Margin of error and sample size: WACE Year 12 Mathematics Methods Unit 4

In a sample of $n=500$ with $\hat{p}=0.5$, find the margin of error at $95\%$ and at $99\%$.

Step 1: Calculate the standard error (shared by both confidence levels)

The standard error depends only on $\hat{p}$ and $n$, not on the confidence level, so it is the same for both intervals.

Step 2: Apply each $z$-multiplier to get the margin of error

A higher confidence level requires a larger $z$-score, which stretches the margin of error. At $95\%$ the multiplier is $1.96$; at $99\%$ it is $2.576$.

At $95\%$: $E=1.96\times 0.02236\approx 0.0438$.

At $99\%$: $E=2.576\times 0.02236\approx 0.0576$.

Final answer: raising the confidence from $95\%$ to $99\%$ widens the margin from about $0.044$ to $0.058$.

How the margin responds to a larger sample

A $95\%$ confidence interval with $\hat{p}=0.5$ and $n=400$ has margin of error $E=1.96\sqrt{\dfrac{0.25}{400}}=1.96\times 0.025=0.049$. Suppose the sample is increased to $n=1600$.

The new standard error is $\sqrt{\dfrac{0.25}{1600}}=0.0125$, exactly half the old $0.025$, because $\sqrt{1600}=2\sqrt{400}$. So the new margin of error is

half the original. Quadrupling the sample halved the margin of error, illustrating the square-root law directly. Markers reward recognising that $n\to 4n$ gives $E\to\tfrac{1}{2}E$.

What sample size gives a $95\%$ margin of error of at most $0.03$, using $\hat{p}=0.5$?

Step 1: Choose the conservative value of $\hat{p}$ and identify the parameters

When no prior estimate of the proportion is available, using $\hat{p}=0.5$ gives the largest possible margin of error for any $n$, so the resulting sample size is guaranteed to be large enough regardless of the true proportion. Here $z=1.96$ for $95\%$ confidence and $E=0.03$.

Step 2: Substitute into the sample-size formula

Rearranging $E = z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ for $n$ gives $n = \dfrac{z^{2}\hat{p}(1-\hat{p})}{E^{2}}$.

Step 3: Round up to ensure the margin is met

Because $n$ must be a whole number and rounding down would give a margin slightly larger than $0.03$, always round up.

Final answer: $n = 1068$.