What is the difference between a confidence interval and a hypothesis test?

A confidence interval estimates a range where the population mean likely lies. A hypothesis test checks whether a specific claimed value for the population mean is plausible. They are complementary: if a value falls outside your 95% confidence interval, a hypothesis test at the 5% significance level would reject it.

Why do I have to look up t^* in a table instead of using the same critical value every time?

The critical value t^* depends on both your chosen confidence level and your sample size (through degrees of freedom). Different confidence levels and sample sizes require different critical values, so you must consult a t-table or calculator each time.

What does 95% confident mean exactly?

It means that if you repeated your sampling procedure many times and constructed a 95% confidence interval each time, approximately 95% of those intervals would contain the true population mean. It does not mean there is a 95% probability that this specific interval contains the mean—the mean either is or is not in this interval.

Why does my confidence interval get narrower when I increase the sample size?

The margin of error includes the term \frac{s}{\sqrt{n}}. As n increases, \sqrt{n} gets larger, so the whole fraction gets smaller. A smaller margin of error produces a narrower interval, reflecting greater precision from a larger sample.

Confidence Interval for a Mean

Estimates the range where a population mean likely falls using sample data when the population standard deviation is unknown.

\bar{x} \pm t^{*}\frac{s}{\sqrt{n}}

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What each symbol means

What Confidence Interval for a Mean takes

Symbol	Meaning
$\bar{x}$	The sample mean, the average of your data values. It estimates the unknown population mean and is located at the center of the confidence interval. Confusing it with individual data points or the population mean $\mu$ leads to wrong interpretations.
$t^{*}$	The critical value from the Student's t-distribution, found using a t-table with your confidence level and degrees of freedom. It tells you how many standard errors to extend from the sample mean. Using a z-critical value instead gives an interval that is too narrow.
$s$	The sample standard deviation, measuring the spread of your sample data around the sample mean. It estimates the unknown population standard deviation $\sigma$ and uses the same units as your data. Using the population standard deviation directly (if you somehow knew it) would change the entire calculation.
$n$	Your sample size, the number of observations in your sample. Larger samples give narrower intervals because the standard error decreases as $\sqrt{n}$ increases, making your estimate more precise.

When to use it

Use this when you have a sample from a population and want to estimate where the true population mean falls, particularly with smaller samples or when population variability is unknown.

Level

Usually taught in: Statistics · Appears on: AP Statistics

Worked examples

1. Estimate population mean height from a small sample

Problem

A school measures the heights of 10 randomly selected students. The sample mean is 65 inches and the sample standard deviation is 2 inches. Construct a 95% confidence interval for the population mean height.

$\bar{x} = 65, \quad s = 2, \quad n = 10, \quad \text{confidence level} = 95\%$
We organize all given values: sample mean, sample standard deviation, sample size, and desired confidence level.
$df = n - 1 = 10 - 1 = 9$
Degrees of freedom for a confidence interval for the mean is $df = n - 1 = 9$ .
$t^{*} = 2.262$
From the t-table with 95% confidence and $df = 9$ , we find $t^* = 2.262$ .
$SE = \frac{s}{\sqrt{n}} = \frac{2}{\sqrt{10}} = \frac{2}{3.162} = 0.632$
The standard error measures the precision of the sample mean: $SE = \frac{s}{\sqrt{n}}$ .
$ME = t^{*} \cdot SE = 2.262 \times 0.632 = 1.430$
The margin of error extends the interval on either side of the sample mean: $ME = t^* \cdot SE$ .
$65 \pm 1.430 = (63.570, 66.430)$
The confidence interval is the sample mean plus and minus the margin of error.

Answer: $(63.57, 66.43) \text{ inches}$

We use the t-distribution because the population standard deviation is unknown and we work with a modest sample size. This interval tells us we can be 95% confident that the true average height of all students at this school falls between 63.57 and 66.43 inches.

2. Estimate population mean calcium intake

Problem

A nutritionist measures daily calcium intake (in mg) for 15 patients and finds a sample mean of 850 mg with a sample standard deviation of 125 mg. Construct a 90% confidence interval for the population mean daily calcium intake.

$\bar{x} = 850, \quad s = 125, \quad n = 15, \quad \text{confidence level} = 90\%$
We list the given values: sample mean in mg, sample standard deviation in mg, sample size, and confidence level.
$df = 15 - 1 = 14$
Degrees of freedom is $df = n - 1 = 14$ .
$t^{*} = 1.761$
Using a t-table for 90% confidence with $df = 14$ , we find $t^* = 1.761$ .
$SE = \frac{125}{\sqrt{15}} = \frac{125}{3.873} = 32.276$
The standard error is $SE = \frac{s}{\sqrt{n}} \approx 32.276$ mg.
$ME = 1.761 \times 32.276 = 56.852$
The margin of error is calculated as $ME = t^* \times SE \approx 56.852$ mg.
$850 - 56.852 = 793.148, \quad 850 + 56.852 = 906.852$
The lower bound is $850 - 56.852$ and the upper bound is $850 + 56.852$ .

Answer: $(793.15, 906.85) \text{ mg}$

A 90% confidence interval is appropriate here because we are estimating the population mean from a sample without knowing the population variability. We conclude that the true mean daily calcium intake for patients likely falls between 793.15 and 906.85 mg.

3. Estimate population mean monthly phone bill

Problem

A cell phone company analyzes monthly bills for 20 randomly selected customers on the same plan. The sample mean bill is $65 with a sample standard deviation of $8. Construct a 95% confidence interval for the population mean monthly bill.

$\bar{x} = 65, \quad s = 8, \quad n = 20, \quad \text{confidence level} = 95\%$
We identify the sample mean in dollars, sample standard deviation in dollars, sample of 20 customers, and 95% confidence level.
$df = 20 - 1 = 19$
Degrees of freedom equals $df = n - 1 = 19$ .
$t^{*} = 2.093$
From the t-table at 95% confidence with $df = 19$ , we find $t^* = 2.093$ .
$SE = \frac{8}{\sqrt{20}} = \frac{8}{4.472} = 1.789$
The standard error is $SE = \frac{s}{\sqrt{n}} \approx 1.789$ dollars.
$ME = 2.093 \times 1.789 = 3.744$
The margin of error is $ME = t^* \times SE \approx 3.744$ dollars.
$65 - 3.744 = 61.256, \quad 65 + 3.744 = 68.744$
Subtract and add the margin of error from the sample mean to get the interval bounds.

Answer: $(61.26, 68.74) \text{ dollars}$

Since the company does not know the population standard deviation of monthly bills, the t-distribution is the correct choice. This 95% confidence interval indicates that the company can be 95% confident the true average monthly bill for customers on this plan is between $61.26 and $68.74.

Common mistakes

Where Confidence Interval for a Mean usually goes wrong

The mistake	Why it is wrong	The fix
Using the z-critical value $z^$ (such as 1.96 for 95% confidence) instead of the t-critical value $t^$ .	The z-distribution is used only when the population standard deviation $\sigma$ is known; here we use the sample standard deviation $s$ , so the t-distribution is correct.	Always use the t-distribution when $\sigma$ is unknown and you're estimating from a sample. Look up $t^*$ in a t-table using your confidence level and $df = n - 1$ .
Forgetting to divide the sample standard deviation by $\sqrt{n}$ and instead writing the interval as $\bar{x} \pm t^* s$ (without the square root term).	The formula requires the standard error $\frac{s}{\sqrt{n}}$ , not the raw sample standard deviation. Omitting $\sqrt{n}$ makes the interval far too wide and gives a meaningless result.	The correct formula is $\bar{x} \pm t^* \frac{s}{\sqrt{n}}$ . Always divide $s$ by $\sqrt{n}$ before multiplying by $t^*$ .
Using $n$ instead of $n - 1$ when calculating degrees of freedom, writing $df = n$ instead of $df = n - 1$ .	One degree of freedom is consumed in computing the sample standard deviation $s$ , so the t-distribution has $n - 1$ degrees of freedom. Using the wrong df gives an incorrect critical value.	Always calculate degrees of freedom as $df = n - 1$ . For a sample of $n = 25$ , use $df = 24$ to look up $t^*$ .

Tips and when to use something else

A confidence interval for the mean estimates a range; it does not test whether a specific value is plausible. If you need to test a claim about the population mean, use a One-Sample t Test instead.
Higher confidence levels (99% vs. 90%) produce wider intervals because they require a larger $t^*$ value. More confidence means less precision.
Larger samples reduce the standard error because of the $\sqrt{n}$ term, making your interval narrower and your estimate more precise. This is why larger samples are generally preferred in statistics.
The confidence level (such as 95%) describes the long-run behavior of the method: if you repeated sampling and constructed intervals many times, about 95% would contain the true population mean.

Related formulas

Standard Error of the MeanThe standard error of the mean measures how much sample means vary around the true population mean; use it for confidence intervals and hypothesis tests.Margin of ErrorMargin of Error measures how far your sample statistic might differ from the true population value, used when estimating populations from sample data.One-Sample t TestA one-sample t test compares a sample mean to a hypothesized population mean when the population standard deviation is unknown.Standard DeviationStandard deviation measures how spread out data points are from the average—use it to understand variability in a dataset.Arithmetic MeanArithmetic Mean is the average value found by adding all numbers in a dataset and dividing by how many numbers there are.

Frequently asked questions

What is the difference between a confidence interval and a hypothesis test?: A confidence interval estimates a range where the population mean likely lies. A hypothesis test checks whether a specific claimed value for the population mean is plausible. They are complementary: if a value falls outside your 95% confidence interval, a hypothesis test at the 5% significance level would reject it.
Why do I have to look up $t^*$ in a table instead of using the same critical value every time?: The critical value $t^*$ depends on both your chosen confidence level and your sample size (through degrees of freedom). Different confidence levels and sample sizes require different critical values, so you must consult a t-table or calculator each time.
What does 95% confident mean exactly?: It means that if you repeated your sampling procedure many times and constructed a 95% confidence interval each time, approximately 95% of those intervals would contain the true population mean. It does not mean there is a 95% probability that this specific interval contains the mean—the mean either is or is not in this interval.
Why does my confidence interval get narrower when I increase the sample size?: The margin of error includes the term $\frac{s}{\sqrt{n}}$ . As $n$ increases, $\sqrt{n}$ gets larger, so the whole fraction gets smaller. A smaller margin of error produces a narrower interval, reflecting greater precision from a larger sample.

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Reviewed 2026-09-18