How to Calculate Margin of Error

The margin of error is a popular statistics error estimation that helps researchers and statisticians discover by how many points a sample will differ from a population parameter based on a particular confidence interval. The margin of error is generally one of the main steps that a statistician or poll researcher will need to determine to predict the result of a particular population parameter.

Before calculating the margin of error of any sample, it is important to understand why it is needed as that is the only way to understand its usefulness in real life. While further look into the article will give better clarity about the use of Margin of Error, the basic statistics, and samples that call for its use will be discussed.

For the margin of error to be calculated, a confidence interval must first have been determined based on a population parameter. Population parameters are what define a population. For example, the number of students in the University of Oxford is a population parameter while the students are the population. Other population parameters are the population mean of the Students and the standard deviation of the students from the population mean.

Now, it is possible to determine the number of students by checking the active students in Oxford digital systems. On the other hand, assuming there is a need to find out the average age of all the students in oxford university. It would be difficult to determine this as it means that all students’ age must be collected and the average calculated. However, opting to use a sample mean would be easier. But as expected, the issue with a sample is that it leaves room for errors.

However, finding the confidence interval of the sample mean can help a researcher get a much better range of the average mean and improve the possibility of getting the population means. The role of the marginal error then shows the possible points in which the confidence interval will not represent the population parameter.

The section below will explain more about the confidence interval and the role of margin of error. Also, students will learn how to calculate the margin of error and how to use it appropriately.

The Margin of Error Formula

The Margin of Error in statistics is a number that highlights the vastness of the random sampling error of a survey.

The error is calculated on a confidence interval, and the larger it is, the less reliable the confidence interval is estimated to cover a population parameter.

The formula for the margin of error is:

Margin of Error (ME) = Z2 n

$MarginofError(ME)=&space;Z_{\frac{a}{2}}&space;\times&space;\frac{\delta&space;}{\sqrt{n}}$

$MarginofError(ME)=&space;t_{\frac{a}{2}}&space;\times&space;\frac{\delta&space;}{\sqrt{n}}$

Where $Z_{\frac{a}{2}}$ is the Z score when $\geq&space;30$

$t_{\frac{a}{2}}$ is the Z score where n $<&space;30$

α = 1 – CI

δ = standard deviation

$\frac{\delta&space;}{\sqrt{n}}$ = Standard Deviation

n = sample value

CI = Confidence Interval

How to Calculate Margin of Error?

This section will outline the steps to find the margin of error.

• Determine the Z-score or t-score
• Determine the standard deviation or standard error. The standard deviation is what is calculated when working with a population parameter, while the standard error is for calculating the sample mean.
• Multiply the Z score with the standard deviation or multiply the t score with the standard deviation.

What Is the Margin of Error?

There are many reasons to calculate the margin of error based on how it relates to a confidence interval. A confidence interval generally has a lower and upper bound that surrounds a sample statistic like the mean. The lower and upper bound ensure that the population mean is within the bound. The two major confidence intervals are 95% and 99% interval.

Therefore, a 99% confidence interval of a sample mean simply means that 99 out of 100 times, the population mean will be accounted for in the sample.

However, it is important to note that since the CI is dependent on a sample, there is a possibility that the CI bounds will not reflect the population mean, and it is in this instance that the margin of error will be important.

The margin of error determines the number of points at which a CI will fall short of representing a population parameter. So assuming that there is a 99% confidence interval, a politician will win a presidential election based on a sample pool with 61% vote. That means there is a 99% confidence that the politician will win the election, with a good possibility of garnering a little below or above 61% votes.

The margin of error then calculates the possible reason, while the 99% confidence interval will not produce such a result. The higher the margin of error after calculation, the less reliable the confidence interval will be.

Subsequent examples on how to calculate the marginal error will be discussed

Example 1

A researcher trying to determine the average days it took all Japanese Affiliate Car companies to meet 100 sales monthly, randomly sampled twenty companies. The researcher found that it took the sampled companies an average of 20 days to sell 100 cars with a standard deviation of 4 days.

The researcher claimed the samples were a good representative of all other companies’ average performance in a month at 95% and 99% confidence intervals. However, after a year of using the information to predict the trend of affiliate companies in Japan, it was difficult to achieve consistency. This led another researcher to check the margin of error for both confidence intervals to know if they were reliable or not.

Solution

To determine the answer to the above problem, the first issue is to identify which statistic fits into the computation. For easy calculation, the first step is determining which formula to use to solve the problem.

The first thing to consider is whether the sampled value fits the T or Z scores. 20 companies were sampled, so n = 20, meaning the T-test is the one to go for.

So the formula for this calculation is;

Margin of Error $(ME)=&space;t_{\frac{a}{2}}&space;\times&space;\frac{\delta&space;}{\sqrt{n}}$

Now you should note that the computation is for two different confidence intervals, so computation will start for the 95% confidence interval to determine its margin of error. Also, since it is the t-test we are computing for, the standard error will apply

α = 1 – CI

= 1 – 95%

= 1 – 0.95 = 0.05

$\frac{a}{2}=\frac{0.05}{2}=0.025$

δ = 4

$\frac{\delta&space;}{\sqrt{n}}=\frac{4}{\sqrt{20}}=0.8944$

n = sample value

Therefore, the

Margin of Error $(ME)=t_{0.025}\&space;\times&space;0.8944$

To get the critical value for the t-test, a t table has to be read. To read this table however, it is important to define the t test mathematically, which is;

$t_{(n-1)}$ where (n-1) is the degree of freedom

Below is a T table, and it will be used to read the $t_{0.0025&space;}&space;=t_{(9)}0.025$

To determine the t score. Check the vertical part of the table and select 9 as it is the d.f. Next, choose 0.025 from the horizontal headings under significance level. Trace both numbers to the value where they intersect. The intersection is the value for t0.025 = 2.262

Margin of Error (ME) = 2.262 x 0.8944 = 2.023

For the 95% confidence interval, the ME is significantly large, which makes its reliability questionable.

Example 2

Now for the 99% confidence interval

Margin of Error $(ME)=t_{\frac{1-0.99}{2}}\&space;\times&space;0.8944$

$(ME)=t_{0.005}\&space;\times&space;0.8944$

Now $t_{0.005}=&space;t_{(9)}0.005$

Using the t -test table $t_{(9)}0.005\&space;=&space;3.250$

Therefore, the Marginal Error (ME) = 3.250 x 0.8944 = 2.9068

The 99% confidence even had a higher significance level with a percentage point close to 3. Obviously, this explains why the confidence interval has failed to determine the average output of affiliate companies in Japan correctly.

Example 3.

Let everything in the first example be true except for the fact that n = 30. Determine the 99% confidence interval.

Since n = 30, the formula would be

Margin of Error $(ME)=&space;Z_{\frac{a}{2}}&space;\times&space;\frac{\delta&space;}{\sqrt{n}}$

Margin of Error $(ME)=Z_{\frac{1-0.99}{2}}\&space;\times&space;\frac{4}{\sqrt{30}}$

Margin of Error $(ME)=Z_{0.005}&space;\times&space;\frac{4}{\sqrt{30}}$

Margin of Error $(ME)=Z_{0.005}&space;\times&space;0.7303$

The next step will be to determine the Z score. The process of determining the Z score is quite easier compared to the T test.

For the $Z_{0.005}$, the table would be read for $Z_{1-0.005}&space;=&space;Z_{0.995}$

This is so because the tail of the distribution is already determined.

Looking it up from the Z table below $Z_{0.005}&space;=&space;Z_{0.995}=2.576$

This means that to determine a Z score from a table, trace the value (0.995) among the vast range of numbers. If it cannot be determined, then pick the value close to it. Trace that value to the horizontal part of the table to get the first 2.5. trace upward to get 7. Join both values together to get the Z value (2.57)

Margin of Error (ME) = 2.576 x 0.7303 = 1.88

Conclusion

The margin of error is well used when certain population parameters are unknown and can help give a whole picture of a sample data.

While closely related methods like the empirical rule emphasizes how confidence interval help in making decision about population parameter, the margin of error determines whether these confidence intervals can be relied on.

Also, It is important to understand that the fact a margin of error is small does not mean that the Confidence Interval is good. Actually small margin of errors (values greater than one) signifies that a confidence interval may not be so reliable even at 99%. An example of this particular situation was the 2012 election that saw Barrack Obama re-elected as the United States president.

Gallup carried out a survey pool on who will emerge the winner of the United States presidential election between President Barrack Obama and opposition Romney. The pool sampled vast respondents from different states in the country, and the computed results had Romney as the winner with 49% votes while Obama had 48% votes.

The confidence interval for the calculation was 95% (95% confidence interval) with a 2 margin of error. What this implied was that the result of the pool, if carried out again, will be accurate 95 out of 100 times with the other 5 possible outcomes no more than a two-point percentage, plus, and minus. This meant that the main election should bring out Romney as the winner, with Obama having a minute chance of coming tops

However, the outcome of the results saw the two candidates amass a marginal error of 4 percentage points. Obama scored 95% vote while Romney scored 47%. The outcome showed that poor randomness and insufficient sampled data could lead to wrong outcomes. Hence the margin of error can only be as accurate as a sample’s randomness and sufficiency.