Construct and interpret confidence intervals for a population mean and find the required sample size

What this dot point is asking

SCSA wants you to build the interval from a sample, choose the right critical value, interpret the level correctly, and rearrange to find the sample size needed for a target margin of error.

The interval

where $z$ is the standard-normal critical value for the chosen level: $z \approx 1.645$ for $90\%$, $z \approx 1.96$ for $95\%$, and $z \approx 2.576$ for $99\%$. The half-width $z\tfrac{\sigma}{\sqrt{n}}$ is the margin of error $E$.

Interpreting the confidence level

Where the interval comes from

The interval is not arbitrary; it falls straight out of the central limit theorem. Because $\bar{X}$ is approximately normal with mean $\mu$ and standard error $\dfrac{\sigma}{\sqrt{n}}$, about $95\%$ of sample means lie within $1.96$ standard errors of $\mu$, that is $P\!\left(\mu - 1.96\dfrac{\sigma}{\sqrt{n}} < \bar{X} < \mu + 1.96\dfrac{\sigma}{\sqrt{n}}\right) \approx 0.95$. Rearranging the inequality to isolate $\mu$ turns this into a statement about an interval centred on the observed $\bar{x}$, namely $\bar{x} \pm 1.96\dfrac{\sigma}{\sqrt{n}}$. Knowing this derivation makes the repeated-sampling interpretation natural: the randomness is in the sample, so it is the interval that varies from sample to sample, not the fixed but unknown $\mu$.

Reading the interval back to front

Many SCSA questions give you the interval and ask you to recover the ingredients. The sample mean is always the midpoint of the interval, and the margin of error is always half the width. From the margin you can recover the standard error by dividing by the critical value, and from the standard error you can recover $n$ if $\sigma$ is known. Being able to move in both directions, from data to interval and from interval back to data, is a reliable source of marks and a useful check on your arithmetic.

Confidence versus precision

A higher confidence level uses a larger $z$, widening the interval: more confidence costs precision. A larger sample reduces the standard error, narrowing the interval. The two competing effects are both visible in the margin of error $z\tfrac{\sigma}{\sqrt{n}}$.

Finding the required sample size

Set the margin of error to the target $E$ and solve for $n$:

Round $n$ up to the next whole number, since a fractional sample is not possible and rounding down would exceed the target margin.

Notice the inverse-square relationship: because $n$ depends on $\dfrac{1}{E^2}$, halving the desired margin of error requires four times the sample size, and reducing it to a tenth requires a hundred times the sample. This is why high-precision estimates can demand very large samples, and it is a common point for a follow-up part asking students to comment on the practicality of a stated precision.