The margin of error is a statistic expressing the amount of random sampling error in the results of a survey. The larger the margin of error, the less confidence one should have that a poll result would reflect the result of a census of the entire population. The margin of error will be positive whenever a population is incompletely sampled and the outcome measure has positive variance, which is to say, whenever the measure varies.
The term margin of error is often used in non-survey contexts to indicate observational error in reporting measured quantities.
Consider a simple yes/no poll
P
n
N,(n\llN)
p
p
N
P
n
N
p1,p2,\ldots
\overline{p}
\overline{p}
Going by the Central limit theorem, the margin of error helps to explain how the distribution of sample means (or percentage of yes, in this case) will approximate a normal distribution as sample size increases. If this applies, it would speak about the sampling being unbiased, but not about the inherent distribution of the data.[1]
According to the 68-95-99.7 rule, we would expect that 95% of the results
p1,p2,\ldots
\plusmn2\sigmaP
\overline{p}
Generally, at a confidence level
\gamma
n
\sigma
MOE\gamma=z\gamma x \sqrt{
\sigma2 | |
n |
where
z\gamma
\sqrt{ | \sigma2 |
n |
We would expect the average of normally distributed values
p1,p2,\ldots
n
n
\sigma\overline{p}
For the single result from our survey, we assume that
p=\overline{p}
p1,p2,\ldots
2=P(1-P) | |
\sigma | |
P |
Standarderror=\sigma\overline{p} ≈ \sqrt{
| |||||||
n |
Note that
p(1-p)
For a confidence level
\gamma
\mu\plusmnz\gamma\sigma
[\mu-z\gamma\sigma,\mu+z\gamma\sigma]
P
\gamma
z\gamma
Note that
z\gamma
|\gamma|\ge1
z1.00
z1.10
\gamma | z\gamma | \gamma | z\gamma | |
---|---|---|---|---|
0.84 | 0.9995 | |||
0.95 | 0.99995 | |||
0.975 | 1.959963984540 | 0.999995 | ||
0.99 | 0.9999995 | |||
0.995 | 0.99999995 | |||
0.9975 | 0.999999995 | |||
0.9985 | 0.9999999995 |
Since
max
2 | |
\sigma | |
P |
=maxP(1-P)=0.25
p=0.5
p=\overline{p}=0.5
\sigmaP
\sigma\overline{p}
z\gamma\sigma\overline{p}
P
\gamma
n
p=0.5,n=1013
MOE95(0.5)=z0.95\sigma\overline{p} ≈ z0.95\sqrt{
| |||||||
n |
MOE99(0.5)=z0.99\sigma\overline{p} ≈ z0.99\sqrt{
| |||||||
n |
Also, usefully, for any reported
MOE95
MOE99=
z0.99 | |
z0.95 |
MOE95 ≈ 1.3 x MOE95
If a poll has multiple percentage results (for example, a poll measuring a single multiple-choice preference), the result closest to 50% will have the highest margin of error. Typically, it is this number that is reported as the margin of error for the entire poll. Imagine poll
P
pa,pb,pc
71%,27%,2%,n=1013
MOE95(Pa)=z0.95\sigma\overline{pa
MOE95(Pb)=z0.95\sigma\overline{pb
MOE95(Pc)=z0.95\sigma\overline{pc
As a given percentage approaches the extremes of 0% or 100%, its margin of error approaches ±0%.
Imagine multiple-choice poll
P
pa,pb,pc
46%,42%,12%,n=1013
MOE95(Pa)
pa
If, hypothetically, we were to conduct a poll
P
n
N
pw=pa-pb
p | |
w1 |
,p | |
w2 |
,p | |
w3 |
,\ldots
\overline{pw}
2 | |
\sigma | |
Pw |
2 | |
\sigma | |
Pa-Pb |
=
2 | |
\sigma | |
Pa |
+
2-2\sigma | |
\sigma | |
Pa,Pb |
=pa(1-pa)+pb(1-pb)+2papb
where
\sigma | |
Pa,Pb |
=-PaPb
Pa
Pb
Thus (after simplifying),
Standarderrorofdifference=\sigma\overline{w
MOE95(Pa)=z0.95
\sigma | |
\overline{pa |
MOE95(Pw)=z0.95\sigma\overline{w
Note that this assumes that
Pc
Pa
Pb
2 | |
\sigma | |
Pw |
The formulae above for the margin of error assume that there is an infinitely large population and thus do not depend on the size of population
N
n
In cases where the sampling fraction is larger (in practice, greater than 5%), analysts might adjust the margin of error using a finite population correction to account for the added precision gained by sampling a much larger percentage of the population. FPC can be calculated using the formula[2]
\operatorname{FPC}=\sqrt{
N-n | |
N-1 |
...and so, if poll
P
MOE95(0.5)=z0.95\sigma\overline{p} ≈
0.98 | |
\sqrt{72,000 |
MOE | |
95FPC |
(0.5)=z0.95
\sigma | ||||
|
Intuitively, for appropriately large
N
\limn\sqrt{
N-n | |
N-1 |
\limn\sqrt{
N-n | |
N-1 |
In the former case,
n