Question 8
In a survey, 80 respondents exercised daily, while 120 did not. When calculating the approximate 95% confidence interval for the proportion of people who exercise daily, the margin of error is
That’s it. Step through the hints below to see why. In 2022, 67.1% of Queensland students chose (A).
Close. $\hat p=0.4$ and $n=200$ are right, but 0.95 is the confidence level, not the multiplier. The multiplier is the $z$-value that leaves 95% of the standard normal curve in the middle, $z=1.96$. Using $z=0.95$ only covers about 66%. In 2022, 6.2% of Queensland students chose (B).
Not quite. $\tfrac{80}{120}\approx0.67$ compares the people who exercise with the people who don’t. The proportion who exercise is out of everyone surveyed: $n=80+120=200$, so $\hat p=\tfrac{80}{200}=0.4$. In 2022, 24.2% of Queensland students chose (C).
Not quite. This has two slips. $\tfrac{80}{120}$ isn’t the proportion who exercise, because the sample is all 200 respondents. And 0.95 is the confidence level, not the multiplier, which is $z=1.96$. In 2022, 2.2% of Queensland students chose (D).
$z=1.96$ leaves 95% of the curve in the middle. That is the multiplier for 95% confidence.
$z=0.95$ only leaves about 65.8% in the middle. 0.95 is the area you want in the middle, not the $z$-value that gives it.
$z=2.58$ leaves 99% in the middle. That is the multiplier for 99% confidence, so the interval would be wider than this question needs.
This $z$ leaves % in the middle. For 95% confidence you need $z=1.96$.
$\hat p=\tfrac{80}{200}$ counts everyone surveyed, so $n=200$.
$\tfrac{80}{120}$ compares exercisers with non-exercisers. The sample is everyone surveyed, so the proportion should be out of 200, not 120.
This is option (A).
This is option (B).
This is option (C).
This is option (D).
The sample is everyone who answered: $80+120=200$. The proportion who exercise daily is out of all of them.
$n=200$$\hat p=\dfrac{80}{200}=0.4$
That rules out (C) and (D).
The multiplier is the $z$-value that leaves 95% of the standard normal curve in the middle, with 2.5% in each tail.
$P(-1.96\lt Z\lt1.96)$$\approx0.95$
So $z=1.96$. The 0.95 is the area, not the multiplier. That rules out (B).
Put the three numbers into the margin of error formula.
$E=z\sqrt{\dfrac{\hat p(1-\hat p)}{n}}$$=1.96\sqrt{\dfrac{0.4(1-0.4)}{200}}$
That is option (A). You don’t need to evaluate it, but it comes to about 0.068.
$n=80+120=200$$\hat p=\dfrac{80}{200}=0.4$$z=1.96$
$$E=1.96\sqrt{\frac{0.4(1-0.4)}{200}}$$
The sample is all 200 respondents, and 95% confidence uses the multiplier $z=1.96$, not 0.95.
How Queensland students answered in 2022
From the QCAA Mathematical Methods subject report for 2022. 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 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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