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Wize AP Statistics Textbook > Confidence Intervals for Mean and Proportion

Margin of Error and Sample Size

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Margin of Error and Sample Size

Recall that the margin of error is:
z∗p^q^n\boxed{\displaystyle{z^{\ast}\sqrt{ \frac{\hat{p}\hat{q}}{n}}}}

Suppose we have a desired margin of error, mm. Given the confidence level, we can find the sample size, n⋆n^\star, to give us the desired margin of error.

Example
If p^=0.4\hat{p}=0.4 and we want a 95% confidence interval with a margin of error of 0.05. What should be our sample size?

Since m=z∗p^q^nm=z^{\ast}\sqrt{ \frac{\hat{p}\hat{q}}{n}}

we use algebra to isolate n and get
n∗=(z∗)2p^q^m2n^{\ast}=\frac{\left(z^{\ast}\right)^2\hat{p}\hat{q}}{m^2}
z∗=1.96z^{\ast}=1.96 (for 95% confidence)

Solve:

n∗=(1.96)2(0.4)(1−0.4)(0.05)2\displaystyle{}n^{\ast}=\frac{\left(1.96\right)^2\left(0.4\right)\left(1-0.4\right)}{\left(0.05\right)^2}

n∗=368.79 ≈369n^{\ast}=368.79\ \approx369 (rounded up!)


Wize Tip
Always round UP to the next whole number!

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How do we reduce the margin of error?

z∗p^q^n\boxed{\displaystyle{z^{\ast}\sqrt{ \frac{\hat{p}\hat{q}}{n}}}}
  1. Increase n; or
  2. Decrease z⋆z^\star (opt for a lower confidence interval)
Suppose the sample proportion is 60% and we want to construct a 95% confidence interval with a margin of error of 3%.