Building a One-Sample t Interval by Hand

Turning a Sample Summary into an Interval

In “Computing the Standard Error of a Sample Mean,” you calculated $s/\sqrt{n}$. Now you will combine that estimated standard error with a t critical value to build a confidence interval for a population mean $\mu$. The focus here is carrying out the calculation step by step, without relying on a one-button interval command.

A one-sample t interval uses the sample mean $\bar{x}$ as its center. Its margin of error is the t critical value $t^*$ multiplied by the estimated standard error. The interval extends that margin of error below and above $\bar{x}$.

Formula: A one-sample t confidence interval for a population mean is $$ \bar{x}\ \pm\ t^*\left(\frac{s}{\sqrt{n}}\right), $$ or, as endpoints, $$ \left(\bar{x}-t^*\frac{s}{\sqrt{n}},\ \bar{x}+t^*\frac{s}{\sqrt{n}}\right). $$ Here $t^*$ depends on the confidence level and the degrees of freedom $df=n-1$.

The formula is familiar from “The Form of a One-Sample t Interval.” The new work is putting its pieces together accurately. First check whether a t interval is appropriate. Then find $df$, use the confidence level to select $t^*$, calculate the standard error and margin of error, and state the endpoints with units.

For a central confidence level $C$, the area outside the interval is $1-C$, split equally between the two tails. Thus the cumulative area to the left of the positive critical value is $(1+C)/2$. As covered in “Finding t* Critical Values With a Table or invT,” use that area and $df$ to find $t^*$. Keep extra digits in intermediate calculations so that rounding the standard error early does not change the final endpoints.

Conditions and a Hand-Calculation Workflow

A t interval is intended to estimate a population mean from sample data, so check how the data were collected and whether the sample is suitable for the t procedure. With a small sample, the shape check matters: the sample data should not show strong skewness or outliers. A roughly symmetric sample without outliers is generally suitable; a Normal population also supports the procedure.

Conditions: Before using a one-sample t interval, check that the data come from a random sample from the population of interest (random assignment alone supports causal inference, not generalization to a population), that observations are independent (use the 10% condition for sampling without replacement from a finite population), and that the data are sufficiently well behaved for a t procedure. For a small sample, look for strong skewness and outliers; a Normal population is also appropriate.

For a simple random sample without replacement, the 10% condition is $n<0.10N$, where $N$ is the population size. This supports treating observations as independent. As noted in “The 10% Condition for Sample Means,” the condition is not a requirement that the population be ten times the sample size; it says the sample must be less than 10% of the population.

State the target and check conditions.
Identify the population mean $\mu$, describe the sampling method, address independence and the 10% condition when needed, and inspect the sample shape when the sample is small.

Find the degrees of freedom and $t^*$.
Calculate $df=n-1$. Use the confidence level and this $df$ to select the positive critical value.

Calculate the standard error and margin of error.
Find $s/\sqrt{n}$, then multiply by $t^*$. Keep unrounded values during the calculation when possible.

Write and interpret the interval.
Subtract and add the margin of error to $\bar{x}$. Include units and interpret the interval as a range of plausible values for the population mean.

The resulting interval does not say that a particular percentage of individual observations falls between its endpoints. It estimates the population mean, not individual values. A confidence level describes the long-run success rate of the interval method: if the method were used repeatedly under the same conditions, about $C$% of the resulting intervals would contain the true population mean.

Worked Examples

Worked Example: Mean Height in a Small Seedling Sample

A fictional greenhouse manager takes a random sample of 15 seedlings from a population of 1,200 seedlings. The sample mean height is $\bar{x}=42.4$ centimeters, and the sample standard deviation is $s=3.0$ centimeters. The sample is roughly symmetric with no apparent outliers. Construct a 95% confidence interval for the population mean height.

State. The parameter is $\mu$, the population mean seedling height. We will estimate it with a 95% one-sample t interval.

Plan. The sample is random. Because it is selected without replacement, check the 10% condition: $15<0.10(1200)=120$, so the condition is met and independence is reasonable. The sample is small, but its shape is roughly symmetric with no apparent outliers. These conditions support using a one-sample t interval.

Do. The sample size is $n=15$, so

$$ df=n-1=15-1=14. $$

For 95% confidence, the cumulative area to the left of the positive critical value is $(1+0.95)/2=0.975$. With $df=14$, a t table or invT gives $t^*\approx2.1448$. The estimated standard error is

SE_{\bar{x}}=\frac{s}{\sqrt{n}} =\frac{3.0}{\sqrt{15}} \approx0.7745967\text{ centimeters}.

Now multiply the unrounded standard-error expression by $t^*$ to find the margin of error:

\text{margin of error} =2.1448\left(\frac{3.0}{\sqrt{15}}\right) \approx1.6614\text{ centimeters}.

Subtract and add this margin to the sample mean:

42.4\pm1.6614 =(40.7386,\ 44.0614)\text{ centimeters}.

Conclude. We are 95% confident that the population mean height of the seedlings is between 40.7386 and 44.0614 centimeters. This interval gives plausible values for the population mean, not a range containing 95% of individual seedling heights.

Worked Example: A 90% Interval for Delivery Time

A fictional delivery service selects a random sample of 10 completed routes from a large set of routes. The mean delivery time is 18.6 minutes, and the sample standard deviation is 2.4 minutes. The observed times are roughly symmetric with no outliers. Construct a 90% confidence interval for the population mean delivery time.

The sample is random. Since the routes are sampled without replacement, independence is reasonable if the set contains more than 100 routes: $10<0.10N$ requires $N>100$. The stated set is large, so this condition is met. The sample is small, and the roughly symmetric shape without outliers supports using a t interval.

Here $n=10$, so $df=10-1=9$. For a 90% interval, the cumulative area to the left of $t^*$ is $(1+0.90)/2=0.95$. The t critical value for $df=9$ is approximately 1.8331. Calculate the standard error and margin of error:

SE_{\bar{x}}=\frac{2.4}{\sqrt{10}} \approx0.7589466\text{ minutes},

\text{margin of error} =1.8331\left(\frac{2.4}{\sqrt{10}}\right) \approx1.3912\text{ minutes}.

Using the unrounded calculation for the margin of error, the endpoints are

18.6\pm1.3912 =(17.2088,\ 19.9912)\text{ minutes}.

We are 90% confident that the population mean delivery time is between 17.2088 and 19.9912 minutes. Compared with a higher confidence level using the same data, a 90% interval uses a smaller critical value and therefore has a smaller margin of error.

Worked Example: A Higher Confidence Level for Battery Life

A fictional lab takes a random sample of 15 rechargeable batteries from a large shipment. Their mean operating time is 7.8 hours, with a sample standard deviation of 1.5 hours. The sample distribution is approximately symmetric, with no outliers. Find a 99% confidence interval for the population mean operating time.

The data come from a random sample. For sampling without replacement, the 10% condition is met if the shipment contains more than 150 batteries, since $15<0.10N$ requires $N>150$. The shipment is large enough to satisfy this condition. The sample is small, but its shape is appropriate for a t procedure.

The degrees of freedom are $df=15-1=14$. For 99% confidence, the left-tail area at the positive critical value is $(1+0.99)/2=0.995$. With $df=14$, $t^*\approx2.9768$. Then

SE_{\bar{x}}=\frac{1.5}{\sqrt{15}} \approx0.3872983\text{ hours},

\text{margin of error} =2.9768\left(\frac{1.5}{\sqrt{15}}\right) \approx1.1529\text{ hours}.

The interval is

7.8\pm1.1529 =(6.6471,\ 8.9529)\text{ hours}.

We are 99% confident that the population mean operating time is between 6.6471 and 8.9529 hours. The higher confidence level produces a wider interval than a lower confidence level would for the same sample statistics because it uses a larger $t^*$.

Common Mistakes and AP Exam Tips

Using $n$ instead of $n-1$ for degrees of freedom. In a one-sample t interval, $df=n-1$. For $n=15$, use $df=14$, not 15.
Using the wrong tail area. A 95% central interval leaves 0.05 outside, with 0.025 in each tail. Use cumulative area 0.975 to find the positive $t^*$, not 0.95.
Using a z critical value. For a one-sample interval for a mean when $\sigma$ is unknown, use $t^*$ and $s$, as explained in “Why We Use t Instead of z for Means.”
Multiplying by $s$ instead of the standard error. The margin of error is $t^*(s/\sqrt{n})$, not $t^*s$. First divide $s$ by $\sqrt{n}$.
Rounding too early. Keep the calculator’s unrounded standard error in the margin-of-error calculation. If you display rounded intermediate values, ensure the reported result agrees with the displayed arithmetic or explicitly show the unrounded expression.
Skipping conditions for a small sample. A random sample alone does not resolve the shape requirement. A complete response also addresses the sample distribution or the population’s shape, as appropriate.
Giving an imprecise interpretation. A full-credit statement identifies the population mean, gives the interval in context and with units, and says “we are [confidence level]% confident.” Do not say there is a [confidence level]% probability that this fixed interval contains $\mu$.

For an AP response, show enough work to make each choice visible: $df$, the selected $t^*$, the standard error, the margin of error, and both endpoints. Then write a contextual confidence statement. The calculation is only part of the answer; condition checks and a correct interpretation explain why the interval is meaningful.

Key takeaway: Build a one-sample t interval by checking conditions, finding $df=n-1$ and the appropriate $t^*$, calculating $s/\sqrt{n}$ and the margin of error, then writing $\bar{x}\pm\text{margin of error}$. Interpret the result as a confidence interval for the population mean in context.

Check Your Understanding

Use the one-sample t interval process. Show your calculations and include units when a context is given.

A random sample of 12 measurements has $\bar{x}=26.0$ centimeters and $s=4.0$ centimeters. What are the degrees of freedom?
For a 95% confidence interval with $n=12$, what cumulative left-tail area should you use to find the positive critical value?
A random sample of 15 values from a population of 900 has no strong skewness or outliers. Check the 10% condition.
A sample of 10 randomly selected bike rides has $\bar{x}=32$ minutes and $s=5$ minutes. For a 90% interval, use $t^*=1.8331$. Calculate the standard error, margin of error, and interval.
In context, what does it mean to be 95% confident in an interval for a population mean? What does it not mean about individual observations?

Building a One-Sample t Interval by Hand

What You'll Learn