Confidence Intervals for Means

When you want to estimate a population mean $\mu$ from a sample, a single number like $\bar{x} = 105$ tells you surprisingly little. Is the true mean likely to be 100? Or 110? Or somewhere else entirely? A confidence interval for the mean answers this question by providing a range of plausible values for $\mu$, together with a stated level of confidence. In this lesson, you will learn two approaches: the z-interval (when the population standard deviation is known) and the t-interval (when it is not).

Z-Interval (Population Standard Deviation Known)

When the population standard deviation $\sigma$ is known — which is rare in practice but foundational for understanding the concept — the confidence interval for the mean is:

$\bar{x} \pm z^* \cdot \frac{\sigma}{\sqrt{n}}$

Here $\bar{x}$ is the sample mean, $\sigma$ is the known population standard deviation, $n$ is the sample size, and $z^*$ is the critical value from the standard normal distribution for your chosen confidence level (1.645 for 90%, 1.960 for 95%, 2.576 for 99%).

The quantity $\frac{\sigma}{\sqrt{n}}$ is the standard error of the mean — it measures how much the sample mean $\bar{x}$ typically varies from sample to sample. Notice that as $n$ increases, the standard error decreases: larger samples produce more precise estimates.

Example 1: IQ Scores

IQ tests are designed so that $\sigma = 15$ (this is known by the test’s construction). A researcher administers the test to a random sample of $n = 50$ college students and finds $\bar{x} = 105$. Construct a 95% confidence interval for the mean IQ of all college students.

Step 1: Calculate the standard error.

$SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{50}} = \frac{15}{7.071} = 2.121$

Step 2: Calculate the margin of error.

$ME = z^* \times SE = 1.960 \times 2.121 = 4.157$

Step 3: Construct the interval.

$105 \pm 4.16 = (100.84,\ 109.16)$

Answer: We are 95% confident that the mean IQ of all college students is between 100.84 and 109.16.

The T-Distribution

In the real world, you almost never know $\sigma$. When you replace $\sigma$ with the sample standard deviation $s$, you introduce additional uncertainty — your estimate of the spread itself has sampling variability. The t-distribution accounts for this extra uncertainty.

The t-distribution looks like the standard normal distribution (symmetric, bell-shaped, centered at 0) but with heavier tails. This means extreme values are more likely under the t-distribution than under the normal, which makes the confidence interval appropriately wider when you are less certain about the spread.

The t-distribution has one parameter: degrees of freedom (df), which equals $n - 1$ for a one-sample interval. Key properties include:

• When df is small (say 5 or 10), the tails are noticeably heavier than the normal curve
• As df increases, the t-distribution gets closer and closer to the standard normal
• For df of 30 or more, the t-distribution and the normal distribution are nearly identical
• At $df = \infty$, the t-distribution is exactly the standard normal

This means the t-distribution automatically adjusts: small samples get wider intervals (more uncertainty), while large samples get intervals similar to the z-interval.

T-Interval (Population Standard Deviation Unknown)

The t-interval is the standard approach for constructing confidence intervals for means in practice:

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

The formula looks just like the z-interval, with two changes: $\sigma$ is replaced by $s$ (the sample standard deviation), and $z^*$ is replaced by $t^*$ (the critical value from the t-distribution with $df = n - 1$).

Example 2: Patient Recovery Time

A hospital records the recovery time (in days) for $n = 25$ randomly selected patients who underwent a particular surgery. The sample mean is $\bar{x} = 4.8$ days and the sample standard deviation is $s = 1.2$ days. Construct a 95% confidence interval for the true mean recovery time.

Step 1: Determine degrees of freedom and the critical value.

$df = 25 - 1 = 24$

From a t-table at 95% confidence with $df = 24$: $t^* = 2.064$.

Step 2: Calculate the standard error.

$SE = \frac{s}{\sqrt{n}} = \frac{1.2}{\sqrt{25}} = \frac{1.2}{5} = 0.24$

Step 3: Calculate the margin of error.

$ME = t^* \times SE = 2.064 \times 0.24 = 0.495$

Step 4: Construct the interval.

$4.8 \pm 0.50 = (4.30,\ 5.30)$

Answer: We are 95% confident that the true mean recovery time is between 4.30 and 5.30 days. A hospital administrator could use this interval to plan staffing and bed availability.

Common $t^*$ Values

This table provides critical values for the most commonly used confidence levels and degrees of freedom. For degrees of freedom not listed, use the nearest smaller df (this gives a slightly wider, more conservative interval).

Notice the bottom row: as degrees of freedom approach infinity, $t^*$ converges to $z^*$. This confirms that the t-distribution becomes the standard normal for large samples.

Conditions for a Valid T-Interval

Before using the t-interval, verify these conditions:

1. Random sample — the data must come from a random sampling process
2. Independence — observations must be independent (the 10% condition: sample size is less than 10% of the population)
3. Normality — the population should be approximately normally distributed, or the sample size should be at least 30 (at which point the Central Limit Theorem ensures the sampling distribution of $\bar{x}$ is approximately normal regardless of the population shape)
4. No strong outliers — for small samples, the t-interval is sensitive to outliers. A boxplot or dotplot can help check for extreme values.

For small samples from clearly non-normal populations (heavily skewed distributions with outliers), consider using a nonparametric method like the bootstrap confidence interval instead.

Z vs T — Which Should You Use?

When in doubt, use the t-interval. It is always valid (even when $\sigma$ is known, the t-interval works — it is just very slightly wider than necessary). The z-interval is really just a special case of the t-interval where df is infinite.

Choosing Sample Size for a Desired Margin of Error

Just as with proportions, researchers often plan studies by specifying the desired margin of error and then determining how large a sample is needed. For means, the sample size formula is:

$n = \left(\frac{z^* \cdot \sigma}{E}\right)^2$

This formula uses $z^*$ (not $t^*$) because at the planning stage, you do not yet know the sample size, and therefore you do not know the degrees of freedom. The z-approximation is standard for sample size planning.

You also need a preliminary estimate of $\sigma$, which typically comes from prior studies, pilot data, or published literature.

Example 3: Blood Pressure Study

A researcher wants to estimate the mean systolic blood pressure of adults in a community with a margin of error of no more than 2 mmHg at 95% confidence. Prior studies suggest $\sigma \approx 12$ mmHg. How many participants are needed?

$n = \left(\frac{1.96 \times 12}{2}\right)^2 = \left(\frac{23.52}{2}\right)^2 = (11.76)^2 = 138.3$

Round up: $n = 139$.

Answer: The researcher needs at least 139 participants. If the preliminary estimate of $\sigma$ turns out to be too low, the actual margin of error will be larger than 2 mmHg — so researchers sometimes add a buffer by rounding up more generously.

Real-World Application: Nursing — Estimating Average Wait Time

An emergency department manager wants to estimate the average patient wait time before triage. She randomly selects $n = 36$ patients over a two-week period and records their wait times in minutes. The results show $\bar{x} = 22.5$ minutes and $s = 8.4$ minutes.

Construct a 95% confidence interval for the true mean wait time.

With $df = 36 - 1 = 35$, the critical value is approximately $t^* = 2.030$.

$SE = \frac{8.4}{\sqrt{36}} = \frac{8.4}{6} = 1.40$

$ME = 2.030 \times 1.40 = 2.842$

$CI: 22.5 \pm 2.84 = (19.66,\ 25.34)$

Interpretation: The hospital is 95% confident that the true mean wait time before triage is between 19.7 and 25.3 minutes. If the department’s target is 20 minutes or less, this interval suggests the target may not be met — the lower bound is barely below 20, and the best estimate is 22.5. This data provides actionable evidence for staffing decisions.

Why the t-interval is appropriate here: The population standard deviation of wait times is unknown (we estimated it with $s = 8.4$), and the sample size of 36 satisfies the Central Limit Theorem requirement even if wait times are somewhat right-skewed (as they often are in healthcare settings).

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• When $\sigma$ is known (rare), use the z-interval: $\bar{x} \pm z^* \cdot \frac{\sigma}{\sqrt{n}}$
• When $\sigma$ is unknown (almost always), use the t-interval: $\bar{x} \pm t^* \cdot \frac{s}{\sqrt{n}}$ with $df = n - 1$
• The t-distribution has heavier tails than the normal, producing wider intervals that account for the uncertainty in estimating $\sigma$ with $s$
• As the sample size grows, the t-distribution approaches the normal distribution — for $df \geq 30$, the difference is small
• Conditions for the t-interval: random sample, independence (10% rule), approximate normality or $n \geq 30$, and no strong outliers for small samples
• When in doubt, use the t-interval — it is always valid and only slightly more conservative than the z-interval
• To plan a study with a desired margin of error, use $n = (z^* \cdot \sigma / E)^2$ and round up
• In healthcare, confidence intervals for means help administrators quantify metrics like average wait times, recovery durations, and dosage responses — turning sample data into actionable estimates with clearly stated precision

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