Mathematical Methods · Past QCAA questions All simple familiar questions

2025 · Paper 2 · Simple familiar

Question 13

[4 marks] Technology-active

Topic guidesPremiumUnit 4 Topic 4 · The sample proportion Unit 4 Topic 5 · Calculating a confidence interval

Game planPremium6.1 Confidence intervals · Calculating a value

The question

A school investigated how many hours students sleep per night. To obtain data, a random sample of students was surveyed. The results are shown.

Use the data to determine the sample proportion of students who had seven or more hours of sleep per night.

[3 marks] b)

Use your result from Question 13a) to determine an 80% approximate confidence interval for the proportion of students who get seven or more hours of sleep per night.

[1 mark]

Watch the situation first

Count everyone, then count the ones you want

Watch the graph turn into two counts. Their ratio is the sample proportion, and part b) builds an interval around it.

The denominator is every student surveyed, 86, not only the ones who slept seven hours or more.

Now for the mathematics

Choose the bars that count as seven or more

Turn bars on and off to choose which students count. The number line shows your sample proportion and its confidence interval in orange. Then try a different confidence level and watch the interval change.

Bars that count

Confidence level

These are the students who slept seven hours or more, and this is the 80% interval from part b).

Right bars, but not the interval in part b). More confidence needs a bigger $z$, so the interval gets wider. Part b) asks for 80%.

Choose the bars that count as seven or more hours of sleep.

“Seven or more” includes seven. Without the 20 students on the 7 bar you get $\frac{32}{86}\approx0.372$, which is the proportion for more than seven.

Seven or more hours means every bar from 7 upwards, and none below it. Check that 7, 8 and 9 are all chosen, and that 5 and 6 are not.

The denominator stays 86 whichever bars you choose. The sample is every student surveyed.

QCAA marking guide · 4 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. Part a) is reading and adding. Part b) is one calculation on your calculator.

Step 1 · part a)1 mark

Each bar is a number of students. Read every bar off the vertical axis and add them. This is the size of the sample.

$$\begin{aligned}n&=15+19+20+27+5\\&=86\end{aligned}$$

Marker: correctly determines the total number of students in the sample

Step 2 · part a)1 mark

“Seven or more” includes the 7 bar, so add the bars at 7, 8 and 9 hours.

$$x=20+27+5=52$$

Marker: correctly determines the total number of students having seven hours of sleep or more

Step 3 · part a)1 mark

The sample proportion is the number with the feature divided by the number in the sample.

$\hat{p}=\dfrac{x}{n}=\dfrac{52}{86}\approx0.605$

Equivalent fractions and any number of decimal places are accepted, such as $\frac{26}{43}$ or 0.61. Follow-through applies if either count in Steps 1 or 2 had a slip.

Marker: determines the sample proportion of students having seven hours of sleep or more

Step 4 · part b)1 mark

Use your calculator’s one-proportion $z$ interval (MENU → STAT → INTR → Z → 1-PROP). Enter a confidence level of 0.8, $x=52$ and $n=86$. The calculator works out $\hat{p}$ from these, so nothing is rounded early.

This is the same as the formula, with $z\approx1.2816$ for 80% confidence. That $z$ leaves 10% in each tail, so it comes from the inverse normal with an area of 0.9 to the left.

$$\begin{aligned}&\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\&=\frac{52}{86}\pm1.2816\\&\quad\times\sqrt{\frac{\frac{52}{86}\left(1-\frac{52}{86}\right)}{86}}\end{aligned}$$

$(0.537,\ 0.672)$

Stating both bounds without brackets is accepted, and so is any number of decimal places. If you use the formula, keep $\frac{52}{86}$ unrounded. Typing in 0.605 moves the upper bound to about 0.6726.

Marker: determines the 80% approximate confidence interval

Putting it all together

Each phrase of the question gave you something

“a random sample of students was surveyed”

The graph is one sample, so the proportion you find is $\hat{p}$, not the population proportion $p$.

“Number of students”

The bar heights are counts, so adding all of them gives $n=86$.

“seven or more hours of sleep”

The bars at 7, 8 and 9, which give $x=52$.

“determine the sample proportion”

Divide the two counts: $\hat{p}=\frac{52}{86}\approx0.605$.

“Use your result from Question 13a)”

Your $\hat{p}$ and $n=86$ go straight into the interval.

“an 80% approximate confidence interval”

Set the confidence level to 0.8, which uses $z\approx1.2816$. The interval is $(0.537,\ 0.672)$.

What makes this simple familiar

Part a) is reading a column graph and dividing one count by another. Part b) is the standard confidence interval for a proportion, which your calculator does in one step once you have $x$ and $n$.

If I had to guess where you might lose marks, it would be leaving out the 7 bar. “Seven or more” includes seven, and dropping it gives 32 students instead of 52. The other risk is the $z$-value in part b). 80% confidence leaves 10% in each tail, so $z\approx1.2816$. Using $z\approx0.8416$, which leaves 20% in one tail, gives an interval that is far too narrow.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and marking-guide steps are from the 2025 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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