Quick questions on Interval estimates and confidence intervals - TCE Mathematics Methods (Tasmania)

A confidence interval takes the point estimate and adds and subtracts a margin of error.

A common design question reverses the formula: given a required margin of error $E$, find the sample size $n$ needed. Rearranging $E = z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$ gives $$n = \frac{z^{2}\,\hat{p}(1-\hat{p})}{E^{2}}.$$When no prior estimate of$\hat{p}$is available, use$\hat{p} = 0.5$, because$\hat{p}(1-\hat{p})$is largest there and so gives the safest (largest) sample size. Always round$n$up to the next whole number, since a fractional person cannot be surveyed and rounding down would leave the margin slightly too wide. For example, a$95\%$interval with margin$0.03$and$\hat{p} = 0.5$needs$n = \dfrac{1.96^{2}\times 0.25}{0.03^{2}} \approx 1067.1$, so survey$1068$ people.

Some questions hand you a completed interval and ask which confidence level produced it. Take half the interval width as the margin $E$, divide by the standard error to recover $z$, then identify the confidence level from the standard $z$-values ($1.645$ for $90\%$, $1.96$ for $95\%$, $2.576$ for $99\%$). This reverses the construction and is a favourite TASC twist on the standard interval question.