Question 17
In a survey of 326 lecturers, 303 said that on at least one occasion a mobile phone had rung in a lecture they were giving.
Determine the sample size required to conduct a follow-up survey that provides 95% confidence that this one-point estimate is correct to within ± 0.02 of the population proportion.
[4 marks]The first survey gives the estimate, the second needs more people
The 326 lecturers give you the proportion. Then watch the interval around it shrink as the sample grows, until it fits inside ± 0.02.
The first survey supplies $\hat p$. The margin of error formula, run backwards, supplies $n$, and you round up so the interval stays inside ± 0.02.
Find the smallest sample that keeps the margin within 0.02
Keep $\hat p=\frac{303}{326}$ and $z=1.96$ fixed, and change the sample size. The curve shows the margin of error for every $n$. The orange line is the 0.02 you are allowed. Below it, the blue bar is the interval around $\hat p$.
This is the first survey. With 326 lecturers the margin of error is about 0.0278, so the blue bar spills past the orange lines. The follow-up survey needs more lecturers.
The margin of error is still bigger than 0.02, so this sample is too small. Move right.
So close. $E\approx0.020012$ is still just over 0.02. This is what you get if you cut 629.78 down to 629, and it falls short. A required sample size always rounds up.
This is the smallest sample that works. $E\approx0.019996$ is at most 0.02, and 629 lecturers would not do it. That is why 629.78 rounds up to 630.
This sample meets the target, but so does a smaller one. The question wants the sample size required, which is the smallest $n$ that works. Move left until the dot reaches the orange line.
The curve flattens out because $n$ sits under a square root. Shrinking the margin from 0.0278 to 0.02 needs almost twice as many lecturers.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The margin of error is the part after the $\pm$ in the approximate confidence interval on your formula sheet. The work is in reading which number from the question goes where.
“This one-point estimate” is the sample proportion from the first survey. 303 of the 326 lecturers said yes.
$$\hat p=\frac{303}{326}\approx0.9294$$
This mark can be implied by your later working, for example if $\frac{303}{326}$ appears in your equation for $n$. Keep it as a fraction in your calculator.
Marker: correctly determines the sample proportion
“Correct to within ± 0.02” is the margin of error, so $E=0.02$. 95% confidence gives $z=1.96$, and $1-\hat p=\frac{23}{326}$. The sample size $n$ is the unknown.
$$\begin{aligned}E&=z\sqrt{\frac{\hat p(1-\hat p)}{n}}\\0.02&=1.96\times\sqrt{\frac{\frac{303}{326}\times\frac{23}{326}}{n}}\end{aligned}$$
If your sample proportion had a slip, you can still earn this mark by follow-through.
Marker: establishes equation in $n$
Use your calculator’s solve function on that equation. Write down the equation you solved, not the buttons you pressed.
$$n\approx629.778$$
Rearranging by hand gets the same value. Divide by 1.96, square both sides, then make $n$ the subject:
$$\begin{aligned}n&=\frac{1.96^2\times\frac{303}{326}\times\frac{23}{326}}{0.02^2}\\&\approx629.778\end{aligned}$$
This mark can be implied by your later working. Rounding $\hat p$ to 0.9294 first gives about 630.17, which would round up to 631, so keep the fraction.
Marker: determines $n$
You can’t survey part of a lecturer, so $n$ has to be a whole number. With 629 lecturers the margin of error is about 0.020012, just over the limit. Round up.
If an earlier slip changed your value of $n$, you can still earn this mark by follow-through, as long as you turn your $n$ into a sensible whole number of lecturers.
Marker: determines reasonable value of $n$
Each phrase of the question gave you something
Nothing in the question names the formula or says which number is which. You have to see that the first survey only supplies $\hat p$, that “within ± 0.02” is the margin of error, and that the unknown is now $n$ inside a square root. Then you run the formula backwards and make a decision about rounding that the context forces on you. It is not complex unfamiliar because “95% confidence”, “sample size” and “within ± 0.02” still point straight at the margin of error inside the confidence interval on the formula sheet.
If I had to guess where you might lose marks, it would be the last one. Here 629.78 rounds to 630 anyway, but the reason matters. A required sample size always rounds up, because 629 lecturers gives a margin just over 0.02. If your working gave something like 629.3, rounding to the nearest whole number would leave you one lecturer short. The other trap is putting 326 in for $n$ and working out an interval nobody asked for. The subject report for this exam also recommends showing the equation your calculator solved rather than a list of keystrokes, so write the equation from step 2 in full before you give $n$.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it