Free worked solutions · 2026 Edition

IB Maths AI HL Chapter 17: Hypothesis tests and confidence intervals worked solutions

Worked solutions to the practice questions in Chapter 17 of Photon Guidebook: Mathematics Applications and Interpretation HL 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 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.
A1
0.037<0.05.
R1

Final answer

reject H_0: 0.037<0.05

Question 2

[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.
M1
E=25.5.
A1
(b) (2-1)(2-1)=1.
A1

Final answer

  • (a) 25.5
  • (b) 1

Question 3

[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.
A1
Enter the observed table in the GDC chi-squared test.
M1
\chi^2=6.82.
A1
(b) p=0.0331.
A1
Reject H_0. Compare p with 0.05.
R1
There is sufficient evidence that visitor status and pass colour are associated.
R1

Final answer

df =2, \chi^2=6.82, p=0.0331: reject H_0; visitor status and pass colour are associated

Question 4

[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.
A1
(b) E=160(0.15).
M1
E=24.
A1

Final answer

  • (a) 4
  • (b) 24

Question 5

[4 marks]
Independent service times have a right-skewed distribution with mean 8.9 minutes and standard deviation 5.1 minutes. A sample contains 100 times; assume this size is adequate for the CLT.
Use a central limit theorem approximation to find P(\overline X>9.7). State the approximating mean and variance.
[4]
Show solutionHide solution
\overline X\approx N(8.9,0.2601) by the CLT, with standard error 0.51. M1 A1 P(\overline X>9.7)\approx0.0583674\approx0.0584.
M1A1

Final answer

\bar X\approx\mathrm{N}(8.9,\,0.2601); P=0.0584

Question 6

[2 marks]
A random sample is used to estimate the unknown population mean \mu. A confidence interval is reported as (42.6,47.8) minutes.
Find the point estimate and the margin of error.
[2]
Show solutionHide solution
The midpoint is (42.6+47.8)/2=45.2 minutes.
A1
The half-width is (47.8-42.6)/2=2.6 minutes.
A1

Final answer

45.2 minutes; margin 2.6 minutes

Question 7

[3 marks]
A random sample of 36 independent rope breaking load measurements has mean 918. The population is normal with known standard deviation 54. Use \overline x\pm z^*\sigma/\sqrt n, where P(-z^*<Z<z^*)=0.99 for standard normal Z.
Calculate a 99% confidence interval for the population mean, giving endpoints to three significant figures.
[3]
Show solutionHide solution
z^*=2.575829 and \operatorname{SE}=54/\sqrt{36}.
M1
The margin is 23.182464.
A1
The interval is (895\,941).
A1

Final answer

(895,\,941)

Question 8

[3 marks]
In an independent random sample of 52 trials, 10 are damaged archival sleeves. The baseline success probability is 0.09. Test H_0:p=0.09 against H_1:p>0.09 at the 5% level using an exact binomial test.
Find the p-value and state a conclusion in context.
[3]
Show solutionHide solution
Under H_0, X\sim B(52,0.09).
M1
The p-value is P(X\ge10)=0.0164.
A1
Reject H_0: there is evidence that the probability of damaged archival sleeves is above 0.09.
R1

Final answer

P(X\ge10)=0.0164: reject H_0

Question 9

[3 marks]
For X\sim B(28,0.22) under H_0, the alternative is that the success probability is higher. Reject when X\ge c.
Find the smallest integer c giving a Type I error probability no more than 5%, and give the actual significance level.
[3]
Show solutionHide solution
P(X\ge10)=0.069101>0.05.
M1
P(X\ge11)=0.029293\le0.05, so c=11.
A1
The actual level is 0.0293, or 2.93%.
A1

Final answer

c=11; actual significance level 0.0293

Tutorial

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

Question 10

[8 marks]
The table shows four hypothesis tests, each with its significance level \alpha and p-value.
\begin{array}{lccc} \hline \text{Test} & \alpha & p \text{-value} & \text{Decision} \\ \hline \text{A} & 0.05 & 0.048 & \\ \text{B} & 0.05 & 0.052 & \\ \text{C} & 0.01 & 0.008 & \\ \text{D} & 0.01 & 0.031 & \\ \hline \end{array}
(a) For each test, write down whether H_0 is rejected or not rejected.
[4]
(b) Test A examines whether the mean daily rainfall in a river valley differs from 6.2 mm. Write down the conclusion of Test A in context.
[2]
(c) Test D examines whether a new fertiliser increases the mean mass of tomatoes grown by Dalia. Write down the conclusion of Test D in context.
[2]
Show solutionHide solution
(a) A: 0.048<0.05, reject H_0 [1] B: 0.052>0.05, do not reject H_0
[1]
C: 0.008<0.01, reject H_0 [1] D: 0.031>0.01, do not reject H_0
[1]
(b) H_0 is rejected: there is sufficient evidence, at the 5% level, that the mean daily rainfall in the valley differs from 6.2 mm. [1] [1] (decision and context each one mark)
(c) H_0 is not rejected: there is insufficient evidence, at the 1% level, that the new fertiliser increases the mean mass of Dalia's tomatoes. [1]
[1]

Final answer

  • (a) A reject, B do not reject, C reject, D do not reject
  • (b) sufficient evidence that the mean daily rainfall differs from 6.2 mm
  • (c) insufficient evidence that the fertiliser increases the mean mass

Question 11

[7 marks]
A researcher records the nightly sleep of a random sample of 12 students. The sample mean is \bar x=7.3 hours and the GDC's standard deviation of the sample is \sigma_x=0.9 hours. Nightly sleep times are assumed to be normally distributed.
(a) Find s_{n-1} for this sample.
[2]
(b) Construct a 95% confidence interval for the mean nightly sleep of students, giving the end-points to 3 significant figures.
[3]
(c) A health leaflet says that the average student sleeps 8 hours each night. Comment on this claim.
[2]
Show solutionHide solution
(a) s_{n-1}=\sqrt{\dfrac{n}{n-1}}\,\sigma_x=\sqrt{\dfrac{12}{11}}(0.9).
[1]
=0.94002\approx0.940 hours.
[1]
(b) t Interval, Stats: \bar x=7.3, s_x=0.94002 (not 0.9), n=12, C-Level =0.95, t^{*}=2.201.
[1]
Margin =2.201\times\dfrac{0.94002}{\sqrt{12}}=0.59726.
[1]
\mathrm{CI}=(6.70,\ 7.90) hours.
[1]
(c) 8 lies outside the interval.
[1]
The claim is not supported: the mean nightly sleep of students is probably less than 8 hours.
[1]

Final answer

  • (a) 0.940 hours
  • (b) (6.70,\,7.90)
  • (c) 8 lies above the interval: evidence that the mean is less than 8 hours

Question 12

[7 marks]
The running time on a full charge of a robot vacuum cleaner model is tested on 9 randomly chosen machines. The running times, in minutes, are 98.2, 104.1, 94.9, 101.5, 105.0, 96.8, 103.4, 99.8, 100.3.
(a) Use your GDC to find \bar x and s_{n-1}.
[2]
(b) Construct a 95% confidence interval for the mean running time \mu of this model, giving the end-points to 3 significant figures.
[3]
(c) State one assumption needed for your interval to be valid, and explain why it is needed here.
[2]
Show solutionHide solution
(a) Enter the list and run One-Variable Statistics: \bar x=100.44\approx100.4 minutes
[1]
and s_x=s_{n-1}=3.418\approx3.418 minutes.
[1]
(b) t Interval, input Data, C-Level =0.95 (\mathrm{df}=8, t^{*}=2.306).
[1]
Margin =2.306\times\dfrac{3.418}{\sqrt9}=2.6273.
[1]
\mathrm{CI}=(97.8,\ 103) minutes.
[1]
(c) Assume the running times are normally distributed (the machines are a random sample from a normal population).
[1]
The sample size 9<30, so the Central Limit Theorem cannot be used to justify normality of \bar X; the t-interval relies on the population being normal.
[1]

Final answer

  • (a) 100.4, 3.418
  • (b) (97.8,\,103)
  • (c) the population is normal; n=9 is too small for the central limit theorem

Question 13

[8 marks]
Ping tests of an online-gaming network, on 36 randomly chosen occasions, give a 95% confidence interval of (84.2,\ 91.6) ms for its mean latency \mu. The interval was calculated as a z-interval using a known population standard deviation \sigma.
(a) The network operator advertises a mean latency of 90 ms, a consumer group claims 80 ms, and a rival says 88 ms. For each of these three values, state whether it is consistent with the interval.
[3]
(b) Find the value of \sigma to 3 significant figures.
[2]
(c) Calculate the 99% confidence interval from the same sample, using your value of \sigma and giving the end-points to 3 significant figures.
[2]
(d) Does the consumer group's claim of 80 remain inconsistent with the data at the 99% level? Justify your answer.
[1]
Show solutionHide solution
(a) 90 lies inside (84.2,\ 91.6): consistent.
[1]
88 lies inside the interval: consistent.
[1]
80 lies outside the interval: not consistent (the data are evidence against it).
[1]
(b) Margin =\dfrac{91.6-84.2}{2}=3.7 and \bar x=87.9.
[1]
1.96\times\dfrac{\sigma}{\sqrt{36}}=3.7\Rightarrow\sigma=\dfrac{3.7\times6}{1.96}=11.327\approx11.3 ms.
[1]
(c) Margin =2.576\times\dfrac{11.3}{6}=4.8515.
[1]
\mathrm{CI}=(83.0,\ 92.8) ms.
[1]
(d) 80 is below the lower end-point 83.0, so it is still inconsistent with the data at the 99% level.
[1]

Final answer

  • (a) 90 and 88 are consistent, 80 is not
  • (b) \sigma=11.3
  • (c) (83.0,\,92.8)
  • (d) yes, 80 is still outside the interval

Question 14

[6 marks]
An online learning platform knows that the time taken by students to finish a course module has standard deviation \sigma=14 hours. It uses a random sample of 49 students to construct confidence intervals for the mean completion time.
(a) Find the margin of error and the width of a 95% confidence interval.
[2]
(b) Find the width of the 95% interval if the sample size were 196.
[1]
(c) Find the width of a 99% interval with n=49.
[2]
(d) Use your answers to (a), (b) and (c) to comment on how to obtain an interval that is both more confident and more precise.
[1]
Show solutionHide solution
(a) Margin =1.96\times\dfrac{14}{\sqrt{49}}=1.96\times2=3.9199\approx3.92 hours.
[1]
Width =2\times3.9199=7.8399\approx7.84 hours.
[1]
(b) n=196 quadruples the sample size, which halves the width: \dfrac{7.8399}{2}=3.9199\approx3.92 hours.
[1]
(c) Margin =2.576\times2=5.1517.
[1]
Width =10.303\approx10.3 hours.
[1]
(d) Raising the level from 95% to 99% widens the interval (from 7.84 to 10.3); only a larger sample (here 196, width 3.92) can make an interval more precise, so a larger sample is needed to gain confidence without losing precision.
[1]

Final answer

  • (a) margin 3.92, width 7.84
  • (b) 3.92
  • (c) 10.3
  • (d) a larger sample narrows the interval, a higher level widens it

Question 15

[6 marks]
Tariq keeps goats and reads that the mean daily milk yield of a goat of this breed is 3.6 litres. He records the yield, in litres, of 8 randomly chosen goats on one day: 3.8,\ 4.4,\ 3.1,\ 4.9,\ 4.2,\ 3.6,\ 4.7,\ 4.0. Assuming that yields are normally distributed, test at the 5% level whether the mean yield of Tariq's goats differs from 3.6 litres. State the hypotheses, the p-value and your conclusion in context.
[6]
Show solutionHide solution
H_0:\ \mu=3.6; H_1:\ \mu≠ 3.6. [1] \sigma unknown, so a t-test with \text{df}=7.
[1]
GDC: \bar{x}=4.09, s_{n-1}=0.591 [1] (summary or data entered into t-Test, \mu_0=3.6, \mu≠\mu_0) t=2.33, [1] p=0.0525.
[1]
0.0525 > 0.05, so do not reject H_0: there is insufficient evidence, at the 5% level, that the mean daily yield of Tariq's goats differs from 3.6 litres.
[1]

Final answer

H_0\!:\mu=3.6, H_1\!:\mu\ne3.6; \bar x=4.09, s_{n-1}=0.591; t=2.33, p=0.0525: do not reject

Question 16

[7 marks]
A baker, Joaquim, compares two yeasts. The times (in minutes) taken for a dough to double in volume are recorded for 8 batches using Yeast X and 9 batches using Yeast Z.
\begin{array}{ll} \text{Yeast X:} & 52,\,47,\,58,\,55,\,49,\,60,\,53,\,56 \\ \text{Yeast Z:} & 48,\,44,\,50,\,51,\,46,\,53,\,45,\,49,\,47 \\ \end{array}
Joaquim believes that dough rises faster with Yeast Z. Assume the times are normally distributed with equal variances.
(a) Find the mean and standard deviation s_{n-1} for each yeast.
[2]
(b) State the hypotheses, in terms of \mu_X and \mu_Z.
[1]
(c) Carry out a test at the 5% level, giving the p-value.
[2]
(d) State the conclusion in context.
[2]
Show solutionHide solution
(a) Yeast X: \bar{x}=53.8, s_{n-1}=4.40. [1] Yeast Z: \bar{x}=48.1, s_{n-1}=2.93.
[1]
(b) Faster rising means a smaller time for Z: H_0:\ \mu_X=\mu_Z; H_1:\ \mu_X>\mu_Z.
[1]
(c) GDC 2-Sample t Test from the two lists, Pooled: Yes, \mu_1>\mu_2: [1] t=3.14, \text{df}=15, p=0.00335.
[1]
(d) 0.00335 < 0.05, so reject H_0. [1] There is sufficient evidence, at the 5% level, that dough rises faster (takes less time) with Yeast Z than with Yeast X.
[1]

Final answer

  • (a) X: 53.8, 4.40; Z: 48.1, 2.93
  • (b) H_0\!:\mu_X=\mu_Z, H_1\!:\mu_X>\mu_Z
  • (c) p=0.00335
  • (d) reject H_0

Question 17

[8 marks]
Mirko manages the cargo terminal at the port of Rijeka. Over several years the number of container-crane faults per week has followed a Poisson distribution with mean 1.8. After a new maintenance contract Mirko records 3 faults in a period of 4 weeks, and he believes that the rate of faults has decreased.
(a) State suitable hypotheses.
[1]
(b) Find the p-value for a test of Mirko's belief.
[3]
(c) State the conclusion at the 5% significance level, in context.
[2]
(d) State the conclusion at the 10% significance level, in context.
[2]
Show solutionHide solution
(a) H_0:\ m=1.8, H_1:\ m<1.8 (faults per week) A1 (b) over 4 weeks, Y\sim\mathrm{Po}(7.2) under H_0 M1 P(Y\le3) M1 =0.0719 A1 (c) 0.0719>0.05: do not reject H_0 A1 insufficient evidence at the 5% level that the rate of faults has decreased R1 (d) 0.0719<0.10: reject H_0 A1 sufficient evidence at the 10% level that the rate of faults has decreased
R1

Final answer

  • (a) H_0\!:m=1.8, H_1\!:m<1.8 per week
  • (b) 0.0719
  • (c) do not reject H_0
  • (d) reject H_0

Question 18

[9 marks]
Aiyana inspects kiwifruit packed by a co-operative near Tauranga. Historically 12% of the packed fruit is bruised. She thinks the proportion has increased. She will inspect a random sample of 40 fruit and reject the claim that the proportion is 0.12 if at least 9 of the fruit are bruised. Let X be the number of bruised fruit in the sample.
(a) State suitable hypotheses.
[1]
(b) Find the probability of a Type I error.
[3]
(c) In fact the proportion of bruised fruit is now 0.25. Find the probability of a Type II error.
[3]
(d) Write down the power of the test when the proportion is 0.25, and interpret it in context.
[2]
Show solutionHide solution
(a) H_0:\ p=0.12, H_1:\ p>0.12 A1 (b) X\sim\mathrm{B}(40,\,0.12) under H_0 M1 P(X\ge9)=1-P(X\le8) M1 =0.0443 A1 (c) Type II error: X\le8 when X\sim\mathrm{B}(40,\,0.25) M1 P(X\le8) M1 =0.300 A1 (d) power =1-0.29983=0.700 A1 if the true proportion is 0.25, the test rejects the claim in about 70% of samples
R1

Final answer

  • (a) H_0\!:p=0.12, H_1\!:p>0.12
  • (b) 0.0443
  • (c) 0.300
  • (d) power 0.700

Question 19

[8 marks]
A lamp manufacturer in Eindhoven states that the mean lifetime of its lamps is 1200 hours. The lifetimes are normally distributed with a known standard deviation of 60 hours. The quality manager, Anouk, suspects that the mean is lower. She will test 25 randomly chosen lamps at the 5% significance level, with H_0:\ \mu=1200 and H_1:\ \mu<1200.
(a) Write down the distribution of the sample mean \bar X under H_0, and find the critical region for \bar X.
[3]
(b) The true mean lifetime is 1170 hours. Find the probability of a Type II error.
[3]
(c) Write down the power of the test when the true mean is 1170 hours.
[1]
(d) Anouk uses a sample of 100 lamps instead. State, with a reason, whether the power at 1170 hours rises or falls.
[1]
Show solutionHide solution
(a) \bar X\sim\mathrm{N}(1200,\,12^2) since 60/\sqrt{25}=12 A1 inverse normal, area 0.05: 1180.26 M1 critical region \bar X<1180 A1 (b) Type II error: \bar X>1180.26 when \bar X\sim\mathrm{N}(1170,\,12^2) M1 P(\bar X>1180.26) M1 =0.196 A1 (c) power =1-0.19624=0.804 A1 (d) it rises: the standard error falls to 6, so the two distributions overlap less
R1

Final answer

  • (a) \bar X\sim\mathrm{N}(1200,\,12^2); \bar X<1180
  • (b) 0.196
  • (c) power 0.804
  • (d) the power rises

Mastery

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

Question 20

[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.
LowMediumHighTotal
Morning205060130
Afternoon304050120
Evening304040110
Total80130150360
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 would make it harder to detect a real association.
[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

  • (a) independence
  • (b) 4
  • (c) shown
  • (d) \chi^2=6.24, p=0.182
  • (e) do not reject
  • (f) 2
  • (g) less power

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.
SiteABCD
Observed pairs46381917
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

Question 22

[8 marks]
Biruk measures the wind speed (km/h) at 8 randomly chosen locations along a coastline: 22.4,\ 19.8,\ 25.1,\ 21.7,\ 23.3,\ 20.6,\ 24.2,\ 22.9. Wind speeds are assumed to be normally distributed.
(a) Use your GDC to find \bar x, s_x and \sigma_x, each to 4 significant figures.
[2]
(b) Construct a correct 95% confidence interval for the mean wind speed, giving the end-points to 3 significant figures.
[3]
(c) Biruk enters \bar x and \sigma_x into the Stats input of t Interval. Find the interval he obtains.
[2]
(d) Explain why Biruk's interval is too narrow.
[1]
Show solutionHide solution
(a) One-Variable Statistics: \bar x=22.5\approx22.50, s_x=1.7728\approx1.773, \sigma_x=1.6583\approx1.658.
[2]
(b) t Interval, Data, C-Level =0.95 (\mathrm{df}=7, t^{*}=2.3646): margin =2.3646\times\dfrac{1.7728}{\sqrt8}=1.4821.
[2]
\mathrm{CI}=(21.0,\ 24.0) km/h.
[1]
(c) Using \sigma_x=1.6583: margin =2.3646\times\dfrac{1.6583}{\sqrt8}=1.3864.
[1]
\mathrm{CI}=(21.1,\ 23.9) km/h.
[1]
(d) \sigma_x uses divisor n and under-estimates the population spread, so it is smaller than s_{n-1}; the margin of error, and so the interval, comes out too small.
[1]

Final answer

  • (a) 22.50, 1.773, 1.658
  • (b) (21.0,\,24.0)
  • (c) (21.1,\,23.9)
  • (d) \sigma_x divides by n, so it is too small

Question 23

[11 marks]
A manufacturer of steel cables states that the mean breaking strength is 200 kN. From long experience the population standard deviation is known to be 8 kN. A safety inspector suspects the mean is actually higher and tests a random sample of 50 cables, obtaining a sample mean of 203 kN. A significance level of 5% is used.
(a) State suitable null and alternative hypotheses.
[2]
(b) Explain why a z-test is appropriate.
[1]
(c) Using your GDC, find the value of the test statistic and the p-value.
[3]
(d) State the conclusion in context.
[2]
(e) Find the smallest sample mean that would lead to rejection of H_0 at the 5% level.
[2]
(f) In this context, describe what a Type I error would mean.
[1]
Show solutionHide solution
(a) H_0: \mu=200 and H_1: \mu>200, where \mu is the mean breaking strength.
[2]
(b) The population standard deviation \sigma=8 is known, so the standardised mean is normal and a z-test applies.
[1]
(c) Using GDC (z-test, \sigma=8, \bar{x}=203, n=50): z=2.65 and p\text{-value}=0.00401.
[3]
(d) Since 0.00401 < 0.05, reject H_0; there is significant evidence that the mean breaking strength exceeds 200 kN.
[2]
(e) Reject when z>1.645, i.e. \bar{x}>200+1.645\times\dfrac{8}{\sqrt{50}}=201.86 kN. The smallest such mean is about 201.9 kN.
[2]
(f) A Type I error would mean concluding the mean strength exceeds 200 kN when in fact it equals 200 kN.
[1]

Final answer

  • (a) \mu>200
  • (b) \sigma known
  • (c) z=2.65, p=0.00400
  • (d) reject H_0
  • (e) 201.9 kN
  • (f) concluding the mean exceeds 200 kN when it is 200 kN

Question 24

[9 marks]
Thandi is a teacher in Cape Town. She gives nine pupils a spelling test out of 50 before a six-week reading programme and the same test again afterwards.
PupilABCDEFGHI
Before312735293326383028
After342938283730403331
Thandi believes that the programme raises the mean score. The differences in score may be assumed to be normally distributed.
(a) Explain why a paired test is appropriate.
[1]
(b) Define a suitable parameter and state the hypotheses.
[2]
(c) Find the test statistic and the p-value.
[3]
(d) State the conclusion at the 1% significance level, in context.
[2]
(e) Thandi concludes that the programme caused the improvement. Give one reason why this test alone does not justify that.
[1]
Show solutionHide solution
(a) the same nine pupils are tested twice, so the two lists are not independent samples R1 (b) \mu_d is the population mean of (after - before) A1 H_0:\ \mu_d=0, H_1:\ \mu_d>0 A1 (c) differences 3,\,2,\,3,\,-1,\,4,\,4,\,2,\,3,\,3; one-sample t-test with \mu_0=0 M1 t=5.08 A1 p=0.000477 A1 (d) 0.000477<0.01: reject H_0 A1 sufficient evidence at the 1% level that the programme raises the mean spelling score R1 (e) there was no comparison group, so the pupils may have improved anyway
R1

Final answer

  • (a) the same pupils are tested twice
  • (b) \mu_d=0 against \mu_d>0
  • (c) t=5.08, p=0.000477
  • (d) reject H_0
  • (e) no comparison group

Question 25

[8 marks]
At a bird observatory, the wingspans of adult male kestrels are normally distributed with mean 74 cm and standard deviation 2.4 cm. Those of adult females are normally distributed with mean 78 cm and standard deviation 2.6 cm. A ringer cannot see the sex of a bird, so she takes the null hypothesis that the bird is male, and classifies it as female if its wingspan is more than 76.5 cm.
(a) Find the probability of a Type I error.
[2]
(b) Find the probability of a Type II error.
[2]
(c) Of the adult kestrels caught, 55% are male. Find the probability that a kestrel caught at random is classified wrongly.
[2]
(d) The ringer wants the probability of a Type I error to be 5%. Find the wingspan she should use as the cut-off.
[2]
Show solutionHide solution
(a) A Type I error is a male classified as female: P(W>76.5) with W\sim\mathrm{N}(74,\,2.4^2) M1 =0.149 A1 (b) A Type II error is a female classified as male: P(W\le76.5) with W\sim\mathrm{N}(78,\,2.6^2) M1 =0.282 A1 (c) 0.55\times0.14878+0.45\times0.28200 M1 =0.209 A1 (d) inverse normal, area 0.95, mean 74, standard deviation 2.4 M1 77.9 cm
A1

Final answer

  • (a) 0.149
  • (b) 0.282
  • (c) 0.209
  • (d) 77.9 cm

Question 26

[16 marks]
Hamid is a quality engineer at a ceramics factory in Valencia. He measures the thickness, x mm, of 100 randomly chosen floor tiles and finds \sum x=810.0 and \sum x^2=6567.2.
(a) Find unbiased estimates of the mean and the variance of the thickness of a tile.
[3]
(b) Find a 95% confidence interval for the mean thickness.
[2]
(c) The factory states that the mean thickness is 8.0 mm. Comment on this statement.
[1]
Hamid groups his measurements. A class such as 7.6–7.8 includes 7.6 but not 7.8.
x (mm)<7.67.6–7.87.8–8.08.0–8.28.2–8.48.4–8.6\ge8.6
Observed18203819113
He tests whether the thickness can be modelled by a normal distribution, using your estimates from part (a) as its mean and variance.
(d) Show that the expected frequency for 8.0\le x<8.2 is 31.1, correct to 3 significant figures.
[2]
The other expected frequencies are 2.29, 9.24, 22.9, 22.9, 9.24 and 2.29, in order.
(e) Explain why some classes must be combined, and write down the observed and expected frequencies that Hamid should use.
[2]
(f) Find the number of degrees of freedom.
[2]
(g) Find the \chi^2 statistic and the p-value.
[2]
(h) State the conclusion of the test at the 5% significance level.
[1]
(i) Explain why a z-interval would give almost the same answer as your interval in part (b).
[1]
Show solutionHide solution
(a) \bar x=\frac{810}{100}=8.1 mm A1 s_{n-1}^2=\frac{6567.2-810^2/100}{99} M1 =0.0626 mm^2 A1 (b) t-interval, \bar x=8.1, s_{n-1}=0.25025\ldots, n=100 M1 (8.050,\ 8.150) mm A1 (c) 8.0 mm is below the interval: evidence that the mean thickness is more than 8.0 mm R1 (d) 100\times P(8.0\le X<8.2) with \mathrm{N}(8.1,\,0.25025^2) M1 =31.1 AG (e) the end classes have expected frequencies 2.29, below 5 R1 combine the first two and the last two classes: observed 9, 20, 38, 19, 14; expected 11.5, 22.9, 31.1, 22.9, 11.5 A1 (f) 5 classes, mean and variance estimated: 5-1-2 M1 =2 A1 (g) \chi^2=3.69 A1 p=0.158 A1 (h) 0.158>0.05: do not reject H_0; insufficient evidence that a normal model is unsuitable for the tile thickness R1 (i) with n=100 the t-distribution with 99 degrees of freedom is very close to the normal: t^*=1.984 against z^*=1.960
R1

Final answer

  • (a) 8.1, 0.0626
  • (b) (8.050,\,8.150)
  • (c) 8.0 is below the interval
  • (d) shown
  • (e) 9, 20, 38, 19, 14 and 11.5, 22.9, 31.1, 22.9, 11.5
  • (f) 2
  • (g) \chi^2=3.69, p=0.158
  • (h) do not reject
  • (i) df 99, t^*\approx z^*

Going further

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

G1 The power of a test

G1 Worked example

[8 marks]
Hine runs a trout hatchery in Rotorua. Historically half of the fish raised in a tank are female. She tests whether a new water-temperature setting raises the proportion of females, using 20 fish from a tank. Let X be the number of females. She will test H_0:\ p=0.5 against H_1:\ p>0.5 at the 5% significance level.
(a) Find the critical region for the test, and its actual significance level.
[3]
(b) The true proportion of females with the new setting is 0.7. Find the probability of a Type II error.
[2]
(c) Write down the power of the test when p=0.7, and interpret it in context.
[2]
(d) Explain why the power of the test when p=0.5 equals the actual significance level.
[1]
Show solutionHide solution
(a) X\sim\mathrm{B}(20,\,0.5) under H_0; P(X\ge14)=0.0577>0.05 M1 P(X\ge15)=0.0207\le0.05 A1 critical region X\ge15, actual significance level 0.0207 A1 (b) Type II error: X\le14 when X\sim\mathrm{B}(20,\,0.7) M1 P(X\le14)=0.584 A1 (c) power =1-0.58363=0.416 A1 if the new setting really gives 70% females, the test detects the improvement in about 42% of tanks R1 (d) at p=0.5 the null hypothesis is true, so the probability of rejecting it is the probability of a Type I error, which is the actual significance level
R1

Final answer

  • (a) X\ge15, level 0.0207
  • (b) 0.584
  • (c) power 0.416
  • (d) at p=0.5, rejecting is a Type I error

G2 Degrees of freedom count free choices

G2 Investigation

[18 marks]
A basketball player scores 70% of her free throws. After coaching she believes her success rate has improved. She will take 25 free throws and test H_0:\ p=0.7 against H_1:\ p>0.7 at the 5% significance level, rejecting H_0 if X\ge22, where X is the number scored. The power of the test at a true value p is the probability that H_0 is rejected.
(a) Explain why the power at p is P(X\ge22) with X\sim\mathrm{B}(25,\,p), and find the power when p=0.75, 0.8, 0.9 and 0.95.
[4]
(b) The power is 0.0332 at p=0.7 and 0.471 at p=0.85. Sketch the power against p for 0.7\le p\le1, marking these values and your values from part (a).
[2]
(c) The coach now uses 50 free throws. Find the critical region at the 5% level and its actual significance level, and compare the power at p=0.85 with the power for 25 throws.
[5]
(d) Kiri tests H_0:\ \mu=250 against H_1:\ \mu<250 for a normal population with \sigma=4 at the 5% level, with a sample of size n. The probability of a Type II error when \mu=248 must be at most 0.1. Show that \sqrt n\ge\frac{(1.645+1.282)\times4}{2} and find the smallest sample size.
[4]
(e) Explain why making both the probability of a Type I error and the probability of a Type II error small needs a larger sample.
[3]
Show solutionHide solution
(a) the test rejects H_0 exactly when X\ge22, so the probability of rejecting is P(X\ge22) R1 powers 0.0962, 0.234, 0.764, 0.966 A1A1A1 (b) an increasing S-shaped curve through (0.7,\,0.0332), (0.75,\,0.0962), (0.8,\,0.234), (0.85,\,0.471), (0.9,\,0.764), (0.95,\,0.966) A1 rising towards 1 at p=1 A1 (c) X\sim\mathrm{B}(50,\,0.7); P(X\ge40)=0.0789>0.05 M1 P(X\ge41)=0.0402\le0.05 A1 region X\ge41, actual level 0.0402 A1 power at 0.85 is P(X\ge41)=0.791 with X\sim\mathrm{B}(50,\,0.85) A1 0.791>0.471: the larger sample is far more likely to detect a rate of 0.85 R1 (d) critical value 250-1.645\frac{4}{\sqrt n} M1 Type II error at most 0.1 needs the critical value at least 248+1.282\frac{4}{\sqrt n} M1 \sqrt n\ge\frac{(1.645+1.282)\times4}{2}=5.853 AG n\ge34.26, so n=35 A1 (e) for a fixed n, moving the cut-off to lower one error raises the other R1 only a smaller standard error \sigma/\sqrt n pulls the two distributions apart R1 so both errors can be small only if n is large
R1

Final answer

  • (a) explained; 0.0962, 0.234, 0.764, 0.966
  • (b) S-shaped curve through the points
  • (c) region X\ge41, level 0.0402; power 0.791 against 0.471
  • (d) \sqrt n\ge5.853, n=35
  • (e) explained

From the book

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

Chapter 17 teaches hypothesis tests and confidence intervals 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

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