Free worked solutions · 2026 Edition

IB Maths AI SL Chapter 12: Statistical tests worked solutions

Worked solutions to the practice questions in Chapter 12 of Photon Guidebook: Mathematics Applications and Interpretation SL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[2 marks]

A curator asks whether preferred gallery theme is independent of membership type.

State the null and alternative hypotheses for a chi-squared test of independence.

[2]

Show solutionHide solution

H_0: preferred theme and membership type are independent.

H_1: preferred theme and membership type are not independent.

Final answer

H_0: preferred theme and membership type are independent; H_1: they are not independent

Question 2

[2 marks]

A test gives p=0.037. The chosen significance level is 5%.

State the statistical decision and justify it numerically.

[2]

Show solutionHide solution

Reject H_0.

0.037<0.05.

Final answer

reject H_0 (0.037<0.05)

Question 3

[3 marks]

A contingency table has row totals 52 and 68, column totals 45 and 75, and grand total 120.

(a) Find the expected count for row 2, column 1 under independence.

[2]

(b) State the degrees of freedom.

[1]

Show solutionHide solution

(a) E=68(45)/120.

E=25.5.

(b) (2-1)(2-1)=1.

Final answer

(a) 25.5
(b) 1

Question 4

[3 marks]

For one cell in a chi-squared calculation, the observed count is 32 and the expected count is 25.

(a) Calculate the contribution (O-E)^2/E to the test statistic.

[2]

(b) Explain why a count below its expectation can also make a positive contribution.

[1]

Show solutionHide solution

(a) (32-25)^2/25.

49/25=1.96.

(b) The difference is squared, so either direction of discrepancy contributes a non-negative amount.

Final answer

(a) 1.96
(b) the difference is squared

Question 5

[6 marks]

Independent randomly selected records give the observed frequencies below. Test independence of visitor status and pass colour at the 5% level. All expected counts exceed 5.

\begin{array}{c|rrr}&Red&Blue&Gold\\New&19&31&24\\Returning&32&18&26\end{array}

Use a GDC chi-squared test without continuity correction.

(a) Find the degrees of freedom and the chi-squared statistic.

[3]

(b) Find the p-value and state a contextual conclusion.

[3]

Show solutionHide solution

(a) df=2.

Enter the observed table in the GDC chi-squared test.

\chi^2=6.82.

(b) p=0.0331.

Reject H_0. Compare p with 0.05.

There is sufficient evidence that visitor status and pass colour are associated.

Final answer

(a) df =2, \chi^2=6.82
(b) p=0.0331; reject H_0; sufficient evidence that visitor status and pass colour are associated

Question 6

[3 marks]

A goodness-of-fit test uses five categories with probabilities specified in advance. No parameters are estimated from the sample.

(a) State the degrees of freedom.

[1]

(b) For a sample of 160, find the expected frequency of a category with probability 0.15.

[2]

Show solutionHide solution

(a) df=5-1=4.

(b) E=160(0.15).

E=24.

Final answer

(a) 4
(b) 24

Question 7

[8 marks]

A random sample of independent outcomes tests the claim that visitors choose each of five paths with equal probability. The category probabilities were specified before collecting data.

\begin{array}{c|rrrrr}&I&II&III&IV&V\\Observed&22&31&17&25&30\\Probability&0.2&0.2&0.2&0.2&0.2\end{array}

Test at the 1% level. No parameters are estimated.

(a) Find the expected frequencies and degrees of freedom.

[3]

(b) Calculate the chi-squared statistic and p-value.

[3]

(c) State the decision in context.

[2]

Show solutionHide solution

(a) E_i=n p_i, with n=125.

Expected counts: 25, 25, 25, 25, 25.

df=4.

(b) Use \sum(O-E)^2/E or the GDC goodness-of-fit test.

\chi^2=5.36.

p=0.252.

(c) Do not reject H_0. Since p >0.01.

There is insufficient evidence against the claim that visitors choose each of five paths with equal probability.

Final answer

(a) 25 for each path; df =4
(b) \chi^2=5.36, p=0.252
(c) do not reject H_0; insufficient evidence against equal popularity

Question 8

[3 marks]

A pooled two-sample t-test is proposed for two small samples of numerical durability scores.

State three assumptions that should be satisfied.

[3]

Show solutionHide solution

The observations are randomly sampled and independent within and between groups.

Each population distribution is approximately normal for these small samples.

The population variances are equal for pooling.

Final answer

independent random samples; normal populations; equal variances

Question 9

[4 marks]

A researcher compares mean drying time for independently sampled coatings C and D without specifying a direction. Population variances are assumed equal.

(a) State the hypotheses.

[2]

(b) Identify the alternative option and pooled setting on a GDC two-sample t-test.

[2]

Show solutionHide solution

(a) H_0:\mu_C=\mu_D.

H_1:\mu_C\ne\mu_D.

(b) Select \mu_1\ne\mu_2.

Select pooled: yes, reflecting the assumed equal population variances.

Final answer

(a) H_0:\mu_C=\mu_D, H_1:\mu_C\neq\mu_D
(b) alternative \neq, pooled yes

Question 10

[3 marks]

Two independent groups use packaging machines A and B. A researcher expects A to have a greater population mean fill mass. Let \mu_A and \mu_B denote these means.

(a) State H_0 and H_1 for a one-tailed two-sample t-test.

[2]

(b) Explain why a two-sample test is appropriate.

[1]

Show solutionHide solution

(a) H_0:\mu_A=\mu_B.

H_1:\mu_A>\mu_B.

(b) The question compares population means using two independent samples.

Final answer

(a) H_0:\mu_A=\mu_B, H_1:\mu_A>\mu_B
(b) two independent samples are compared

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 11

[6 marks]

Moana drops phone cases of two designs from increasing heights and compares the mean height (in cm) at which they crack. She tests at the 5% level whether the mean heights differ, and the GDC gives p=0.029.

(a) Explain, in the context of this test, what the value 0.029 measures.

[2]

(b) State the conclusion of the test, in context.

[2]

(c) Moana says, “There is a 2.9% chance that the two designs have the same mean.” Explain why this statement is not correct.

[2]

Show solutionHide solution

(a) It is the probability of obtaining sample means at least as far apart as those observed R1 [1], assuming that the two designs have the same mean cracking height (H_0 true).

R1[1]

(b) 0.029<0.05, so reject H_0. M1 [1] There is sufficient evidence, at the 5% level, that the mean cracking heights of the two designs differ.

A1[1]

(c) The p-value is calculated assuming H_0 is true; it is a probability about the data, not the probability that H_0 is true. R1 [1] A hypothesis is either true or false, so a test cannot assign it a probability of 2.9%.

R1[1]

Final answer

(a) the probability of sample means at least this far apart if H_0 is true
(b) 0.029<0.05: reject H_0; sufficient evidence that the mean heights differ
(c) p is not the probability that H_0 is true

Question 12

[6 marks]

A school tests, at the 5% level, whether students taught with a new method have a higher mean score than students taught with the old method. The p-value is 0.021. Four students write conclusions.

A: “H_0 is true.” B: “There is sufficient evidence, at the 5% level, that the mean score is higher with the new method.”

C: “Reject H_0.” D: “The new method is proved to be better.”

(a) State which conclusion is the best answer.

[1]

(b) For each of the other three conclusions, explain what is wrong with it.

[3]

(c) Another school finds p=0.21 from the same kind of test. Write down a correct conclusion in context.

[2]

Show solutionHide solution

(a) Conclusion B.

A1[1]

(b) A: p<\alpha, so H_0 is rejected, not supported; and a test never states that H_0 is true.

R1[1]

C: this is the correct decision but it is not interpreted in context; the question asks about the mean scores.

R1[1]

D: a test gives evidence at a stated significance level; it does not prove anything.

R1[1]

(c) 0.21>0.05, so do not reject H_0. M1 [1] There is insufficient evidence, at the 5% level, that the mean score is higher with the new method.

A1[1]

Final answer

(a) B
(b) A: H_0 is rejected; C: no context; D: a test proves nothing
(c) 0.21>0.05: insufficient evidence that the new method gives a higher mean score

Question 13

[6 marks]

A trial compares the mean recovery times of patients given two diets. The test gives p=0.18 and Dr Lopez concludes, “The two diets are equally effective.”

(a) Explain why this conclusion is not justified by the test.

[2]

(b) Give a reason, other than H_0 being true, why a test might fail to reject H_0 even though the two diets do differ.

[2]

(c) Write down a correct conclusion at the 5% level, in context.

[2]

Show solutionHide solution

(a) Failing to reject H_0 only means there is insufficient evidence of a difference. R1 [1] It does not show that H_0 is true: a real difference may exist but the data did not reveal it.

R1[1]

(b) For example, the samples may be too small, or the variability of recovery times within each diet group may be large, so that a real difference in means is hidden.

R2[2]

(c) 0.18>0.05, so do not reject H_0. M1 [1] There is insufficient evidence, at the 5% level, that the mean recovery times differ between the two diets.

A1[1]

Final answer

(a) not rejecting H_0 does not show it is true
(b) small samples or large spread
(c) insufficient evidence of a difference in recovery times

Question 14

[8 marks]

Sana expects that a new bus route has a shorter mean travel time than the old route. She times 18 trips on the new route (mean 31.4 minutes, standard deviation 4.1) and 20 trips on the old route (mean 29.8 minutes, standard deviation 3.8).

(a) Write down the null and alternative hypotheses for Sana's test, where \mu_1 and \mu_2 are the mean travel times on the new and old routes.

[2]

(b) Without calculating, state whether the sample means support H_1.

[1]

(c) Sana runs a pooled test with the alternative set to match her expectation. Use your GDC to find the p-value.

[2]

(d) State the conclusion at the 5% level, in context.

[1]

(e) Explain why the p-value in part (c) is greater than 0.5.

[2]

Show solutionHide solution

(a) H_0: \mu_1=\mu_2 A1 [1] H_1: \mu_1<\mu_2

A1[1]

(b) No: the new-route sample mean (31.4) is greater than the old-route mean (29.8), the opposite of H_1.

A1[1]

(c) GDC: 2-SampTTest from statistics, \bar x_1=31.4, s_1=4.1, n_1=18, \bar x_2=29.8, s_2=3.8, n_2=20, \mu_1<\mu_2, Pooled: Yes. (M1) [1] t=1.25, df=36, p=0.890

A1[1]

(d) 0.890>0.05, so do not reject H_0: there is insufficient evidence, at the 5% level, that the mean journey time on the new route is shorter.

A1[1]

(e) t>0, so the observed statistic lies in the upper tail, while the test asks about the lower tail. R1 [1] The lower-tail area is therefore more than half the distribution: it equals 1-\tfrac12(0.220)=0.890.

R1[1]

Final answer

(a) H_0:\mu_1=\mu_2, H_1:\mu_1<\mu_2
(b) no
(c) p=0.890
(d) do not reject H_0
(e) t>0, so the sample difference lies in the other tail

Question 15

[6 marks]

Kenji compares the mean sprint times of athletes using two types of shoe. His first test, with H_1: \mu_1\neq\mu_2, gives p=0.072. Wanting a significant result, he repeats the calculation with H_1: \mu_1<\mu_2, obtains p=0.036, and reports that he has “rejected H_0 at the 5% level”.

(a) State when a one-tailed test is appropriate.

[2]

(b) Explain why Kenji's procedure is not valid.

[2]

(c) State what Kenji should have done before collecting the data.

[2]

Show solutionHide solution

(a) When the claim or research question concerns a difference in one specific direction only (e.g. “greater than” or “less than”).

R2[2]

(b) He chose the direction of H_1 after seeing which way the data pointed, only to obtain a smaller p-value. R1 [1] This makes the stated 5% risk of a false rejection meaningless.

R1[1]

(c) He should have decided, from the research question alone, whether to test for a difference or for a difference in a stated direction, and fixed H_1 and \alpha before collecting the data.

R2[2]

Final answer

(a) when the claim has one stated direction
(b) he chose the tail after seeing the data
(c) fix H_1 and \alpha before collecting the data

Question 16

[6 marks]

Elias plans to compare two population means using a pooled two-sample t-test on the GDC.

(a) Three of the following five statements are conditions for the test to be valid. State which three.

(i) The two samples are independent random samples. (ii) Both populations are approximately normally distributed.

(iii) Both populations have the same variance. (iv) The two samples must be the same size.

(v) The two sample means must be equal.

[2]

(b) The samples have sizes 9 and 14. Find the degrees of freedom.

[1]

(c) Ana measures the grip strength of 10 climbers before and after a training course, and proposes to compare the two sets of ten measurements with a two-sample test. Explain why this is not appropriate.

[1]

(d) In another study the sample standard deviations are 2.1 and 6.8. State which condition in part (a) is in doubt.

[2]

Show solutionHide solution

(a) (i), (ii) and (iii).

A2[2]

(b) df=9+14-2=21

A1[1]

(c) The two sets of measurements come from the same 10 climbers, so the samples are not independent.

R1[1]

(d) Condition (iii), equal population variances. A1 [1] The sample standard deviations differ by a factor of more than 3, which suggests that the population variances are not equal.

R1[1]

Final answer

(a) (i), (ii), (iii)
(b) 21
(c) the samples are not independent
(d) equal variances

Question 17

[7 marks]

Two bakeries supply loaves labelled 250 g. Roza weighs random samples of loaves from Bakery 1 and Bakery 2 and tests, at the 5% level, whether the mean masses differ. Her GDC shows

\begin{array}{l} \text{2-SampTTest} \mu_1\neq\mu_2 \text{Pooled: Yes} \\ \text{t = 2.28 df = 28 p = 0.0303} \\ \bar x_1 \text{= 250.8 Sx1 = 3.1 n1 = 14} \\ \bar x_2 \text{= 248.3 Sx2 = 2.9 n2 = 16} \end{array}

(a) Write down the null and alternative hypotheses.

[2]

(b) Show how the value of df is obtained from the sample sizes.

[1]

(c) State the conclusion of the test, in context,

(i) at the 5% level, (ii) at the 1% level.

[4]

Show solutionHide solution

(a) H_0: \mu_1=\mu_2 A1 [1] H_1: \mu_1\neq\mu_2 (\mu_1,\mu_2 the mean masses of loaves from Bakery 1 and Bakery 2)

A1[1]

(b) df=n_1+n_2-2=14+16-2=28

A1[1]

(c) (i) p=0.0303 and 0.0303<0.05, so reject H_0. M1 [1] There is sufficient evidence, at the 5% level, that the mean masses of loaves from the two bakeries differ.

A1[1]

(ii) 0.0303>0.01, so do not reject H_0. M1 [1] There is insufficient evidence, at the 1% level, that the mean masses differ.

A1[1]

Final answer

(a) H_0:\mu_1=\mu_2, H_1:\mu_1\neq\mu_2
(b) 14+16-2=28
(c) (i) reject H_0 at 5% (ii) do not reject H_0 at 1%

Question 18

[8 marks]

A courier company tries a new delivery route. The times (minutes) of 10 random deliveries on each route are

New route	41	38	45	36	40
Old route	43	46	40	44	47
New route	43	39	44	37	42
Old route	42	44	45	41	43

Assume the times are normally distributed with equal variances. The manager tests, at the 5% level, whether the mean time on the new route is shorter.

(a) Write down the null and alternative hypotheses.

[2]

(b) Use your GDC to find the p-value of the test.

[3]

(c) Write down the conclusion at the 5% level, in context.

[2]

(d) State whether H_0 would be rejected at the 1% level.

[1]

Show solutionHide solution

(a) H_0: \mu_1=\mu_2 A1 [1] H_1: \mu_1<\mu_2 (\mu_1,\mu_2 the mean times on the new and old routes)

A1[1]

(b) GDC: Stats \to Tests \to 2-Sample t-Test, Data input, \mu_1<\mu_2, Pooled: Yes.

M1[1]

t=-2.55, df=18 A1 [1] p=0.0101

A1[1]

(c) 0.0101<0.05, so reject H_0. M1 [1] There is sufficient evidence, at the 5% level, that the mean delivery time is shorter on the new route.

A1[1]

(d) 0.0101>0.01, so H_0 would not be rejected at the 1% level.

A1[1]

Final answer

(a) H_0:\mu_1=\mu_2, H_1:\mu_1<\mu_2
(b) p=0.0101
(c) reject H_0
(d) H_0 is not rejected at 1%

Question 19

[8 marks]

Imani compares the typing speeds (words per minute) of two classes. Class 1: n=16, mean 48.3, standard deviation 6.1. Class 2: n=15, mean 44.2, standard deviation 5.4. Assume the speeds are normally distributed with equal variances. Use a pooled t-test at the 5% level.

(a) Imani first tests whether the mean speeds of the classes differ. Write down the hypotheses, the p-value and the decision.

[3]

(b) Her teacher had already decided to test whether Class 1 types faster. Write down the hypotheses, the p-value and the decision.

[3]

(c) Explain why the two tests lead to different decisions, and state which test is appropriate for the teacher's claim.

[2]

Show solutionHide solution

(a) H_0: \mu_1=\mu_2, H_1: \mu_1\neq\mu_2 A1 [1] p=0.0577 A1 [1] 0.0577>0.05, do not reject H_0

A1[1]

(b) H_0: \mu_1=\mu_2, H_1: \mu_1>\mu_2 A1 [1] p=0.0289 A1 [1] 0.0289<0.05, reject H_0

A1[1]

(c) The one-tailed p-value is half the two-tailed p-value (the sample difference is in the direction of H_1), and 0.05 lies between the two values. R1 [1] The one-tailed test is appropriate, because the claim is directional and was fixed in advance.

R1[1]

Final answer

(a) p=0.0577: do not reject H_0
(b) p=0.0289: reject H_0
(c) the claim is one-tailed, so the one-tailed test fits

Question 20

[10 marks]

Tunde records the battery life (hours) of nine phones of Model A and nine of Model B:

Model A	24.1	25.3	23.8	26.0	24.7
Model B	22.9	23.8	22.4	24.1	23.2
Model A	25.5	24.2	25.8	38.9
Model B	23.9	22.7	24.4	23.5

He tests, at the 5% level, whether the mean battery lives differ, assuming normal populations with equal variances.

(a) Find the p-value using all the data and state the conclusion, in context.

[3]

(b) Find the quartiles of the Model A data and use the rule “outlier if more than 1.5\timesIQR above Q_3” to show that 38.9 is an outlier.

[3]

(c) Tunde finds that the reading 38.9 was caused by a faulty timer, and removes it. Find the new p-value and state the new conclusion.

[2]

(d) Explain why the two conclusions differ and say which Tunde should report.

[2]

Show solutionHide solution

(a) H_0: \mu_1=\mu_2, H_1: \mu_1\neq\mu_2. GDC (Data input, Pooled: Yes): t=1.91, df=16, p=0.0737

A1[1]

0.0737>0.05, so do not reject H_0 M1 [1]: insufficient evidence, at the 5% level, that the mean battery lives differ.

A1[1]

(b) Ordered: 23.8,\,24.1,\,24.2,\,24.7,\,25.3,\,25.5,\,25.8,\,26.0,\,38.9. Q_1=24.15, Q_3=25.9

A1[1]

IQR=25.9-24.15=1.75, upper fence =Q_3+1.5\times IQR=28.525 M1 [1] 38.9>28.525, so 38.9 is an outlier.

R1[1]

(c) Without 38.9 (n_1=8): t=4.06, df=15, p=0.00104 A1 [1] 0.00104<0.05, so reject H_0: there is sufficient evidence, at the 5% level, that the mean battery lives differ.

A1[1]

(d) The outlier greatly inflates the Model A sample standard deviation (4.72 against 0.84 without it) and so reduces t and raises p, hiding the real difference between the means. R1 [1] The faulty reading is not a genuine battery life, so Tunde should report the analysis without it (and say that he removed it and why).

R1[1]

Final answer

(a) p=0.0737: do not reject H_0
(b) Q_1=24.15, Q_3=25.9, fence 28.525
(c) p=0.00104: reject H_0
(d) the outlier inflates the spread

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 21

[9 marks]

A conservation group studies whether the number of returning storks to four nesting sites fits an expected ratio of 4:3:2:1 predicted by an old habitat model. One spring they count the returning pairs at each site.

Site	A	B	C	D
Observed pairs	46	38	19	17

They test, at the 5% significance level, whether the data fit the 4:3:2:1 model using a \chi^2 goodness-of-fit test.

(a) State the null and alternative hypotheses.

[2]

(b) Show that the expected number of pairs at site A is 48.

[2]

(c) Find the expected number of pairs at each of sites B, C and D.

[2]

(d) Using your GDC, find the \chi^2 statistic and the p-value.

[2]

(e) State the conclusion of the test.

[1]

Show solutionHide solution

(a) H_0: the data fit the 4:3:2:1 ratio.

H_1: the data do not fit the 4:3:2:1 ratio.

[2]

(b) Total =46+38+19+17=120; ratio parts sum to 4+3+2+1=10. Site A expected =\dfrac{4}{10}\times120=48.

[2]

(c) B =\dfrac{3}{10}\times120=36, C =\dfrac{2}{10}\times120=24, D =\dfrac{1}{10}\times120=12.

[2]

(d) \chi^2=\dfrac{2^2}{48}+\dfrac{2^2}{36}+\dfrac{5^2}{24}+\dfrac{5^2}{12}=0.0833+0.1111+1.0417+2.0833=3.32 with 3 d.f., giving p=0.345.

[2]

(e) Since p=0.345>0.05, do not reject H_0: the data are consistent with the 4:3:2:1 model.

[1]

Final answer

(b) 48
(c) 36, 24, 12
(d) \chi^2=3.32, p=0.345
(e) do not reject H_0

Question 22

[13 marks]

A café owner records how satisfied customers are (Low, Medium, High) and which of three service times of day they visited (Morning, Afternoon, Evening). She wants to know whether satisfaction is independent of the time of day. A survey of 360 customers gives the table below.

	Low	Medium	High	Total
Morning	20	50	60	130
Afternoon	30	40	50	120
Evening	30	40	40	110
Total	80	130	150	360

A \chi^2 test for independence is performed at the 5% significance level.

(a) State the null and alternative hypotheses.

[2]

(b) Find the number of degrees of freedom.

[1]

(c) Show that the expected number of Morning customers reporting Low satisfaction is approximately 28.9.

[2]

(d) Using your GDC, find the \chi^2 statistic and the p-value.

[3]

(e) State, with a reason, the conclusion of the test.

[2]

(f) The owner then removes the Evening row and repeats the test on Morning and Afternoon only. State the new number of degrees of freedom.

[1]

(g) Explain why a smaller sample makes a real association harder to detect.

[2]

Show solutionHide solution

(a) H_0: satisfaction is independent of time of day.

H_1: satisfaction is not independent of time of day.

[2]

(b) Degrees of freedom =(3-1)(3-1)=4.

[1]

(c) Expected =\dfrac{130\times80}{360}=\dfrac{10\,400}{360}=28.9 (3 s.f.).

[2]

(d) Using GDC, \chi^2=6.24 and p=0.182.

[3]

(e) Since p=0.182>0.05, do not reject H_0: no significant evidence that satisfaction depends on time of day.

[2]

(f) New degrees of freedom =(2-1)(3-1)=2.

[1]

(g) With a smaller sample the expected frequencies and observed counts are smaller, so random variation has a larger relative effect; the test has less power, making a genuine association harder to detect as statistically significant.

[2]

Final answer

(b) 4
(c) 28.9
(d) \chi^2=6.24, p=0.182
(e) do not reject H_0
(f) 2
(g) a smaller sample makes a real association harder to detect

Question 23

[11 marks]

An agronomist compares the yields (in kg per plot) of a crop grown with two different fertilisers. Random plots are assigned to each fertiliser and the yields recorded.

Fertiliser P	Fertiliser Q
42	47
45	51
39	49
44	46
41	53
47	48
43	50

A two-sample t-test for a difference of means (assuming equal variances) is performed at the 5% significance level to test whether fertiliser Q gives a higher mean yield than fertiliser P.

(a) Calculate the mean yield for each fertiliser.

[2]

(b) State the null and alternative hypotheses, defining your symbols.

[3]

(c) Using your GDC, find the t-statistic and the p-value for this one-tailed test.

[3]

(d) State the conclusion of the test in context.

[2]

(e) Give one reason why using randomly assigned plots strengthens the conclusion.

[1]

Show solutionHide solution

(a) P mean =\dfrac{42+45+39+44+41+47+43}{7}=\dfrac{301}{7}=43 kg.

Q mean =\dfrac{47+51+49+46+53+48+50}{7}=\dfrac{344}{7}=49.14 kg.

[2]

(b) Let \mu_P and \mu_Q be the mean yields with fertilisers P and Q.

H_0: \mu_P=\mu_Q. H_1: \mu_Q>\mu_P.

[3]

(c) Entering both lists into the GDC pooled two-sample t-test with H_1:\mu_Q>\mu_P gives t=4.54 and p=0.000338.

[3]

(d) Since p=0.000338<0.05, reject H_0. There is significant evidence that fertiliser Q gives a higher mean yield than fertiliser P.

[2]

(e) Random assignment reduces bias, so any difference in yield is more likely due to the fertiliser rather than to systematic differences between the plots.

[1]

Final answer

(a) 43 kg, 49.1 kg
(b) H_0:\mu_P=\mu_Q, H_1:\mu_Q>\mu_P
(c) t=4.54, p=0.000338
(d) reject H_0
(e) random assignment removes bias

Question 24

[10 marks]

A sports scientist compares the mean fitness scores of two groups of volunteers, group 1 with mean 52.0 and standard deviation 9.0, and group 2 with mean 47.5 and standard deviation 8.0. She tests, at the 5% level with a pooled t-test, whether the population means differ. Both samples have the same size n.

(a) For n=15, find the p-value and state the conclusion, in context.

[3]

(b) Keeping the means and standard deviations unchanged, find the smallest value of n for which H_0 is rejected. Show the p-values for the two consecutive values of n that decide your answer.

[4]

(c) Now suppose the means were 52.0 and 43.0, with the same standard deviations. Find the smallest value of n for which H_0 is rejected.

[2]

(d) Comment on how the smallest sample size changes when the difference between the means doubles.

[1]

Show solutionHide solution

(a) H_0: \mu_1=\mu_2, H_1: \mu_1\neq\mu_2. GDC (summary statistics, Pooled: Yes): p=0.159

A1[1]

0.159>0.05, so do not reject H_0 M1 [1]: insufficient evidence, at the 5% level, that the mean fitness scores of the two groups differ.

A1[1]

(b) Repeat the test for increasing n (using a table or trial and improvement).

M1[1]

n=28: p=0.0531>0.05 A1 [1] n=29: p=0.0490<0.05

A1[1]

Smallest n=29

A1[1]

(c) n=8: p=0.0529>0.05; n=9: p=0.0395<0.05 M1 [1] smallest n=9

A1[1]

(d) Doubling the difference between the means reduces the required sample size from 29 to 9, to less than one third (roughly in proportion to 1/\text{difference}^2, so about one quarter).

R1[1]

Final answer

(a) p=0.159: do not reject
(b) n=29 (p=0.0531 at 28, 0.0490 at 29)
(c) n=9
(d) about a quarter of the sample size

Question 25

[9 marks]

Kamau is a librarian in Nairobi. For 240 loans he records the type of borrower and the format of the book borrowed.

	Print	E-book	Audio	Total
Student	60	30	10	100
Adult	60	64	16	140
Total	120	94	26	240

(a) Find the probability that a loan chosen at random is an e-book.

[1]

(b) An audio loan is chosen at random. Find the probability that it was borrowed by a student.

[2]

Kamau uses a \chi^2 test to decide whether the format borrowed is independent of the type of borrower.

(c) Write down the null hypothesis.

[1]

(d) Find the p-value.

[2]

(e) State the conclusion at (i) the 5% level, (ii) the 1% level.

[3]

Show solutionHide solution

(a) \dfrac{94}{240}=0.392 A1 (0.39166…)

(b) \dfrac{10}{26} M1 =0.385 A1 (0.38461…)

(c) H_0: the format borrowed is independent of the type of borrower.

(d) Two-way \chi^2 test on the 2\times3 counts M1 \chi^2=7.22, df =2, p=0.0271 A1 (0.027102…)

(e) (i) 0.0271<0.05: reject H_0; sufficient evidence that the format borrowed depends on the type of borrower.

A1R1

(ii) 0.0271>0.01: do not reject H_0; insufficient evidence at the 1% level.

Final answer

(a) 0.392
(b) 0.385
(c) the format borrowed is independent of the type of borrower
(d) p=0.0271
(e) (i) reject H_0 (ii) do not reject H_0

Question 26

[16 marks]

Ama runs a plant nursery in Kumasi. A supplier claims that each seed of a pepper variety germinates with probability 0.7, independently of the others. Ama sows 4 seeds in each of 150 trays and records how many germinate in each tray.

Seeds germinating	0 or 1	2	3	4
Number of trays	18	44	55	33

(a) Show that, if the supplier's claim is true, the probability that exactly 3 seeds in a tray germinate is 0.4116.

[2]

(b) Find the expected number of trays in each of the four categories.

[3]

Ama tests the supplier's claim at the 5% level, using a \chi^2 goodness of fit test.

(c) Write down the null hypothesis.

[1]

(d) Write down the number of degrees of freedom.

[1]

(e) Find the p-value.

[2]

(f) State the conclusion of the test.

[1]

Ama grows the seedlings in two composts, X and Y, and measures their heights, in cm, after three weeks.

Compost X	14.2	15.8	13.9	16.4	15.1	14.7	16.0
Compost Y	13.1	14.5	12.8	14.9	13.6	14.0	13.3	14.2

She believes that seedlings grow taller in compost X. The heights may be assumed to be normally distributed with equal variances.

(g) Write down suitable hypotheses, defining your symbols.

[2]

(h) Find the p-value for a test of Ama's belief.

[2]

(i) State the conclusion at the 5% significance level, in context.

[2]

Show solutionHide solution

(a) P(X=3)=\binom43(0.7)^3(0.3) M1 =4\times0.343\times0.3=0.4116

M1AG

(b) P(0\text{ or }1)=0.0081+0.0756=0.0837, P(2)=0.2646, P(4)=0.2401 M1 Expected 150\times each: 12.6, 39.7, 61.7, 36.0 A1A1 (12.555, 39.69, 61.74, 36.015; total 150)

(c) H_0: the number of seeds germinating in a tray follows \mathrm{B}(4,0.7).

(d) 4-1=3

(e) Goodness of fit test with df 3 M1 \chi^2=3.82, p=0.282 A1 (0.28184…)

(f) 0.282>0.05: do not reject H_0; insufficient evidence against the supplier's claim.

(g) \mu_X, \mu_Y: population mean heights in composts X and Y A1 H_0:\mu_X=\mu_Y, H_1:\mu_X>\mu_Y

(h) Pooled two-sample t-test, X first, alternative > M1 p=0.00390 A1 (t=3.14, df 13, 0.0038984…)

(i) 0.00390<0.05: reject H_0 A1 sufficient evidence that seedlings grow taller on average in compost X.

Final answer

(a) shown
(b) 12.6, 39.7, 61.7, 36.0
(c) the number germinating follows B(4, 0.7)
(d) 3
(e) p=0.282
(f) do not reject H_0
(g) H_0:\mu_X=\mu_Y, H_1:\mu_X>\mu_Y
(h) p=0.00390
(i) reject H_0; sufficient evidence of taller seedlings in compost X

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Why expected = row \times column \div total

G1 Worked example

[5 marks]

A café in Cape Town records the orders of 200 customers. Of these, 120 order a hot drink, and 50 of the 200 customers are under 25 years old. Assume that the type of drink is independent of the age group.

(a) Find the probability that a customer chosen at random is under 25 and orders a hot drink.

[2]

(b) Hence find the expected number of customers who are under 25 and order a hot drink.

[1]

(c) Use the formula E=\dfrac{rc}{n} to check your answer to part (b).

[2]

Show solutionHide solution

(a) P(\text{hot})=\dfrac{120}{200}=0.6, P(\text{under }25)=\dfrac{50}{200}=0.25 M1 0.6\times0.25=0.15

(b) 200\times0.15=30

(c) E=\dfrac{120\times50}{200} M1 =30

Final answer

(a) 0.15 (b) 30 (c) 30

G2 The same proportions in a bigger sample

G2 Worked example

[6 marks]

Asha asks 50 students whether they walk or cycle to school, and records their year group. She tests whether the way students travel is independent of year group.

	Walk	Cycle
Year 10	15	10
Year 11	10	15

(a) Find the \chi^2 statistic and the p-value, and state the conclusion at the 5% level.

[3]

(b) Asha repeats the survey with 100 students and every count doubles. Find the new \chi^2 statistic and p-value, and state the conclusion at the 5% level.

[3]

Show solutionHide solution

(a) Every expected frequency is \dfrac{25\times25}{50}=12.5 M1 \chi^2=4\times\dfrac{2.5^2}{12.5}=2, p=0.157 A1 0.157>0.05: do not reject H_0; insufficient evidence of a link between year group and the way students travel.

(b) \chi^2=2\times2=4 M1 p=0.0455 A1 0.0455<0.05: reject H_0; sufficient evidence of a link.

Final answer

(a) chi-squared 2, p 0.157, do not reject (b) chi-squared 4, p 0.0455, reject

G3 Investigation

[12 marks]

Tomasz, the climbing gym manager, records the membership type and the preferred session time of 160 members.

	Morning	Afternoon	Evening	Total
Monthly	18	22	40	80
Ten-visit pass	22	30	28	80
Total	40	52	68	160

A \chi^2 test for independence gives \chi^2=3.748 and p=0.153 with 2 degrees of freedom. Tomasz surveys twice as many members, and every count doubles.

(a) Show that every expected frequency also doubles, and that the contribution of each cell to \chi^2 doubles. Hence write down the new value of \chi^2.

[4]

(b) If every count is multiplied by k, write down \chi^2 in terms of k.

[1]

(c) The critical value for 2 degrees of freedom at the 5% level is 5.991. Find the smallest whole number k for which Tomasz would reject the null hypothesis, and find the p-value for this value of k.

[3]

(d) Comment on what this shows about a “significant” result from a very large sample.

[2]

(e) The table has 2 rows and 3 columns, and its totals are fixed. Explain why only 2 of the 6 counts can be chosen freely, and link this to the degrees of freedom.

[2]

Show solutionHide solution

(a) E=\dfrac{(2r)(2c)}{2n}=2\times\dfrac{rc}{n} M1A1 \dfrac{(2O-2E)^2}{2E}=2\times\dfrac{(O-E)^2}{E} M1 New \chi^2=2\times3.748=7.497

(b) \chi^2=3.748k

(c) 3.748k>5.991, so k>1.598 M1 k=2 A1 p=0.0236

(d) With enough data even a small difference in proportions gives a significant result. R1 A significant result is not necessarily a large or important one.

(e) Choose two cells of the first row (for example 18 and 22); the third cell of the row and the whole second row are then forced by the totals. R1 So the degrees of freedom are (2-1)(3-1)=2.

Final answer

(a) E and each contribution double; chi-squared 7.50 (b) 3.75k (c) k = 2, p = 0.0236 (d) a significant result need not be large or important (e) two cells fix the rest; df = 2

From the book

Photon Guidebook: Mathematics Applications and Interpretation SL for the IB Diploma

Chapter 12 teaches statistical tests from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book