Margin of Error
In survey sampling, the margin of error is a measure of the uncertainty in a survey outcome, given a particular sampling plan. In combination with a confidence level, the margin of error defines a range within which the true population value may be found.
For example, if a survey reports a margin of error of ±3% at a 95% confidence level for a sample statistic, we can expect the sampling plan (sample size, sampling method, etc.) used by the researcher to yield a sample statistic within 3 percentage points of the true population parameter, 95% of the time.
Prerequisite
This lesson assumes that you are familiar with the standard deviation of a sampling distribution and the standard error, topics that we covered in the previous lesson.
If you are not familiar with these topics, click the link below for a quick review:
- Lesson 49: What is the Standard Error?
How to Compute the Margin of Error
The margin of error is calculated from either of the following equations:
ME = CV x σs
ME = CV x SE
where ME is the margin of error, CV is a critical value (a z-score or a t-score) corresponding to the desired confidence level, σs is the standard deviation for the sampling distribution of a statistic, and SE is the standard error.
Note: We seldom know the values of population parameters required to compute the standard deviation for the sampling distribution of a statistic (σs); whereas the standard error (SE) can be easily calculated from sample data, so we use the second equation ( ME = CV x SE ) to compute margin of error most of the time.
How to Express Critical Value as z-Score
The critical value is a factor used to compute the margin of error. To express the critical value as a z-score, follow these steps.
- Compute alpha (α): α = 1 - (confidence level / 100)
- Find the critical probability (p*): p* = 1 - α/2
- Find the z-score having a cumulative probability equal to the critical probability (p*).
To find the critical z-score, use an online calculator (e.g, Stat Trek's Normal Distribution Calculator), a graphing calculator, or a normal distribution statistical table (found in the appendix of most introductory statistics texts).
How to Express Critical Value as t-Score
To express the critical value as a t-score, follow these steps.
- Compute alpha (α): α = 1 - (confidence level / 100)
- Find the critical probability (p*): p* = 1 - α/2
- Find the degrees of freedom (df): df = n - 1 (for a mean score or proportion from a single sample)
- Find the t-score having degrees of freedom equal to df and a cumulative probability equal to the critical probability (p*).
When estimating a mean score or a proportion from a single sample, degrees of freedom are equal to the sample size (n) minus one. For other applications, the degrees of freedom may be calculated differently. (Elsewhere on this site, we explain how to calculate degrees of freedom for different applications.)
To find the critical t-score, use an online calculator (e.g.,Stat Trek's t Distribution Calculator), a graphing calculator, or a t-distribution statistical table (found in the appendix of most introductory statistics texts).
t-Score vs. z-Score
Should you express the critical value as a t-score or as a z-score? One way to answer this question focuses on the population standard deviation.
- If the population standard deviation is known, use the z-score.
- If the population standard deviation is unknown, use the t-score.
Another approach focuses on sample size.
- If the sample size is large, use the z-score. (The central limit theorem provides a useful basis for determining whether a sample is "large".)
- If the sample size is small, use the t-score.
In practice, researchers employ a mix of the above guidelines. On this site, we use z-scores when the population standard deviation is known and the sample size is large. Otherwise, we use t-scores, unless the sample size is small and the underlying distribution is not normal.
Warning: If the sample size is small and the population distribution is distinctly not normal (e.g., badly skewed, with outliers), we cannot be confident that the sampling distribution of the statistic will be normal. In this situation, neither the t-score nor the z-score should be used to compute critical values.
Test Your Understanding
This problem demonstrates each step required to compute the margin of error.
Problem 1
Nine hundred (900) high school freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point average (GPA) was 2.7, and the sample standard deviation was 0.4. What is the margin of error, assuming a 95% confidence level?
(A) 0.013
(B) 0.025
(C) 0.500
(D) 1.960
(E) None of the above.
Solution
The correct answer is (B). To compute the margin of error, we need to find the critical value and the standard error of the mean. To find the critical value, we take the following steps.
- Compute alpha (α):
α = 1 - (confidence level / 100)
α = 1 - 0.95 = 0.05
- Find the critical probability (p*):
p* = 1 - α/2
p* = 1 - 0.05/2 = 0.975
- Find the degrees of freedom (df):
df = n - 1 = 900 -1 = 899
- Find the critical value. Since we don't know the population standard deviation, we'll express the critical value as a t-score. For this problem, it will be the t-score having 899 degrees of freedom and a cumulative probability equal to 0.975. Using the t Distribution Calculator, we find that the critical value is about 1.96.
Next, we find the standard error of the mean, using the following equation:
SEx = s / sqrt( n )
SEx = 0.4 / sqrt( 900 ) = 0.4 / 30 = 0.013
And finally, we compute the margin of error (ME).
ME = Critical value x Standard error
ME = 1.96 * 0.013 = 0.025
So, for this study, we report that the average GPA is 2.7 ± 0.025.
Note: The larger the sample size, the more closely the t-distribution looks like the normal distribution. For this problem, since the sample size is very large, we would have found the same result with a z-score as we found with a t-score. That is, the critical value would still have been about 1.96. The choice of t-score versus z-score does not make much practical difference when the sample size is very large.