The Margin of Error (ME) is often expressed in the following formula as the Standard Error (SE) times an adjustment value which is a z-critical or t-critical value:
ME = critical value * Standard Error (SE)
The ME is also sometimes called the maximum error of estimate or error tolerance. The ME is the "radius" or half the width of the confidence interval for a distribution. See below.
The ME is closely related to the calculation of a Confidence Interval (CI). The CI is:
Ways to REDUCE the Margin of Error:
The larger the ME, the less confidence that the results represent those of the entire population. The larger the sample size, the smaller ME.
NOTE: When calculating the ME (and Confidence Intervals), the sample size should be <5% of the true population or a correction factor is needed which isn't discussed here (trying to keep it simple). These examples assume a large population that is not finite.
The most common calculations for the ME are shown below:
EXAMPLE:
if there were thousands of customers surveyed for a particular contractor and we sifted through a sampling of 125 of them and found the average rating was 4.26. Also, we know from the past that the population standard deviation is 0.21.
Calculate the Margin of Error for a desired 90% confidence level (or alpha risk of 10%)?
SOLVING:
In this case, we know the population standard deviation so we will use the z-critical value at risk of 0.10 or CL of 90% (same thing).
The z-critical value is 1.645 for 90% CL
The standard deviation of the sample = 0.21.
SE = the sample standard deviation / sqrt n = 0.21 / sqrt 125 = 0.21 / 11.18 = 0.01878 = SE
Therefore, the ME = 1.645 * 0.01878 = 0.0309 = ME
EXAMPLE:
if there were thousands of customers surveyed for a particular contractor and we sifted through a sampling of 125 of them an found the average rating was 4.26 with a sample standard deviation of 0.21.
Calculate the Margin of Error for a desired 90% confidence level (or 0.10 alpha-risk = 10%)?
Now we need to use the two-tailed t-table for 90% CL (or 0.10 risk) at a Degrees of Freedom of n-1 which = 124.
The t-critical value at 90% CL, dF = 124, is 1.657
(we used the calculator found here and a picture of it is below or you can use a table that shows the most commonly used values and interpolate from there)
The standard deviation of the sample = 0.21.
SE = the sample standard deviation / sqrt n = 0.21 / sqrt 125 = 0.21 / 11.18 = 0.01878 = SE
Therefore the ME = 1.657 * 0.01878 = 0.0311 = ME
Notice how close the answer is to the first example. The ME's are almost exactly the same. This is because the sample size is large and the t-critical approaches z-critical value as the sample size increases.
In this example, we will take it a step further and calculate the Confidence Interval (CI).
The center of the confidence interval, x-bar, is the point estimate. Take the point estimate and subtract the ME for the lower bound of the CI and add the ME to the point estimate to the upper bound of the CI.
It's that simple.
The x-bar value is 4.26, then the CI = {4.26 - 0.0311, 4.26 + 0.0311} = {4.2289, 4.2911}
In practical terms this means that you can be 90% certain that the population would choose a rating between 4.2289 to 4.2911 for that contractor.
If all things remain the same, the next result from the survey is 90% to have a result between 4.2289 and 4.2911. Obviously, most surveys don't ask for that level of resolution so there needs to be some reality check to the statistical answer....and this reality check applies to any statistical answer.
If you wanted a CL of 95% or 99%, the then margin for error increases since the critical value increase (in other words you want more confidence than 90% therefore you need to accept wider interval if you're going to keep the number of samples the same at 125).
EXAMPLE:
If there were a sampling of 100 people and 20 of them voted yes on a new regulation, calculate the Margin of Error for a 95% confidence level (or 0.10 alpha-risk = 10%)?
The following formula assumes that nP(1-P) > 5 (which essentially says that the sample size is <5% of the total true population).
The z-critical value is 1.645 for 95% CL
p-hat = 0.20 = the sample proportion
n = 100
SE = sqrt ((0.20(1-0.20)) / 100) =sqrt ((0.20*0.80)/100) = sqrt (0.16/100) = sqrt 0.0016 = 0.04 = SE
Therefore, the ME = 1.645 * 0.04 = 0.0658
NOTE: In this case the ME is a percentage but not in the previous two examples. Be careful when expressing ME as a percentage.
Therefore, the Margin of Error = 6.58%
The Confidence Interval (CI) is calculated as follows:
p-hat = 20/100 = 0.20
CI = 0.20 +/- 0.0658 = {0.1342 , 0.2658}
CI = 20% +/- 6.58% = {13.42% , 26.58%)
You can be 95% confident that the true percentage of people that will vote yes is between 13.42% and 26.58%.
Although there is chance the the interval of 13.42% to 26.58% may not contain the true proportion, 95% of confidence intervals created will contain the true proportion.
To test the MEANS of two related populations these are considered dependent samples. This takes away the variation among the subjects. This section assumes:
The following formulas apply:
the i^{th} paired difference is: d_{i} = x_{i} - y_{i}
d-bar is the point estimate used to calculate the confidence interval.
n = number of matched pairs in the sample.
The sample standard deviation in this case is:
The formula is the same as the Means for a sample except the standard deviation is considerate of both samples instead of one sample and denoted s_{D}
For the Confidence Interval, the point estimate is d-bar +/- ME.
This Margin of Error calculator is available, along with several others, with examples. Click here to see all the Templates and Calculators that can help further understand these topics and accelerate the execution of a Six Sigma Project. |
You can see how many common terms are related now. Once you have the ME, the Confidence Interval calculation is easy. The hardest part is calculating the ME.
Sometimes you may have a desired ME or may be given a directive not to exceed a certain ME. In that case you can adjust scenarios to find out what Sample Size is necessary at various Confidence Levels.
Rearrange the formula and solve for n. Then plug in the desired ME and Confidence Level and you'll get the minimum sample size required....and round up the answer to the nearest whole number.
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