Unit 4 · Topic 4 · Interval estimates for proportions
It is rarely practical to survey a whole population, so instead we take a sample and use it to estimate what the population looks like. This topic and Topic 5 are both about that process. Here you will learn how to take a fair sample, how to work out a sample proportion, and how sample proportions behave from one sample to the next. Tick off each skill once you can do it without help.
Part 1
The language matters here. A number that describes the whole population is a parameter. The same kind of number worked out from a sample is a statistic, and it is our estimate of the parameter.
| Measure | Population | Sample |
|---|---|---|
| General term | Parameter | Statistic |
| Size | $N$, the population size | $n$, the sample size |
| Proportion | $p$, the population proportion | $\hat{p}$, the sample proportion |
A sample is only useful if it looks like the population. If the way you choose it favours some people over others, the sample is biased. There are three kinds of sample you need to recognise:
Every member of the population has the same chance of being chosen, for example by using a random number generator on a full list. This is the gold standard.
Every $k$th member is chosen, such as every 100th name on a list. It is convenient and often close enough to random, but the order of the list can still build in bias.
People opt in, such as by clicking a survey link. The people who choose to answer usually feel strongly, so this is almost never representative.
Also watch where a sample is taken. Asking people at a dog park about a new dog park will over-represent dog owners, however randomly you pick the day.
The syllabus also expects you to read graphs of real or simulated samples from the distributions you know: uniform, Bernoulli, binomial and normal. A sample's histogram will not match its distribution exactly, but the shape should be recognisable, and it gets closer as the sample gets bigger. You will see this happen in Part 3.
Watch a sample go wrong
Practise
A council wants to survey residents about a new dog park. Which sampling method would best minimise bias in the survey?
(A) is systematic, but only reaches people near an existing dog park. (B) is self-selected. (C) picks the day at random, but everyone asked is already at a dog park. Only (D) gives every resident the same chance of being chosen.
A survey plans to draw conclusions based on a random sample of 1% of Queensland's adult population. To be regarded as a random sample, every
This is the definition of a random sample. (A) describes a systematic sample and (B) a self-selected one. (C) sounds sensible, but matching subgroups is a different method, and it does not make the sample random.
Part 2
The sample proportion, $\hat{p}$ (read "p hat"), is the fraction of the sample that has the feature you are interested in. It is usually worth a mark on its own, and later parts of the question build on it.
$$\hat{p}=\frac{\text{number in the sample with the feature}}{\text{sample size}}=\frac{X}{n}$$
When the data come from a graph or table, add up every group to get $n$ first. It is easy to divide by the wrong total.
Practise
For the written-answer questions, the solutions show where each mark goes, based on the QCAA marking guides. Many marks allow follow-through: if an earlier step is wrong but you use your wrong answer correctly, you can still get the later marks. That is another reason to show every step.
A school investigated how many hours students sleep per night. To obtain data, a random sample of students was surveyed. The results are shown.
Part (b), worth 1 mark, asks for a confidence interval. That is in Topic 5.
Read every column, then add them all to find the sample size:
$$n=15+19+20+27+5=86$$
"Seven or more" includes 7, so count the 7, 8 and 9 hour columns:
$$X=20+27+5=52$$
$$\hat{p}=\frac{52}{86}\approx0.605$$
With 3 marks on offer, show the column readings, the total and the fraction.
A researcher found that 17 out of 50 randomly selected people had used public transport in the past week.
Parts (b) and (c), worth 3 marks, use a confidence interval. They are in Topic 5.
$$\hat{p}=\frac{17}{50}=0.34$$
Part 3
Take a different sample and you will almost certainly get a different $\hat{p}$. That makes $\hat{p}$ a random variable, with its own distribution. Values near the population proportion $p$ are the most likely, and values further away in either direction get less and less likely.
For large samples, the distribution of $\hat{p}$ is approximately normal, centred on $p$. Bigger samples also give a narrower spread, so a single $\hat{p}$ from a big sample is more likely to land close to $p$. How good the normal approximation is depends on both $n$ and $p$: it is worst when the sample is small and $p$ is close to 0 or 1.
Try it yourself
Each bar shows how often a value of $\hat{p}$ came up across 1000 random samples. The black curve is the normal distribution that Part 4's formulas predict. Try a small sample with $p$ near 0 or 1, then a big one.
Practise
A box contains 100 balls, each of which is either green, blue or red. A teacher asks students to select a ball at random, record its colour and return it to the box. This process is repeated until each student has selected and recorded the colours of a sample of balls. Each student chooses their own sample size.
The sample size, $n$, and sample proportion of green balls, $\hat{p}$, are shown for four students.
| $n$ | $\hat{p}$ |
|---|---|
| 10 | $\frac{1}{2}$ |
| 15 | $\frac{1}{3}$ |
| 20 | $\frac{1}{4}$ |
| 80 | $\frac{1}{5}$ |
Which sample proportion is expected to best approximate the proportion of green balls in the box?
Every student sampled randomly, so none of the methods is biased. The difference is sample size. The standard deviation of $\hat{p}$ shrinks as $n$ grows, so the biggest sample, $n=80$, is the one most likely to be close to $p$.
Part 4
Both formulas are on your formula sheet. They use the population proportion $p$. They do not use $\hat{p}$.
$$\mu_{\hat{p}}=p$$
$$\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$$
Once you have these, $\hat{p}$ is just a normal distribution, and everything from Topic 3.2 works. You can use normal CD directly with $\mu_{\hat{p}}$ and $\sigma_{\hat{p}}$, or standardise first:
$$z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$$
In a town, 40% of households have solar panels. Samples of 150 households are taken. Find the mean and standard deviation of the sample proportion.
$$\mu_{\hat{p}}=0.4\qquad\sigma_{\hat{p}}=\sqrt{\frac{0.4(0.6)}{150}}=\sqrt{0.0016}=0.04$$
The population size does not appear in either formula. If a question gives you $N$, you can usually ignore it.
Watch the spread shrink
Practise
An online retailer claims that 90% of all orders are shipped within 12 hours of being received. On a particular day, 121 orders were received and 102 orders were shipped within 12 hours.
The distribution of the sample proportion of all orders that are shipped within 12 hours of being received on any day is approximately normal.
(a) $\hat{p}=\frac{102}{121}\approx0.843$
(b) Assume the claim is true, so $p=0.9$ and $n=121$:
$$\mu_{\hat{p}}=0.9$$
$$\sigma_{\hat{p}}=\sqrt{\frac{0.9(0.1)}{121}}\approx0.02727$$
Normal CD with a very small lower bound, an upper bound of 0.85, and that mean and standard deviation:
$$P(\hat{p}<0.85)\approx0.0334$$
(c) If the claim were true, a day as bad as this would happen only about 3% of the time. That is unlikely, and on this day only 84.3% were shipped in time, which is even lower than 85%. So the data cast doubt on the claim. It is one day of data, though, so it does not prove the claim is wrong.
For the 2 marks in (c), quote the probability and then make a decision about the claim in context.
Part 5
Sometimes you are given the mean and standard deviation of $\hat{p}$ and asked for the sample size. The mean gives you $p$. Then rearrange the standard deviation formula by squaring both sides:
$$\sigma_{\hat{p}}^{2}=\frac{p(1-p)}{n}\ \Rightarrow\ n=\frac{p(1-p)}{\sigma_{\hat{p}}^{2}}$$
The same equation can be solved for $p$ instead, if you know $n$ and the standard deviation. That gives a quadratic, so expect two answers.
Practise
The distribution of a certain sample proportion has a mean of 0.70 and a standard deviation of 0.02. Determine the sample size.
$$n=\frac{0.7(0.3)}{0.02^{2}}=\frac{0.21}{0.0004}=525$$
The distribution for a sample proportion $\hat{p}$ has a mean of 0.15 and a standard deviation of 0.0345.
The sample size is
$$n=\frac{0.15(0.85)}{0.0345^{2}}=\frac{0.1275}{0.00119025}\approx107.1$$
The standard deviation was rounded, so $n$ does not come out as a whole number. The nearest option is 107. Round to the nearest whole number only at the end.
At council meetings in a particular town, new proposals are only discussed if more than 80% of the community are in favour of the proposal.
To discover community opinion on a new bus route proposal, the council conducted several surveys, each with a sample size of 120. The distribution of the sample proportions from the surveys had a standard deviation of 0.04.
Make a justified decision as to whether the new bus route proposal would be discussed at a council meeting.
This time $n$ and the standard deviation are known, and $p$ (the proportion in favour) is the unknown:
$$0.04=\sqrt{\frac{p(1-p)}{120}}\ \Rightarrow\ p(1-p)=120(0.0016)=0.192$$
$$p^{2}-p+0.192=0\ \Rightarrow\ p\approx0.259\ \text{ or }\ p\approx0.741$$
Check both solutions against the context. Either way, the proportion in favour is at most about 74%, which is below the 80% needed.
The full walkthrough of 2024 Paper 2, Question 16 goes through it mark by mark.
Game plan
Practice
Eight questions of my own, covering every skill. 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.
A school wants to survey its students about changing the start time of the school day. Which sampling method would best minimise bias?
Look for the option where every student has the same chance of being chosen.
Option (A) only reaches early bus travellers, who may feel differently about start times. Option (B) is self-selected, so students with strong opinions are more likely to reply. Option (C) is one class from one year level.
A council wants to know whether residents support a new skate park. It posts a poll on its social media page and uses the replies. Explain why this sample is likely to be biased, and describe a better way to choose the sample.
Say why it is biased in the context of the question. Just writing "self-selected" usually is not enough.
A survey asked 80 people how many pets they own. 22 had none, 30 had one, 18 had two, and 10 had three or more. Determine the sample proportion of people with at least two pets.
$$18+10=28$$
$$\hat{p}=\frac{28}{80}=0.35$$
As the sample size increases, the distribution of the sample proportion
The mean of $\hat{p}$ is always $p$, so it does not move. The standard deviation $\sqrt{\frac{p(1-p)}{n}}$ gets smaller as $n$ grows, and the shape gets closer to normal. That is what the "Watch the spread shrink" animation shows.
At a large school, 10% of students are left-handed. Random samples of 100 students are taken, and the proportion of left-handed students in each sample is recorded. Determine the mean and standard deviation of the distribution of the sample proportion.
$$\mu_{\hat{p}}=p=0.1$$
$$\sigma_{\hat{p}}=\sqrt{\frac{0.1\times0.9}{100}}=\sqrt{0.0009}=0.03$$
The distribution of a certain sample proportion has a mean of 0.8 and a standard deviation of 0.05. Determine the sample size.
$$0.05=\sqrt{\frac{0.8\times0.2}{n}}$$
$$0.0025=\frac{0.16}{n}\quad\Rightarrow\quad n=\frac{0.16}{0.0025}=64$$
In a large city, 30% of adults cycle to work. A random sample of 200 adults is taken. Determine the probability that more than 35% of the sample cycle to work.
$$\mu_{\hat{p}}=0.3\qquad\sigma_{\hat{p}}=\sqrt{\frac{0.3\times0.7}{200}}\approx0.0324$$
$$P(\hat{p}>0.35)\approx0.061$$
A pollster expects about half of voters to support a policy. Determine the smallest sample size that gives a standard deviation of the sample proportion of at most 0.02.
$$\sqrt{\frac{0.5\times0.5}{n}}\le0.02$$
$$\frac{0.25}{n}\le0.0004\quad\Rightarrow\quad n\ge\frac{0.25}{0.0004}=625$$
A bigger sample gives a smaller standard deviation, so 625 is the smallest size that works.
Past questions and marking-guide steps are from the 2020–2025 QCAA Mathematical Methods 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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