How to Compute the Margin of Error



Margin of Error Calculator

Enter the sample size n. Enter a value between 0 and 1 for p, or if p is unknown, use p = 0.5. Enter the population size N, or leave blank if the total population is large.

            n                      p                     N
                  
  

In statistics, the margin of error represents the approximate amount of variance you can expect in polls and surveys. For example, suppose you conduct a poll that indicates 40% of people will vote 'no' on a proposition, and the margin of error is 3%. This means that if you were to conduct the same poll with another random sample of similar size, you could expect 37%-43% of the respondents in the second survey to also vote 'no.' The margin of error tells you how accurate poll results are; the smaller the margin of error, the greater the accuracy.

There are two main formulas for calculating the margin of error, each explained below. In each formula, the sample size is denoted by n, the proportion of people responding a certain way is p, and the size of the total population is N. For some margin of error formulas, you do not need to know the value of N.

95% Confidence Interval Margin of Error

If you have a sample that is drawn from a very large population (N is larger than 1,000,000), then you can compute the "95% confidence interval margin of error" with the formula

MOE = (1.96)sqrt[p(1-p)/n].

If you perform 100 surveys with the same sample size drawn from the same poplulation, then 95% of the time you can expect the margin of error to fall within the bound above. As you can see, N does not factor into this equation for margin of error. If the total population is large enough, only the size of the random sample matters, not the total population. If the survey has multiple questions and there are several possible values for p, pick the value that is closest to 0.5. Here is an example:

In a random survey of 1,000 Texans, 48% of the respondents liked chocolate ice cream more than vanilla, 46% liked vanilla more than chocolate, and 6% had no preference. First, set n = 1,000 and p = 0.48. Then

(1.96)sqrt[(0.48)(0.52)/1000] = 0.03096,

or 3.096%. This means that if you perform the same survey 100 more times, then 95% of the time the number of people who like chocolate more than vanilla should be between 44.9% and 51.1%.

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Maximum Margin of Error

People often use a simplified margin of error formula, given by the equation

MOE = (0.98)sqrt(1/n).

This simplified formula is obtained by replacing p with 0.5, which gives the maximum possible value for the numerator of the fraction. You can use this simplified formula if you do not know the proportion of respondents giving a particular answer. Here is the derivation of the maximum margin of error formula:

MOE = (1.96)sqrt[p(1-p)/n]
= (1.96)sqrt[(0.5)(0.5)/n]
= (1.96)sqrt[(0.25)/n]
= (1.96)(0.5)sqrt[1/n]
= (0.98)sqrt(1/n)


If we use n = 1,000 from the previous example, we get MOE = (0.98)sqrt(0.001) = 0.03099, or 3.099%.

Finite Population Margin of Error

The two formulas above are accurate if the random samples are drawn from extremely large populations. However, when the total population for a survey is much smaller, or the sample size is more than 5% of the total population, you should multiply the margin of error by the Finite Population Correction Factor (FPCF). The forumula is

FPCF = sqrt[(N-n)/(N-1)].

For example, suppose a small college has 2,401 students and 865 of them participate in a survey. Using the maximum margin of error formula above, we calculate

MOE = (0.98)sqrt[1/865] = (0.98)(0.034001) = 0.033321 or 3.3321%.

If we multiply this result by the FPCF, we get

MOE with FPCF = sqrt[(2401-865)/(2401-1)]*(0.033321)
= sqrt[1536/2400]*(0.033321)
= (0.8)(0.033321)
= 0.026657

So these survey results have a maximum margin of error of 2.6657%.

If 67% of the respondents gave the same answer, then the margin of error would be

sqrt[(2401-865)/(2401-1)]*(1.96)sqrt[(0.67)(0.33)/865]
= sqrt[1536/2400]*(1.96)sqrt[0.000255607]
= (0.8)(1.96)(0.0159877)
= 0.025069, or 2.5069%.


© Had2Know 2010

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