms.

Perform a suitable test, at the 5%5\% significance level, to test the engineer's claim that microchips from Supplier Y have a lower response time than those from Supplier X. You may assume the response times of microchips from each supplier are normally distributed with equal population variance.

[6]

Question 15

MediumPaper 1 · calculator6 marks
(a)

Dr. Anya Sharma is investigating the effectiveness of a new 'active recall' study technique compared to a 'traditional review' method for improving student performance on a challenging physics exam. She believes that students using the active recall technique will achieve higher mean scores.

Dr. Sharma conducts an experiment with two random samples of students. The results are summarized below:

  • Active Recall Group (Group A): Sample size nA=15n_A = 15, mean score xˉA=78\bar{x}_A = 78, sample standard deviation sA=9s_A = 9.
  • Traditional Review Group (Group T): Sample size nT=12n_T = 12, mean score xˉT=72\bar{x}_T = 72, sample standard deviation sT=8s_T = 8.

Dr. Sharma performs a one-tailed t-test at a 5% level of significance. It is assumed that the exam scores are normally distributed and the samples have equal variances.

State the null and alternative hypotheses for this test.

[2]
(b)

Calculate the p-value for this test.

[2]
(c)

State the conclusion of the test in the context of the question. Justify your answer.

[2]

Question 16

HardPaper 2 · calculator21 marks
(a)(i)

(a) The scores, ss, of 200 students on a mathematics test are recorded in the following table.

Score (ss)Frequency
20≤s<4020 \le s < 4015
40≤s<6040 \le s < 6035
60≤s<8060 \le s < 8060
80≤s<10080 \le s < 10050
100≤s<120100 \le s < 12030
120≤s<140120 \le s < 14010

(i) Write down the mid-interval value of 60≤s<8060 \le s < 80.

[3]
(a)(ii)

(ii) Calculate an estimate of the mean score of the 200 students.

[3]
(b)

(b) The data from this table is used to create a cumulative frequency graph. From this graph, the first quartile (Q1Q_1) is estimated to be 6060 and the third quartile (Q3Q_3) is estimated to be 9696.

Use these values to estimate the interquartile range (IQR).

[2]
(c)

(c) A student, Elara, scored 155155 on the test.

Use your answer to part (b) to estimate whether Elara's score is an outlier for this data. Justify your answer.

[3]
(d)

(d) It is believed that the scores of students on this mathematics test follow a normal distribution with mean 77.577.5 and standard deviation 2020.

It is decided to perform a χ2\chi^2 goodness of fit test on the data to determine whether this sample of 200 students could have plausibly been drawn from an underlying distribution N(77.5,202)N(77.5, 20^2).

Write down the null and the alternative hypotheses for the test.

[2]
(e)

(e) As part of the test, the following table is created, where some categories have been combined to ensure expected frequencies are not too low.

Score (ss)Observed FrequencyExpected Frequency
s<40s < 40156.08
40≤s<6040 \le s < 603532.08
60≤s<8060 \le s < 8060a
80≤s<10080 \le s < 1005063.99
s≥100s \ge 10040b

(i) Find the value of aa and the value of bb.

(ii) Hence, perform the test to a 5% significance level, clearly stating the conclusion in context.

[8]

Question 17

MediumPaper 1 · calculator6 marks
(a)

[Maximum mark: 6]

A university is investigating the relationship between student engagement in extracurricular activities and their academic performance. A random sample of 470 students was selected, and their engagement level (Low, Medium, High) and academic performance (Below Average, Average, Above Average) were recorded. The data is summarized in the following table.

Low EngagementMedium EngagementHigh EngagementTotal
Below Average45301590
Average609070220
Above Average255085160
Total130170170470

An item of food is chosen at random from these 500.

(a) Find the probability that a randomly chosen student has 'Below Average' academic performance, given that they have 'Low' engagement.

[2]
(b)

A χ2\chi^2 test at the 5% significance level is carried out to determine if there is a significant relationship between student engagement and academic performance.

The critical value for this test is 9.488.

The hypotheses for this test are:

H0H_0: Student engagement in extracurricular activities and academic performance are independent.

H1H_1: Student engagement in extracurricular activities and academic performance are not independent.

(b) Find the χ2\chi^2 statistic.

[2]
(c)

(c) State, with justification, the conclusion for this test.

[2]

Question 18

HardPaper 3 · calculator27 marks
(a)(i)

(a) Mr. Lee, the owner of "Sweet Delights" bakery, recorded the number of Mooncakes sold each day for a sample of 3030 days. The results are shown in the table below.

Number of Mooncakes soldFrequency
01
12
24
35
47
56
63
72

(a.i) Find the mean and variance for this sample data.

[2]
(a)(ii)

(a.ii) Hence, state why Mr. Lee might believe that the daily sales of Mooncakes follow a Poisson distribution.

[1]
(b)

(b) State one assumption that Mr. Lee needs to make about the sales of Mooncakes to support his belief that it follows a Poisson distribution.

[1]
(c)

(c) Mr. Lee knows from his historic sales records that the bakery sells an average of 3.93.9 Mooncakes each day. The following table shows the expected frequency of Mooncakes sold each day during a 100100-day period, assuming a Poisson distribution with mean 3.93.9.

Number of Mooncakes sold<11234567≥8\ge 8
Expected frequencya7.8947.89415.39415.39420.01220.012b15.21915.2199.8939.8935.5125.512c

Find the value of a, of b, and of c. Give your answers to 3 decimal places.

[5]
(d)(i)

(d) Mr. Lee decides to carry out a χ2\chi^2 goodness of fit test at the 5%5\% significance level to see whether the daily sales of Mooncakes follow a Poisson distribution with mean 3.93.9. He collects observed frequencies for 100100 days, which are given in the table below.

Number of Mooncakes sold<223456≥7\ge 7
Observed frequency12132317121013
Expected frequency9.9189.91815.39415.39420.01220.01219.51219.51215.21915.2199.8939.89310.05210.052

(d.i) Write down the number of degrees of freedom for his test.

[1]
(d)(ii)

(d.ii) Perform the χ2\chi^2 goodness of fit test and state, with reason, a conclusion.

[7]
(e)(i)

(e) Mr. Lee claims that a new social media advertising campaign, costing 250250 THB per day, will increase the number of Mooncakes sold. However, his business partner, Ms. Chen, claims that the advertising will not increase the bakery's overall profit.

Ms. Chen agrees to run the campaign for the next 4040 days. During that time, Mr. Lee records that the bakery sells a total of 180180 Mooncakes, with a profit of 4545 THB on each Mooncake sold.

Mr. Lee wants to carry out an appropriate hypothesis test to determine whether the number of Mooncakes sold during the 4040 days increased when compared with the historic sales records (mean 3.93.9 Mooncakes per day).

By finding a critical value, perform this test at a 5%5\% significance level.

[6]
(f)

(f) Hence state the probability of a Type I error for this test.

[1]
(g)

(g) By considering the claims of both Mr. Lee and Ms. Chen, explain whether the advertising campaign was beneficial to the bakery.

[3]

Question 19

MediumPaper 1 · calculator7 marks
(a)

A tech company launched two new smartphone models, "Voyager" and "Explorer". They collected customer satisfaction scores (out of 100) from a large sample of users for both models. The results are summarized in the following box and whisker diagram.

Box and whisker diagram comparing customer satisfaction scores for Model Voyager and Model Explorer. X-axis from 50 to 100. Model Voyager: min 50, Q1 60, median 70, Q3 90, max 100. Model Explorer: min 55, Q1 70, median 80, Q3 90, max 100.

Identify which two of the following statements must be true according to the box and whisker diagram. Indicate your choices by placing tick marks in the second column of the following table.

Statement | True (✓)

---|---

The satisfaction scores for Model Voyager are normally distributed. |

A higher percentage of customers gave a score less than 70 for Model Voyager than for Model Explorer. |

A higher percentage of customers gave a score greater than 90 for Model Explorer than for Model Voyager. |

The interquartile range for Model Explorer is less than the interquartile range for Model Voyager. |

[2]
(b)

A product manager believes there is no significant difference in the average customer satisfaction scores between the two models. She plans to conduct a t-test at the 10% significance level. Write down the null and alternative hypotheses for her test.

[2]
(c)

The t-test yielded a p-value of 0.0783. Find the p-value for her test.

[1]
(d)

Write down the conclusion to the test. Give a reason for your answer.

[2]

Question 20

MediumPaper 1 · calculator7 marks
(a)

A candy manufacturer claims that a bag of their mixed candies contains four different colors (Red, Green, Blue, Yellow) in specific proportions: 20% Red, 30% Green, 25% Blue, and 25% Yellow. A consumer group suspects this claim is inaccurate. They randomly select a large bag and count the number of candies of each color, recording the observed frequencies in the following table:

ColorRedGreenBlueYellowTotal
Observed Frequency35654852200

The consumer group decides to carry out a χ2\chi^2 goodness of fit test at a 5% significance level to investigate the manufacturer's claim.

Write down the null and alternative hypotheses for this test.

[1]
(b)

Write down the degrees of freedom for this test.

[1]
(c)

Write down the expected frequency of Red candies.

[1]
(d)

Find the p-value for the test.

[2]
(e)

State the conclusion of the test. Give a reason for your answer.

[2]

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What does Chi GOF, Independence, T-test (Null / Alternative hypothesis, S.L. / p-values) cover in IB Maths AI?

Hypothesis Test: Uses sample data to test a statement about a population. Null Hypothesis (H_0): Assumes no difference, no change, no association, or data follows a specific distribution. Alternative Hypothesis (H_1): Opposes H_0, claims a difference, association, or data does not follow a distribution.

Is Chi GOF, Independence, T-test (Null / Alternative hypothesis, S.L. / p-values) SL or HL?

Both. SL and HL students study Chi GOF, Independence, T-test (Null / Alternative hypothesis, S.L. / p-values) to the same depth.

How do I revise Chi GOF, Independence, T-test (Null / Alternative hypothesis, S.L. / p-values) for IB Maths AI?

Start from the core idea: hypothesis Test: Uses sample data to test a statement about a population. In the exam: the longest question in the SL statistics section, usually five to eight marks across several parts: state the hypotheses, give the degrees of freedom, produce the p-value, compare it to the significance level, then state the conclusion in context. The conclusion mark is only earned by referring back to the situation, not by writing "reject H_0". Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

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