Question 3
A sample of size $n$ can be used to obtain a sample proportion $\hat{p}$.
An approximate margin of error for the population proportion can be obtained using the formula
$$E=z\sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}$$
If the level of confidence is increased from 95% to 99%, then
Not quite. $E$ is $z$ multiplied by a part that doesn’t change here. If $z$ went down, $E$ would go down with it. And to be more confident you need to capture more of the normal curve, so $z$ goes up, not down. In 2024, 3.0% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2024, 66.3% of Queensland students chose (B).
Not quite. The two halves agree with each other, but the direction is wrong. For 99% you need to capture more of the normal curve, so the cut-off moves further from the centre. $z$ rises from 1.96 to about 2.58. In 2024, 7.8% of Queensland students chose (C).
Not quite. $z$ does increase, but $E$ is $z$ times $\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$, which stays the same. So a bigger $z$ makes a bigger $E$. In 2024, 22.6% of Queensland students chose (D).
This is the 95% level. The middle 95% of the curve lies between $-1.96$ and $1.96$. Go to 99% and watch both $z$ and $E$.
At 99% the shaded middle has to hold more of the curve, so $z$ moves out to about $2.58$. $E$ is $z$ times $0.05$, so the interval gets wider too.
Raise the level and $z$ moves out, so $E$ grows. Lower it and both shrink. They always move the same way, because the square root part of $E$ doesn’t change.
A 95% interval uses the middle 95% of the standard normal curve. A 99% interval needs the middle 99%, so the cut-off has to move further from the centre.
$\displaystyle 95\%:\ z=1.96$$\displaystyle \qquad 99\%:\ z\approx2.58$
So the $z$-value increases. That rules out (A) and (C).
In the formula, $z$ multiplies a part that doesn’t depend on the confidence level.
$$E=z\times\sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}$$
The same sample gives the same $\hat{p}$ and the same $n$, so the square root doesn’t change. A bigger $z$ means a bigger $E$. That rules out (D).
Try the figure’s example, where $\sqrt{\dfrac{0.5\times0.5}{100}}=0.05$.
$\displaystyle 95\%:\ E=1.96\times0.05$$\displaystyle \approx0.098$
$\displaystyle 99\%:\ E\approx2.58\times0.05$$\displaystyle \approx0.129$
Higher confidence costs you a wider interval.
Increasing the level of confidence from 95% to 99% increases $z$ from 1.96 to about 2.58. In $E=z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$, the square root stays the same, so $E$ increases too.
How Queensland students answered in 2024
From the QCAA Mathematical Methods subject report for 2024. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2024 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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