Unit 4 · Topic 5.2 · Statistical inference
A confidence interval turns one sample into a range of likely values for the population mean. It builds on Topic 5.1, Sample means. The interval is centred on $\bar x$, and its width comes from $\frac{s}{\sqrt n}$, the same spread you used for sample means. Tick off each skill once you can do it without help.
Only tick a skill when you could do it in an exam.
Part 1
A sample mean $\bar x$ is a point estimate of $\mu$. It is almost never exactly right, so you give a range of values instead. That range is an interval estimate, called a confidence interval. It is centred on $\bar x$ and is likely to contain $\mu$.
In Methods you built confidence intervals for a proportion. In Specialist you build them for a mean, and the formula has the same shape:
$$\left(\bar x-z\frac{s}{\sqrt n},\ \bar x+z\frac{s}{\sqrt n}\right)$$
The margin of error is half the width of the interval:
$$E=z\frac{s}{\sqrt n}$$
Use the sample standard deviation $s$ unless the question tells you to use the population standard deviation $\sigma$. The value of $z$ depends on the confidence level:
| Confidence level | 90% | 95% | 99% |
| $z$ | $1.645$ | $1.96$ | $2.576$ |
Learn 1.96 by heart. Marking guides also accept 1.64 or 1.65 for 90% and 2.58 for 99%. On Technology-free papers the question usually gives you a friendlier value, such as $z=2$. On Technology-active papers, your calculator's one-sample z interval does the whole calculation.
If you took many samples and built a 95% interval from each one, about 95% of the intervals would contain $\mu$. Any one interval either contains $\mu$ or it doesn't. The interval is always centred on $\bar x$, so $\bar x$ is always inside it, but $\mu$ might not be.
The width is $2z\frac{s}{\sqrt n}$. Raising the confidence level raises $z$ and widens the interval, while a larger sample makes it narrower. Multiplying $n$ by 4 halves the width.
The idea
Practise
Which statement regarding sample means is true?
$\bar x$ is a statistic, so it changes from one sample to the next. $\mu$ is a parameter of the population, so it stays fixed, which rules out (D).
$X$ can have any distribution, which rules out (A). $\bar X$ is only normal when the population is normal, and only approximately normal when $n\ge30$, which rules out (B).
Four random samples of different sizes were taken to estimate a certain population mean, given a known population standard deviation. A 95% confidence interval was calculated for each sample.
Which sample used the largest sample size?
The standard deviation and the confidence level are the same for all four samples, so the width $2z\frac{\sigma}{\sqrt n}$ only changes with $n$. The largest sample gives the narrowest interval.
Sample 1 runs from about 21 to 24.6, a width of about 3.6. The other three are all wider.
A confidence interval for a parameter is a range of values within which the
Every confidence interval is centred on the sample estimate, $\bar x$, so the sample estimate is always inside it.
(C) is tempting, but some intervals miss the parameter, as the animation above shows. Most intervals do contain it, which rules out (D).
Repeated random samples will be used to calculate a large number of 90% confidence intervals for a population mean $\mu$.
Which statement best describes the possible outcomes?
Each interval either contains $\mu$ or it doesn't. Over a large number of intervals the proportion that contain $\mu$ is close to 90%, but random sampling means it won't be exactly 90%.
The mass of a particular variety of cake is claimed to be normally distributed with a mean of 660 grams. A random sample of five of these cakes is found to have a mean mass of 600 grams.
Which option represents an approximate confidence interval for $\mu$ based on this sample?
A confidence interval is centred on the sample mean, 600 g. The claimed mean of 660 g plays no part in building it.
Only (B) is centred on 600. In (A), 600 is the upper end, and in (C) it is the lower end.
The time taken to complete orders at a pizza store is normally distributed with a mean time $(\mu)$ of 10 minutes.
The owner of the pizza store records the time taken to complete orders for a random sample of 20 pizzas each day over a 30-day period. From this data, an approximate 90% confidence interval for $\mu$ is calculated at the end of each day.
How many of these confidence intervals would be expected to contain $\mu$?
Each day gives one 90% interval, so about 90% of the 30 intervals should contain $\mu$.
$$0.9\times30=27$$
A random variable is normally distributed with a mean $\mu$. An approximate 95% confidence interval for $\mu$ from a sample from this distribution is $(209.7,\ 221.9)$.
An approximate confidence interval for $\mu$ based on the same sample, using a confidence level greater than 95%, could be
The sample is the same, so the centre stays at $\bar x$:
$$\bar x=\frac{209.7+221.9}{2}=215.8$$
A higher confidence level uses a larger $z$, so the interval is wider. (B) is centred on 215.8 and is wider than the original. (A) is centred on 214.9, (C) is narrower and (D) is centred on 221.9.
The scores on a test are assumed to be normally distributed.
Researchers use the results from a random sample of scores to calculate a confidence interval for the population mean. However, a shorter confidence interval width is required so the researchers decide to use a second sample for their calculations.
Assuming that the standard deviations for both samples are the same, the researchers can ensure that a shorter confidence interval width is produced by
The width is $2z\frac{s}{\sqrt n}$. A larger $n$ makes it shorter. A lower confidence level gives a smaller $z$, which also makes it shorter. Doing both is the only option that guarantees a shorter interval.
Part 2
To check a claimed value of $\mu$, build the interval and see whether the claimed value is inside it. If it is inside, the claim is reasonable. If it is outside, the claim is not reasonable, or at least there is reason to doubt it. Quote the interval and the claimed value in your answer, because the reasoning mark depends on them.
This uses the smartphones in 2024 Paper 2, Question 11: $\bar x=9.31$ hours and $s=0.52$ hours, with a claimed mean of 9.5 hours. Change the confidence level and the sample size, and watch whether the interval reaches the claim.
The claimed 9.5 hours is inside this interval, so this interval supports the claim.
The claimed 9.5 hours is outside this interval, so this interval doesn't support the claim.
At $n=40$, the 95% and 99% intervals are parts a) and b) of the question.
Practise
Rounded to two decimal places, the $z$-value used in the calculation of an approximate 95% confidence interval for $\mu$ is
The middle 95% of the standard normal distribution lies between $-z$ and $z$, which leaves 2.5% in each tail.
$$P(Z\le z)=0.975$$
$$z\approx1.96$$
A factory produces cans of juice. Each can has a labelled volume of 500 mL.
The factory manager conducted a random sample of 100 cans to assess whether the labelled volume was being met satisfactorily in production.
The mean volume of the sample was 498.9 mL with a sample standard deviation of 4.0 mL.
Industry regulation requires that the mean volume must meet or exceed the labelled volume.
Part a)
$$n=100,\quad\bar x=498.9,\quad s=4.0,\quad z=2$$
$$\left(498.9-2\times\frac{4}{\sqrt{100}},\ 498.9+2\times\frac{4}{\sqrt{100}}\right)$$
$$=(498.9-0.8,\ 498.9+0.8)$$
$$=(498.1,\ 499.7)\text{ mL}$$
Part b)
The labelled volume of 500 mL is above the upper end of the interval, 499.7 mL.
So the current production does not meet the regulation.
This is Technology-free, so the numbers are chosen to be friendly: $\sqrt{100}=10$ and $z=2$. Saying the whole interval lies below 500 mL also gets the first mark of part b).
A company claims that the mean battery life of their latest model of smartphone is 9.5 hours.
To test this claim, the battery lives of a random sample of 40 of the smartphones were measured.
A sample mean of 9.31 hours and a standard deviation of 0.52 hours were calculated from this data.
A manager comments that either confidence interval could be used to support the company's claim.
Part a)
$$n=40,\quad\bar x=9.31,\quad s=0.52,\quad z=1.96$$
$$\text{CI}\approx(9.15,\ 9.47)\text{ hours}$$
Part b)
$$z=2.576$$
$$\text{CI}\approx(9.10,\ 9.52)\text{ hours}$$
Part c)
The 95% interval doesn't contain the claimed 9.5 hours, but the 99% interval does.
So the manager's comment is not reasonable. Only the 99% interval supports the claim.
Your calculator's one-sample z interval does parts a) and b) directly. Enter 0.52 where it asks for the standard deviation. The statement in part c) only gets its mark if it follows from your reasoning about both intervals.
The wait time for customers put on hold when calling complaint departments is assumed to be normally distributed. A company claims that the mean wait time for their customers is 7.6 minutes.
The following data represents the wait time (minutes) from a random sample of 12 customers who called the complaint department of this company.
The standard deviation of this data is calculated to be 2.384 minutes.
Part a)
$$\bar x=\frac{110.8}{12}\approx9.23\text{ minutes}$$
Part b)
$$\bar x\pm1.96\times\frac{2.384}{\sqrt{12}}$$
$$\text{CI}\approx(7.88,\ 10.58)\text{ minutes}$$
The company's claim is not reasonable.
The claimed mean of 7.6 minutes lies below the lower end of the interval, 7.88 minutes.
The guide also accepts $(7.9,\ 10.6)$ from using $z=2$. With only 12 customers, this interval relies on the wait times being normally distributed, which the question tells you to assume.
A random sample of the petrol price per litre at 50 petrol stations produced a sample mean of $\text{\textdollar}1.52$ and a standard deviation of $\text{\textdollar}0.14$.
Based on this sample and using a $z$-value of 1.5, an approximate confidence interval for $\mu$ is
$$E=1.5\times\frac{0.14}{\sqrt{50}}\approx0.0297$$
$$(1.52-0.03,\ 1.52+0.03)=(1.49,\ 1.55)$$
The heights of all students at a school were measured. A mean height of 157.0 cm was calculated from this data.
A random sample of 35 students from this school was selected. The mean height of this sample was 159.7 cm with a standard deviation of 8.7 cm.
The smallest confidence level that could be used to produce a confidence interval that contains $\mu$, based on this sample, is
157.0 cm is the mean of every student at the school, so $\mu=157.0$. The interval has to reach from 159.7 down to 157.0.
$$z\times\frac{8.7}{\sqrt{35}}\ge159.7-157.0$$
$$z\ge\frac{2.7}{1.4706}\approx1.836$$
$$P(-1.836\lt Z\lt1.836)\approx0.934$$
So the confidence level must be at least 93.4%. Of the options, 85% and 90% are too small, and 95% is the smallest that works.
The time taken by the Year 7 students at a particular school to complete a standardised test is known to be normally distributed. A researcher claims that the population mean is 8.2 minutes.
The mean time taken to complete this test by a sample of 10 of these students is 8.1 minutes with a standard deviation of 1.2 minutes.
The 95% confidence interval for $\mu$ based on this sample is
$$E=1.96\times\frac{1.2}{\sqrt{10}}\approx0.744$$
$$(8.1-0.744,\ 8.1+0.744)\approx(7.36,\ 8.84)$$
The interval is centred on the sample mean, 8.1. (B) and (D) are centred on the claimed 8.2, and (C) divides by $10$ in place of $\sqrt{10}$.
Part 3
Work backwards from the interval $(a,b)$. Its centre is $\bar x$ and half its width is $E$:
$$\bar x=\frac{a+b}{2}\qquad E=\frac{b-a}{2}$$
Then solve $E=z\frac{s}{\sqrt n}$ for whichever of $z$, $s$ or $n$ is missing. Round a sample size to a whole number.
Practise
An approximate confidence interval for a population mean is calculated based on a sample size of 16 and is found to have width, $w$.
A second confidence interval is calculated from another sample with the same sample standard deviation.
Given that the same $z$-value was used for both intervals and the width of the second interval is $2w$, what is the size of the second sample?
The width is $2z\frac{s}{\sqrt n}$. With $z$ and $s$ unchanged, doubling the width means halving $\sqrt n$.
$$\sqrt{n_2}=\frac{\sqrt{16}}{2}=2$$
$$n_2=4$$
The masses (grams) of a random sample of 40 chocolate muffins produced by a local bakery are recorded.
Using the sample standard deviation of 0.845 grams, an approximate confidence interval for the population mean mass of chocolate muffins produced by this bakery is $(149.68,\ 150.12)$ grams.
The $z$-value used in this calculation is
$$E=\frac{150.12-149.68}{2}=0.22$$
$$z\times\frac{0.845}{\sqrt{40}}=0.22$$
$$z\approx1.65$$
This is the $z$-value for a 90% interval.
$T$ is a random variable. A random sample of four values of $T$ is collected and used to produce an approximate confidence interval for the population mean of $(3.3,\ 4.1)$.
Given that three of the sample values are 3.4, 3.6 and 3.9, the remaining sample value is
The centre of the interval is the sample mean.
$$\bar t=\frac{3.3+4.1}{2}=3.7$$
$$3.4+3.6+3.9+t=4\times3.7=14.8$$
$$t=3.9$$
The life span of batteries manufactured by a company is assumed to be normally distributed with an unknown mean and standard deviation.
A supervisor at the company randomly selects $n$ batteries and uses the life spans from this sample to calculate an approximate 95% confidence interval for the population mean of $(2321.4,\ 2423.6)$ hours.
The standard deviation of the life spans of batteries in this sample is 125 hours.
Part a)
$$\bar x=\frac{2321.4+2423.6}{2}=2372.5\text{ hours}$$
Part b)
Use the upper end of the interval.
$$2372.5+1.96\times\frac{125}{\sqrt n}=2423.6$$
$$\sqrt n\approx4.79$$
$$n\approx22.98$$
The sample used 23 batteries.
Part c)
Yes, the assumption is required.
The sample size is less than 30, so the distribution of the sample mean can't be assumed to be normal unless the population is normal.
An article claims that the mean starting salary of graduates in Australia is currently $\text{\textdollar}64\,800$ with a standard deviation of $\text{\textdollar}4500$.
To check the validity of this claim, an employment agent intends to collect data on the starting salaries of a random sample of 360 graduates.
From the data, the agent calculates a confidence interval for the population mean starting salary of $(\text{\textdollar}64\,589,\ \text{\textdollar}65\,811)$.
Part a)
$n=360\ge30$, so $\bar X$ is approximately normal.
$$\mu_{\bar X}=64\,800$$
$$\sigma_{\bar X}=\frac{4500}{\sqrt{360}}\approx237.2$$
$$P(64\,000\le\bar X\le65\,000)\approx0.80$$
Part b)
$$\bar x=\frac{64\,589+65\,811}{2}=65\,200$$
Part c)
The claimed mean of $\text{\textdollar}64\,800$ lies inside the interval $(\text{\textdollar}64\,589,\ \text{\textdollar}65\,811)$.
So the article's claim is reasonable.
Part a) is about the sample mean before any data is collected, so it uses the claimed $\mu$ and $\sigma$. Parts b) and c) use the interval the agent calculated afterwards.
The mass of a population of elephants is known to be normally distributed.
A biologist randomly selects a number of elephants from this population and measures their masses. The mean mass of the sample is 5206 kg with a standard deviation of 356 kg.
The biologist uses the data to calculate a 90% confidence interval for the population mean mass of $(5159.1,\ 5252.9)$ kg.
Determine a 99% confidence interval for the population mean mass based on the same data.
The sample size isn't given. Find it from the 90% interval, then use it for the 99% interval.
$$\bar x=5206,\quad s=356,\quad z_{90\%}=1.645$$
$$5206-1.645\times\frac{356}{\sqrt n}=5159.1$$
$$n\approx155.9$$
The sample has 156 elephants.
$$5206\pm2.576\times\frac{356}{\sqrt{156}}$$
$$\text{CI}\approx(5132.6,\ 5279.4)\text{ kg}$$
Define each value before you use it, and link the steps with short statements such as "solving for $n$".
The guide accepts $n=155$ or $157$ as well, and the matching intervals. The 99% interval is wider than the 90% one and has the same centre, which is a quick check on your answer.
The full walkthrough of 2022 Paper 2, Question 17 goes through it mark by mark.
Practise
A variable, $X$, is assumed to be normally distributed with $\mu=24.311$ and $\sigma=5.102$.
Two 90% confidence intervals for $\mu$ were calculated from two different random samples from $X$, with the second sample being smaller than the first sample by 60.
Both confidence intervals were calculated using the population standard deviation rather than their respective sample standard deviations.
The confidence interval produced from the first sample was $(23.560,\ 25.498)$.
Determine the probability that the confidence interval produced from the second sample overlaps the confidence interval from the first sample.
Start with the first interval. Its centre is the first sample mean, and half its width is the margin of error.
$$\bar x_1=\frac{23.560+25.498}{2}=24.529$$
$$E_1=25.498-24.529=0.969$$
$$1.645\times\frac{5.102}{\sqrt{n_1}}=0.969$$
$$n_1\approx75$$
$$n_2=75-60=15$$
$$E_2=1.645\times\frac{5.102}{\sqrt{15}}\approx2.167$$
The second interval overlaps the first when its centre is within $E_2$ of the first interval.
$$23.560-2.167\le\bar x_2\le25.498+2.167$$
$$21.393\le\bar x_2\le27.665$$
For samples of 15, the sample mean is normal because $X$ is normal.
$$\bar X_2\sim N\left(24.311,\ \left(\frac{5.102}{\sqrt{15}}\right)^2\right)$$
$$P(21.393\le\bar X_2\le27.665)\approx0.98$$
The second interval is about twice as wide as the first, because the second sample is much smaller. This question is worked step by step in the complex unfamiliar collection.
The height of Year 9 students at a school is assumed to be normally distributed with a population mean height of $\mu$ cm.
A teacher at the school measured the height of all the students in her Year 9 class. This data was used to calculate an approximate 95% confidence interval for $\mu$ of $(163.7,\ 166.9)$ cm.
The teacher repeated the procedure using data from another Year 9 class. Although this class had the same number of students, its data produced an approximate 95% confidence interval for $\mu$ of $(167.8,\ 172.4)$ cm.
Using the same data, the teacher recalculated the approximate confidence intervals for $\mu$ for each class using a confidence level of $x\%$. She observed that the upper bound of the confidence interval from her Year 9 class now equalled the lower bound of the confidence interval from the other Year 9 class.
Determine the value of $x$. Give your answer rounded to one decimal place.
$$z_{95\%}=1.96$$
$$\bar x_1=\frac{163.7+166.9}{2}=165.3$$
$$\bar x_2=\frac{167.8+172.4}{2}=170.1$$
Both classes have the same $n$, so write each margin of error in terms of $\frac{s}{\sqrt n}$.
$$1.96\frac{s_1}{\sqrt n}=1.6$$
$$\frac{s_1}{\sqrt n}\approx0.816$$
$$1.96\frac{s_2}{\sqrt n}=2.3$$
$$\frac{s_2}{\sqrt n}\approx1.173$$
With a new $z$-value, $z_x$, the upper end for the teacher's class equals the lower end for the other class.
$$165.3+0.816z_x=170.1-1.173z_x$$
$$z_x=\frac{4.8}{1.990}\approx2.412$$
$$P(-2.412\lt Z\lt2.412)\approx0.984$$
So $x=98.4$.
The guide also accepts 98.3 from early rounding. You never need $n$ or either $s$ on its own. This question is worked step by step in the complex unfamiliar collection.
The mass of a certain species of kangaroo is known to be normally distributed with a mean mass of $\mu$ kg and standard deviation of $\sigma$ kg.
When one of the kangaroos is randomly selected, the probability that its mass is greater than 83.2 kg is 0.145.
When a sample of 12 kangaroos is randomly selected, the probability that the sample mean mass is less than 74.1 kg is 0.079.
A 90% approximate confidence interval for $\mu$ is calculated using a random sample of $n$ of the kangaroos that has a sample mean mass of 79.1 kg and a sample standard deviation equal to $\sigma$.
Determine the possible range of values that $n$ could have been, given that the confidence interval did not contain $\mu$.
Use the two probabilities to find $\mu$ and $\sigma$.
$$P(X\gt83.2)=0.145$$
$$\frac{83.2-\mu}{\sigma}=1.058\quad(1)$$
$$P(\bar X\lt74.1)=0.079$$
$$\frac{74.1-\mu}{\sigma/\sqrt{12}}=-1.412\quad(2)$$
$$\mu\approx76.63,\quad\sigma\approx6.21$$
$\bar x=79.1$ is above $\mu$, so the interval misses $\mu$ when its lower end is above 76.63. Find where the lower end equals $\mu$.
$$79.1-1.645\times\frac{6.21}{\sqrt n}=76.63$$
$$n\approx17.1$$
A larger $n$ makes the interval narrower, which pushes its lower end up. So the interval misses $\mu$ when $n\ge18$, with $n$ a whole number.
Label each equation and say which sample it comes from, so the marker can follow your steps.
The guide uses $z=1.64$, which gives the same range. This question is worked step by step in the complex unfamiliar collection.
Game plan
For the whole topic on one page, see Game plan 9 · Statistical inference in the External Game plan.
Practice
Ten questions of my own, worth 52 marks in total. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to. Check whether each one is Technology-free or Technology-active before you start.
A company produces cans of energy drink. Each can is labelled as containing 80 mg of caffeine.
A food inspector selects a random sample of 64 cans to assess whether the caffeine content stated on the label is being satisfied in production. The mean caffeine content of the sample is 78.5 mg with a sample standard deviation of 4.0 mg.
Food standards require that the mean caffeine content must not exceed the labelled amount.
Part a)
$n=64$, $\overline{x}=78.5$, $s=4.0$, $z=2$
$\displaystyle \left(\overline{x}-z\frac{s}{\sqrt{n}},\ \overline{x}+z\frac{s}{\sqrt{n}}\right)=\left(78.5-2\times\frac{4}{\sqrt{64}},\ 78.5+2\times\frac{4}{\sqrt{64}}\right)$
$=(78.5-1,\ 78.5+1)=(77.5,\ 79.5)$ mg
Part b)
The labelled caffeine content of 80 mg lies above the upper value of the confidence interval, so the whole interval lies below 80 mg.
Current production meets the food standards.
The standard here is a maximum, so an interval below 80 mg is good news. Saying production fails because 80 mg is outside the interval loses the second mark of part b).
A council claims that the mean time its residents spend travelling to work each day is 35 minutes. Travel times are assumed to be normally distributed.
A researcher selects a random sample of 120 residents and uses their travel times to calculate an approximate 95% confidence interval for the population mean travel time of $(31.62,\ 34.18)$ minutes.
Part a)
$\displaystyle \overline{x}=\frac{31.62+34.18}{2}=32.9$ minutes
Part b)
$\displaystyle E=\frac{34.18-31.62}{2}=1.28$
$\displaystyle 1.96\times\frac{s}{\sqrt{120}}=1.28\Rightarrow s=\frac{1.28\sqrt{120}}{1.96}\approx7.15$ minutes
Part c)
The claimed mean of 35 minutes lies above the upper value of the confidence interval, $34.18$ minutes.
The council's claim is not reasonable.
Using the full width, 2.56, as $E$ gives $s\approx14.31$ and loses the second mark of part b).
The time taken to fully charge the battery of a particular model of electric scooter is assumed to be normally distributed. The manufacturer claims that the mean charging time is 4.5 hours.
The following data represents the charging time (hours) for a random sample of 10 of these scooters.
| $4.2$ | $4.9$ | $4.6$ | $4.1$ | $4.8$ | $4.4$ | $5.0$ | $4.3$ | $4.7$ | $4.5$ |
The standard deviation of this data is calculated to be 0.303 hours.
Part a)
Using technology: $\overline{x}=4.55$ hours
Part b)
$n=10$, $s=0.303$, $z=1.645$
Using technology: CI(90%) $=(4.39,\ 4.71)$ hours
The claimed mean charging time of 4.5 hours lies within the confidence interval.
The manufacturer's claim is reasonable.
Part c)
A 99% interval from the same data is wider than the 90% interval and has the same centre, so it also contains 4.5 hours. The evaluation would not change.
Calculating the 99% interval, $(4.30,\ 4.80)$ hours, is also fine for part c). Using your calculator's $s\approx0.3028$ gives the same interval.
The time spent on homework each night by a random sample of 36 students is recorded. Using a $z$-value of 2, an approximate confidence interval for the population mean time is calculated to be $(41.2,\ 44.8)$ minutes.
A second approximate confidence interval is calculated from the same sample using a $z$-value of 2.5.
Part a)
$\displaystyle \overline{x}=\frac{41.2+44.8}{2}=43$ minutes
Part b)
$\displaystyle E=\frac{44.8-41.2}{2}=1.8$
$\displaystyle 2\times\frac{s}{\sqrt{36}}=1.8\Rightarrow\frac{s}{3}=1.8\Rightarrow s=5.4$ minutes
Part c)
$\displaystyle E=2.5\times\frac{5.4}{6}=2.25$; interval $(43-2.25,\ 43+2.25)=(40.75,\ 45.25)$ minutes
Part d)
The interval using $z=2.5$, because a larger $z$-value gives a wider interval that captures $\mu$ for a greater proportion of samples.
You can do part c) without $s$: $E=1.8\times\frac{2.5}{2}=2.25$. Using $n$ in place of $\sqrt n$ in part b) gives $s=32.4$, but part c) can still follow through.
Two inspectors at a cereal factory independently select random samples of boxes from the same production line and record the mass of cereal in each box. The population mean mass of cereal in a box is $\mu$ grams.
| Sample size | Sample mean (g) | Sample standard deviation (g) | |
| Inspector A | $45$ | $507.8$ | $6.3$ |
| Inspector B | $80$ | $504.1$ | $7.0$ |
Each week for 40 weeks, both inspectors repeat this procedure using new, independent random samples.
Part a)
Using technology: A: $(505.96,\ 509.64)$ g
Using technology: B: $(502.57,\ 505.63)$ g
Part b)
The intervals do not overlap: the upper value of B's interval ($505.63$) is below the lower value of A's interval ($505.96$). $\mu$ is a single fixed value, so it cannot lie in both intervals.
Part c)
Width is proportional to $\dfrac{s}{\sqrt{n}}$: A: $\dfrac{6.3}{\sqrt{45}}\approx0.94$; B: $\dfrac{7.0}{\sqrt{80}}\approx0.78$. B's larger sample size more than offsets its larger standard deviation.
Part d)
$P(\text{both contain }\mu)=0.95\times0.95=0.9025$
Expected number $=40\times0.9025=36.1$ weeks
Working out $40\times0.95=38$ for one inspector only loses the mark in part d). In part b), saying the inspectors made a mistake doesn't get the mark.
The masses of eggs produced at a farm are assumed to be normally distributed. A random sample of 64 eggs has a mean mass of 58.4 g and a sample standard deviation of 4.0 g.
A second random sample, of 256 eggs, is planned.
A shopper states that approximately 90% of eggs from this farm have a mass between 57.58 g and 59.22 g.
Part a)
Using technology: CI(90%) $=(57.58,\ 59.22)$ g
Part b)
$58.4\pm1.645\times4.0=(51.82,\ 64.98)$ g
Part c)
The confidence interval width $2\times1.645\frac{s}{\sqrt n}$ halves (from $1.65$ g to $0.82$ g), because $\sqrt{256}=2\sqrt{64}$.
The interval for individual eggs, $\overline{x}\pm1.645s$, does not depend on $n$, so its width (about $13.16$ g) is unchanged.
Part d)
$(57.58,\ 59.22)$ estimates the population mean, not individual masses. About 90% of eggs lie in the much wider interval $(51.82,\ 64.98)$, so the shopper's statement is not reasonable.
Only about 16% of single eggs have a mass in $(57.58,\ 59.22)$, which is why the interval in part b) is so much wider. Saying the interval for single eggs also halves loses the second mark of part c).
A teacher uses software to select five random samples, of different sizes, from a normally distributed population with mean $\mu=50$ and standard deviation $\sigma=6$. For each sample, an approximate confidence interval for $\mu$ is calculated using
$\left(\overline{x}-2\,\frac{\sigma}{\sqrt{n}},\ \overline{x}+2\,\frac{\sigma}{\sqrt{n}}\right).$
The five intervals are shown.
A sixth random sample, of size 64, is selected from the same population and an interval is calculated in the same way.
Part a)
Samples 1, 3 and 4 contain $\mu=50$: 3 intervals.
Part b)
Sample 4: width $=51.5-49.5=2$, so $E=1$
$\displaystyle 2\times\frac{6}{\sqrt{n}}=1\Rightarrow\sqrt{n}=12\Rightarrow n=144$
Part c)
$\displaystyle E=2\times\frac{6}{\sqrt{64}}=1.5$
The interval contains $\mu$ when $|\overline{x}-50|<1.5$, that is, $48.5<\overline{x}<51.5$.
Writing $48.5\le\overline{x}\le51.5$ is also accepted in part c). Using the width, 2, as $E$ in part b) gives $n=36$ and loses the second mark.
A caf\'e chain claims that the mean time customers wait for a coffee is 6.0 minutes. Waiting times are assumed to be normally distributed.
A random sample of 50 customers has a mean waiting time of 6.42 minutes with a standard deviation of 1.35 minutes.
Part a)
Using technology: CI(95%) $=(6.05,\ 6.79)$ minutes
Part b)
$\displaystyle 6.42-z\times\frac{1.35}{\sqrt{50}}=6.0$
$z\approx2.20$
Level $=P(-2.20<Z<2.20)\approx0.972$, that is, $97.2\%$
Part c)
Levels of confidence greater than $97.2\%$ (e.g.\ 99%)
Using one tail only, $P(Z\lt2.20)\approx98.6\%$, loses the third mark of part b). Part c) can still follow through.
A water authority estimates the mean daily water use of households in a town. From a random sample of $n$ households, and using a $z$-value of 2, an approximate confidence interval for the population mean daily water use is calculated to be $(232,\ 248)$ litres.
A further 28 households are then randomly selected and added to the sample. The sample mean and the sample standard deviation are unchanged, and an approximate confidence interval calculated from the enlarged sample, again using a $z$-value of 2, is $(234,\ 246)$ litres.
Determine the value of $n$ and the sample standard deviation.
Both intervals are centred at $\overline{x}=240$.
Original: $E_1=8$; enlarged: $E_2=6$
$\displaystyle 2\times\frac{s}{\sqrt{n}}=8\quad\dots(1)$
$\displaystyle 2\times\frac{s}{\sqrt{n+28}}=6\quad\dots(2)$
(1)$\div$(2): $\displaystyle \sqrt{\frac{n+28}{n}}=\frac43\Rightarrow\frac{n+28}{n}=\frac{16}{9}\Rightarrow9n+252=16n\Rightarrow n=36$
$\displaystyle \frac{2s}{6}=8\Rightarrow s=24$ litres
$n=36$ and $s=24$ litres
Check: $\frac{2\times24}{\sqrt{64}}=6$. Using $\sqrt{28}$ in equation (2) treats 28 as the new sample size and loses the third mark.
An environmental laboratory monitors the concentration of nitrate in a river. Each week, it collects a random sample of 60 water specimens and reports an approximate 99% confidence interval for the population mean nitrate concentration for that week.
To reduce costs, the laboratory manager proposes reporting approximate 90% confidence intervals instead, using the smallest weekly sample size that keeps the margin of error no larger than it is now. Assume that the sample standard deviation is unchanged.
The manager claims that this change would more than halve the weekly sample size, and that over a year of 52 weeks it would be expected to produce fewer than 4 additional intervals that do not contain the population mean.
Evaluate the reasonableness of the manager's claim. Justify your decision using mathematical reasoning.
Current margin of error: $\displaystyle E=2.576\frac{s}{\sqrt{60}}$
Require $\displaystyle 1.645\frac{s}{\sqrt{n}}\le2.576\frac{s}{\sqrt{60}}$
$\displaystyle \sqrt{n}\ge\frac{1.645\sqrt{60}}{2.576}\Rightarrow n\ge60\left(\frac{1.645}{2.576}\right)^2\approx24.47$, so $n=25$
$25<30$, so the weekly sample size would be more than halved; this part of the claim is correct.
Expected number of intervals not containing $\mu$ over 52 weeks:
99%: $52\times0.01=0.52$; 90%: $52\times0.10=5.2$
Additional: $5.2-0.52=4.68$
$4.68>4$, so the change would be expected to produce more than 4 additional intervals that do not contain the population mean.
The manager's claim is not reasonable.
Using 1.65 and 2.58 gives $n\ge24.54$, so the answer is still 25. Forgetting to square the ratio of $z$-values gives $n\ge38.32$ and loses the second mark. Checking only the sample size part of the claim loses the last three marks.
Keep practising
2025 Paper 2, Question 17, 2023 Paper 2, Question 19 and 2020 Paper 2, Question 18 are the hardest confidence interval questions so far. All three are in the complex unfamiliar collection, along with the others.
Past questions and marking-guide steps are from the 2020–2025 QCAA Specialist Mathematics external assessments, © State of Queensland (QCAA) 2020–2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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