Margin of error

A news report says "$52\%$ of voters support the measure, with a margin of error of plus or minus $3$ percentage points." The $52\%$ comes from a sample, and a different random sample would have given a slightly different number. The margin of error tells you how far the sample result is likely to be from the true population value.

Samples vary

Imagine a large population in which exactly $50\%$ of people like a new song. If you survey $100$ random people, you probably won't get exactly $50$ who like it. You might get $46$, or $53$, or $49$. Survey another $100$ and you'll get another number. This natural wobble from sample to sample is called sampling variability.

If you took thousands of random samples of size $100$ and made a histogram of the sample proportions, it would look approximately normal, centered at the true proportion $0.50$. Its standard deviation, called the standard error, is

$$\sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{100}} = 0.05.$$

By the 68–95–99.7 rule, about $95\%$ of samples give a proportion within $2$ standard errors of the truth, here between $0.40$ and $0.60$. That "$2$ standard errors" is exactly what a margin of error measures.

Notice that $n$ is in the denominator under a square root. Bigger samples vary less, but you need to quadruple the sample size to cut the variability in half.

The margin of error for a proportion

In practice you don't know the true proportion $p$ (that's what you're trying to estimate), so you use the sample proportion $\hat{p}$ ("p-hat") in its place.

The quick formula comes from the first one: the value $\hat{p}(1 - \hat{p})$ is largest when $\hat{p} = 0.5$, and then $2\sqrt{\dfrac{0.25}{n}} = \dfrac{2(0.5)}{\sqrt{n}} = \dfrac{1}{\sqrt{n}}$. So $\dfrac{1}{\sqrt{n}}$ is the largest the margin of error can be, a safe upper estimate. (Many statisticians use $1.96$ instead of $2$; the difference is tiny.)

The margin of error for a mean

The same idea works when you estimate a population mean. If a random sample of size $n$ has mean $\bar{x}$ and standard deviation $s$, then

$$\text{ME} \approx \frac{2s}{\sqrt{n}},$$

and the interval $\bar{x} \pm \text{ME}$ likely contains the population mean.

Choosing a sample size

Pollsters often decide in advance how precise they want to be. Solve $\text{ME} = \dfrac{1}{\sqrt{n}}$ for $n$:

$$n = \frac{1}{\text{ME}^2}.$$

For a margin of error of $2\%$: $n = \dfrac{1}{0.02^2} = \dfrac{1}{0.0004} = 2500$ people. For $1\%$, you'd need $10{,}000$, four times as many, to halve the margin again.

Practice

Practice 1

A random sample of $625$ people is surveyed. Use $\text{ME} \approx \dfrac{1}{\sqrt{n}}$ to find the margin of error, as a percent.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 2

A poll reports that $45\%$ of residents favor a new tax, with a margin of error of $3\%$. What is the upper end of the interval that likely contains the true percent, as a percent?

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 3

In a random sample of $700$ adults, $30\%$ say they exercise daily. Use $\text{ME} \approx 2\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}$ to find the margin of error, as a percent rounded to the nearest tenth.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 4

A random sample of $100$ light bulbs has a mean lifetime of $1{,}200$ hours with a standard deviation of $15$ hours. Use $\text{ME} \approx \dfrac{2s}{\sqrt{n}}$ to find the margin of error, in hours.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 5

A pollster wants a margin of error of $5\%$. Using $\text{ME} \approx \dfrac{1}{\sqrt{n}}$, how many people should be surveyed?

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 6

A survey of $500$ people has a certain margin of error. If the survey had used $2{,}000$ people instead, what would happen to the margin of error?

AIt would be one fourth as large.BIt would stay the same.CIt would be half as large.DIt would be four times as large.

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Practice 7

A random poll shows Candidate Lee with $51\%$ of the vote and a margin of error of $3\%$. Which conclusion is best?

AThe poll cannot tell whether Lee has a majority, because the interval 48% to 54% includes values below 50%.BLee definitely has a majority, because 51% is more than 50%.CLee definitely does not have a majority.DThe true support for Lee is exactly 51%.

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Practice 8

A magazine asks readers to vote online on whether they like its new design. Of $10{,}000$ votes, $80\%$ are "no," and the magazine reports a margin of error of $1\%$. What is wrong with this report?

AThe margin of error should be $10\%$ because the sample is large.BThe margin of error should be $0.1\%$.CNothing; a large sample guarantees an accurate result.DIt is a voluntary response sample, so the result may be biased, and the margin of error doesn't account for bias.

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