What the Margin of Error Measures
In “How Sample Size Changes Interval Width,” we saw that sample size affects the width of a one-sample t interval. This tutorial focuses on the margin of error: the distance from the sample mean to either endpoint of the interval. It summarizes how far the interval extends from its center to allow for sampling variability, at the selected confidence level.
For a one-sample t interval, the margin of error depends on the critical value \(t^*\) and the estimated standard error \(s/\sqrt{n}\). The critical value reflects the confidence level and degrees of freedom; the estimated standard error reflects the sample’s variability and size.
Here, \(t^*\) is the positive t critical value for the chosen confidence level and \(df=n-1\), \(s\) is the sample standard deviation, and \(n\) is the sample size. The quantity \(s/\sqrt{n}\) is the estimated standard error of the sample mean. As described in “The Form of a One-Sample t Interval,” the interval can be written as \(\bar{x}\pm\text{margin of error}\).
Because \(t^*\) has no units and \(s/\sqrt{n}\) has the same units as the measured variable, the margin of error also has those units. For example, if measurements are in minutes, the margin of error is in minutes. The full interval width is twice the margin of error; do not confuse the two.
What Influences the Margin of Error?
There are three linked influences to keep in view: the confidence level, the sample standard deviation, and the sample size. The first affects \(t^*\); the latter two affect the estimated standard error. Sample size also affects \(df=n-1\), and therefore has a smaller additional effect on \(t^*\).
- Confidence level: For the same sample, a higher confidence level requires a larger \(t^*\). That increases the margin of error and makes the interval wider. “How Confidence Level Changes Interval Width” develops this relationship.
- Sample variability: A larger \(s\) makes \(s/\sqrt{n}\) larger, so the margin of error increases when the other quantities are held fixed. More variation among individual observations makes the mean less precise.
- Sample size: A larger \(n\) reduces \(s/\sqrt{n}\) when \(s\) is comparable. It also increases \(df\), which makes \(t^*\) a little smaller at a fixed confidence level. Together these changes generally reduce the margin of error.
These are comparisons that hold other quantities fixed or similar. In real samples, \(s\) may change as \(n\) changes, and different samples can produce different estimates. So a larger sample does not guarantee that every realized margin of error is smaller than every margin of error from a smaller sample.
A Reliable Calculation Process
Use the following steps when a question asks for a mean’s margin of error. “Building a One-Sample t Interval by Hand” gives the fuller conditions and interval workflow; here, the calculation is centered on the margin of error itself.
Record the confidence level, \(n\), and \(s\). Find the response variable’s units, and calculate \(df=n-1\).
Use the positive t critical value for the specified confidence level and degrees of freedom, as in “Finding \(t^*\) Critical Values With a Table or invT.”
Compute \(s/\sqrt{n}\), keeping its units.
Calculate \(t^*(s/\sqrt{n})\), round appropriately, and state what that distance means in context.
For a confidence interval, the sample mean is needed to find the endpoints, but not to calculate the margin of error. If you are given \(\bar{x}\), \(s\), \(n\), and the confidence level, the margin of error comes from \(s\), \(n\), and \(t^*\).
Worked Examples
Worked Example: Estimating a Mean Water-Use Time
A fictional facilities team takes a random sample of 16 irrigation cycles from a population of 500 cycles. The sample mean duration is 42.6 minutes, and the sample standard deviation is 5.2 minutes. The sample data are roughly symmetric with no outliers. Find the margin of error for a 90% one-sample t interval.
State. The parameter is \(\mu\), the true mean duration of irrigation cycles in the stated population. We want the 90% interval’s margin of error, in minutes.
Plan. The cycles were randomly sampled. Because \(16<0.10(500)=50\), the 10% condition is met, supporting independence for sampling without replacement. The data are roughly symmetric with no outliers, which supports using a t procedure with this sample size. Random sampling supports generalizing to the stated population of cycles.
Do. The degrees of freedom are \(df=16-1=15\). For 90% confidence and 15 degrees of freedom, \(t^*\approx1.753\). The estimated standard error is:
Now multiply the critical value by the estimated standard error:
As a check, \(1.753\times1.3=1.753+0.526=2.279\), to three decimal places. If we use this margin with the sample mean, the interval endpoints are \(42.6-2.279=40.321\) and \(42.6+2.279=44.879\) minutes.
Conclude. The 90% t interval extends about 2.279 minutes on either side of the sample mean of 42.6 minutes. This is the margin of error for estimating the population’s mean irrigation-cycle duration; it is not a guaranteed maximum error for every possible sample.
Worked Example: A Margin of Error for Package Fill Amount
A fictional random sample of 36 sealed packages has a mean fill amount of 508 grams and a sample standard deviation of 12.6 grams. The packages were randomly selected without replacement from a production lot of 1,000. The sample distribution is reasonably symmetric with no outliers. Calculate the margin of error for a 95% one-sample t interval.
State. The parameter is the true mean fill amount, \(\mu\), for packages in this lot. We will calculate the 95% margin of error in grams.
Plan. The sample is random, and \(36<0.10(1000)=100\), so the 10% condition supports treating observations as independent. The sample distribution is reasonably symmetric with no outliers, supporting a t procedure. Random sampling supports generalizing to the packages in this lot.
Do. Here \(df=36-1=35\), and the 95% critical value is \(t^*\approx2.030\). First calculate the estimated standard error:
Then calculate the margin of error:
Multiplying in parts gives \(2.030(2)+2.030(0.1)=4.060+0.203=4.263\), confirming the result. The interval’s full width would be \(2(4.263)=8.526\) grams.
Conclude. The 95% interval extends 4.263 grams in each direction from the sample mean of 508 grams. The margin of error is in grams, just like the fill amount, and is half the interval’s full width.
Worked Example: Separating the Effects of Confidence, Variability, and Sample Size
A fictional random sample of 25 rechargeable lights has an average operating time of 18.4 hours and a sample standard deviation of 8 hours. The sample is from a population of 1,000 lights, and the data are reasonably symmetric with no outliers. Compare the margin of error at 90% and 95% confidence. Then consider how it would change if the sample standard deviation were 10 hours, or if a comparable sample had 100 lights with \(s=8\) hours.
State. The parameter is the true mean operating time, \(\mu\), for the population of lights. All margins of error are measured in hours.
Plan. The initial sample is random, and \(25<0.10(1000)=100\), so the 10% condition supports independence. The data shape supports a t procedure. For \(n=25\), \(df=24\); the 90% and 95% critical values are approximately 1.711 and 2.064. To compare the effects, change one input at a time and hold the others fixed.
Do. For the initial sample, the estimated standard error is:
At 90% confidence, the margin of error is \(1.711(1.6)=2.7376\), or about 2.738 hours. At 95% confidence, it is \(2.064(1.6)=3.3024\), or about 3.302 hours. The higher confidence level increases \(t^*\), and therefore increases the margin of error.
If \(n=25\) and the confidence level remains 95%, but \(s\) is 10 hours, the standard error becomes \(10/\sqrt{25}=2\) hours. The margin of error is \(2.064(2)=4.128\) hours. The larger \(s\) increases the estimated standard error and margin of error.
For a comparable sample of 100 lights with \(s=8\) hours at 95% confidence, \(df=99\), \(t^*\approx1.984\), and the standard error is \(8/\sqrt{100}=0.8\) hours. The margin of error is \(1.984(0.8)=1.5872\), or about 1.587 hours. Here both the standard error and \(t^*\) are smaller than in the \(n=25\) comparison.
Conclude. Holding other inputs fixed, raising confidence or increasing sample variability raises the margin of error, while increasing sample size generally lowers it. These comparisons isolate each influence; in actual samples, the observed \(s\) can also change.
Common Mistakes and AP Exam Tips
- Using \(s\) instead of \(s/\sqrt{n}\). The sample standard deviation describes the spread of individual observations. The estimated standard error describes the estimated spread of sample means. The margin of error uses the standard error.
- Forgetting the multiplier \(t^*\). The standard error alone is not the margin of error. Multiply it by the critical value for the stated confidence level and \(df=n-1\).
- Using \(t^*\) as though it has units. It is a multiplier, so the margin of error takes its units from \(s/\sqrt{n}\), such as minutes, grams, or hours.
- Reporting the full width as the margin of error. The margin of error is the distance from the center to one endpoint. The full interval width is twice that distance.
- Claiming the margin of error is a guaranteed bound on the estimation error. It describes the interval produced by the procedure, but it does not guarantee that a particular sample’s interval contains \(\mu\).
- Saying a larger sample always gives a smaller margin of error. A larger \(n\) generally reduces the standard error when \(s\) is comparable, but actual samples can have different standard deviations. State what is being held fixed.
- Skipping conditions because the question gives summary statistics. When making an inference, still address random sampling or an appropriate sampling process, independence and the 10% condition when relevant, and whether the data support a t procedure.
For full-credit communication, show the \(t^*\), the standard error, the multiplication, and the result with units. Then describe the result as a distance from the sample mean to either endpoint. Keep the margin of error distinct from both the interval’s full width and the confidence level.
Check Your Understanding
Use the margin-of-error formula and explain each result in context.
- A 90% t interval has \(s=6\) seconds, \(n=9\), and \(t^*=1.860\). Calculate the estimated standard error and margin of error, including units.
- For the same sample, what happens to the margin of error if the confidence level is raised and \(t^*\) increases? Explain why.
- With \(n\) and confidence level fixed, how would increasing \(s\) affect the margin of error?
- A student reports a margin of error of 3 kilograms and says the interval is 3 kilograms wide. Identify the mistake and give the correct width.
- Why is it not correct to say that a particular 95% interval has a 95% probability of containing the true population mean?