Choosing a sample size

Three quick questions from earlier lessons.

A survey of 100 has margin of error 0.10. About how many people give a margin of 0.05?

Plan the sample size before collecting data.

To get margin of error at most

ME

for a proportion:

n \ge \left(\dfrac{z^*}{ME}\right)^2 p(1 - p)

, using a prior estimate of

p

or

p = 0.5

(the safest choice), then round up.

95% confidence, ME = 0.03, $p = 0.5$ .

\approx 4268.4

(1.96/0.03)².

\approx 1067.1

× 0.25.

n = 1068

Round up.

Your turn: with $z^* = 2$ , ME = 0.1, $p = 0.5$ : $(2/0.1)^2 \times 0.25$ .

n = 100

400 × 0.25.

A survey of 264 people gives a margin of error you want to halve. About how many people are needed?
1. 1056
2. 528
3. About half as many

A margin of error is 0.08. If the sample size is multiplied by 9, what is the new margin of error?
A margin of error is 0.07. If the sample size is multiplied by 25, what is the new margin of error?

Using p = 0.5 and z* = 1.96, find the value of n (before rounding up) needed for a margin of error of 0.04.

Three questions to check you have got it.

A margin of error is 0.08. If the sample size is multiplied by 9, what is the new margin of error?
A margin of error is 0.07. If the sample size is multiplied by 25, what is the new margin of error?
A margin of error is 0.07. If the sample size is multiplied by 4, what is the new margin of error?

Sample size

$n \ge (z^*/ME)^2 p(1 - p)$ , $p = 0.5$ if unknown; round up.

Watch out: halving ME needs 4 times the sample.

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