Plan a sample size from a margin of error

Practise plan a sample size from a margin of error with an original VectorPaths question, worked explanation and related VCE Methods practice.

3 practice marks

Original practice question

A survey uses the approximate 95% margin of error $E=1.96\sqrt{p(1-p)/n}$. Use the conservative planning value $p=0.5$. The required margin is at most 0.04. (a) Form and solve the inequality for n. [2 marks] (b) State the smallest whole-number sample size and explain the rounding. [1 marks]

Worked explanation

• Part (a)

  $1.96\sqrt{0.25/n}\leq0.04$. Squaring positive quantities and rearranging gives $n\geq(1.96\times0.5/0.04)^2=600.25$.

• Part (b)

  Round upwards to 601 because n must be an integer and a smaller n would give a margin greater than 0.04.

(a) $n\geq600.25$.; (b) $601$.

Original VectorPaths practice; not an official VCAA question or marking rubric.