AP Statistics Unit 7: Confidence Intervals for Quantitative Means (One-Sample and Two-Sample)

Introduction to t-Distributions

Why you need something other than the normal distribution

When you build a confidence interval for a population mean, you’re trying to estimate an unknown parameter: the true mean of a population, written as $\mu$. In the “best case,” you would know the population standard deviation $\sigma$, which tells you how spread out individual observations are around $\mu$. If $\sigma$ were known, the standardized statistic

$z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$

would follow a standard normal distribution (under the usual random sampling conditions), and you could build a confidence interval using a normal critical value.

In real life (and on AP Statistics), $\sigma$ is almost never known. Instead, you estimate it with the sample standard deviation $s$. That seemingly small substitution creates extra variability: different random samples produce different $s$ values, and that uncertainty must be reflected in your model for the standardized statistic.

That is exactly why the t-distribution exists.

What a t-distribution is

A t-distribution is a family of distributions used to model the standardized sample mean when $\sigma$ is unknown and you use $s$ instead. The corresponding standardized statistic is

$t = \frac{\bar{x}-\mu}{s/\sqrt{n}}$

This statistic follows a t-distribution (assuming the conditions for inference are met). Unlike the standard normal distribution, the t-distribution depends on degrees of freedom.

Degrees of freedom (df): what they mean here

For the one-sample mean setting, the degrees of freedom are

$df = n - 1$

Conceptually, degrees of freedom measure how much “independent information” is available to estimate variability. Because $s$ is computed using $\bar{x}$ (the sample mean), you lose one degree of freedom—hence $n-1$.

How t-distributions compare to the standard normal

All t-distributions are:

• Centered at 0 and symmetric like the standard normal.
• More spread out than the standard normal, especially in the tails.

Those heavier tails matter: they make confidence intervals wider (more cautious) to account for the extra uncertainty from estimating $\sigma$ with $s$.

As $n$ increases, $s$ becomes a better estimate of $\sigma$, and the t-distribution approaches the standard normal distribution. In practice, for large $df$, t critical values are close to z critical values.

Critical values and notation

For confidence intervals, you use a critical value from a t-distribution. The notation

$t^*$

means “the t critical value that captures the middle C of the distribution,” where $C$ is your confidence level (like 0.95 for a 95% interval), using the correct degrees of freedom.

For example, for a 95% confidence interval, $t^*$ is chosen so that 95% of the t-distribution lies between $-t^*$ and $t^*$.

Notation reference (common symbols you must read fluently)

Example: seeing how df changes the critical value

Suppose you want a 95% confidence interval.

• With $df = 9$, $t^*$ is noticeably larger than 1.96 (the z critical value).
• With $df = 99$, $t^*$ is only slightly larger than 1.96.

The key takeaway is not memorizing specific numbers—it’s understanding the direction: smaller samples mean larger $t^*$, which means wider intervals.

Exam Focus

• Typical question patterns:
  • Identify whether to use z or t (AP almost always expects t for means because $\sigma$ is unknown).
  • Determine degrees of freedom and select/interpret $t^*$.
  • Compare the shape/spread of t-distributions as $df$ changes.
• Common mistakes:
  • Using a normal critical value (like 1.96) when $\sigma$ is not given.
  • Using $df = n$ instead of $df = n-1$ for one-sample mean procedures.
  • Thinking the t-distribution is skewed; it is symmetric—its difference is heavier tails.

Constructing a Confidence Interval for a Population Mean

What a confidence interval for a mean is (and what it is not)

A confidence interval for a population mean is a range of plausible values for $\mu$, based on sample data. The interval is built from:

• a point estimate (usually $\bar{x}$), and
• a margin of error that accounts for sampling variability.

A 95% confidence interval does not mean there is a 95% probability that $\mu$ is in your computed interval. After you compute it, the interval either contains $\mu$ or it doesn’t. The “95%” refers to the long-run success rate of the method: if you repeated the sampling process many times, about 95% of those intervals would capture $\mu$.

Why the t-interval works

Because you don’t know $\sigma$, you use $s$ to estimate the spread of the sampling distribution of $\bar{x}$. The estimated standard deviation of $\bar{x}$ is called the standard error:

$SE = \frac{s}{\sqrt{n}}$

Then you take your point estimate $\bar{x}$ and go out $t^*$ standard errors in both directions.

The one-sample t confidence interval formula

A one-sample t interval for the population mean $\mu$ is

$\bar{x} \pm t^*\left(\frac{s}{\sqrt{n}}\right)$

Equivalently, you can write it as

$\left(\bar{x} - t^*\frac{s}{\sqrt{n}},\ \bar{x} + t^*\frac{s}{\sqrt{n}}\right)$

Where:

• $\bar{x}$ is the sample mean.
• $s$ is the sample standard deviation.
• $n$ is the sample size.
• $t^*$ comes from the t-distribution with $df = n-1$ at the chosen confidence level.

Conditions for using a one-sample t interval (what AP wants you to check)

You’re expected to justify inference with conditions. A common AP-friendly structure is:

1. Random: Data come from a random sample or a randomized experiment.
2. Normal (or approximately normal sampling distribution of $\bar{x}$):
   • If the population is approximately normal, you’re fine.
   • If $n$ is large, the Central Limit Theorem supports that $\bar{x}$ is approximately normal.
   • If $n$ is small, you should check that the sample distribution looks roughly symmetric with no strong outliers.
3. Independence: Observations are independent. If sampling without replacement, a common check is the 10% condition:

$n \le 0.10N$

where $N$ is the population size.

These conditions matter because the t-interval is derived assuming the t statistic behaves like a t-distribution. Strong skewness with a small sample or extreme outliers can break that approximation.

How confidence level, sample size, and variability affect the interval

It helps to predict what happens before calculating.

• Increasing the confidence level (like 90% to 95%) increases $t^*$, so the interval gets wider.
• Increasing sample size $n$ decreases $s/\sqrt{n}$, so the interval gets narrower.
• Increasing variability $s$ increases the standard error, so the interval gets wider.

This is the logic behind margin of error:

$ME = t^*\left(\frac{s}{\sqrt{n}}\right)$

Interpreting the interval in context (the wording AP expects)

A correct interpretation:

• Names the confidence level.
• Refers to the parameter $\mu$ (not $\bar{x}$).
• Uses the context (what the mean represents).

Template you can adapt:

“We are $C$% confident that the true mean (context) for (population) is between (lower) and (upper).”

Worked example: one-sample t interval

A nutrition researcher takes a random sample of $n = 25$ energy bars of a certain brand and measures calories per bar. The sample mean is $\bar{x} = 214.0$ calories and the sample standard deviation is $s = 10.0$ calories. Construct a 95% confidence interval for the true mean calories per bar $\mu$.

Step 1: Identify the procedure and check conditions

• We want $\mu$ and $\sigma$ is not given, so we use a one-sample t interval.
• Random: stated random sample.
• Independence: reasonable if the sample is less than 10% of all bars produced (assume yes).
• Normal: $n = 25$ is moderately sized; if no strong skew/outliers are indicated, t procedures are typically considered reasonable.

Step 2: Degrees of freedom

$df = n - 1 = 24$

Step 3: Find the critical value
For 95% confidence with $df = 24$, use $t^*$ from a t table or technology. (You do not need to memorize it; you must select it appropriately.)

Step 4: Compute the standard error

$SE = \frac{s}{\sqrt{n}} = \frac{10.0}{\sqrt{25}} = 2.0$

Step 5: Compute the interval

$\bar{x} \pm t^*SE = 214.0 \pm t^*(2.0)$

If $t^*$ is approximately 2.064 for $df = 24$, then

$ME = 2.064(2.0) = 4.128$

So the interval is

$\left(214.0 - 4.128,\ 214.0 + 4.128\right) = \left(209.872,\ 218.128\right)$

Interpretation (in context)
“We are 95% confident that the true mean calories per bar for this brand is between about 209.9 and 218.1 calories.”

What can go wrong (common conceptual traps)

• Confusing standard deviation and standard error: $s$ describes spread of individual data; $s/\sqrt{n}$ describes spread of $\bar{x}$ across samples.
• Ignoring outliers: A single extreme value can inflate $s$ and distort $\bar{x}$, making the interval less meaningful.
• Thinking higher confidence means ‘more accurate’: Higher confidence increases reliability but widens the interval. You gain certainty by giving a wider range.

Exam Focus

• Typical question patterns:
  • “Construct and interpret a 95% confidence interval for $\mu$” given $\bar{x}$, $s$, and $n$.
  • “Check conditions” using a graph (dotplot/histogram/boxplot) and sampling description.
  • Compute or interpret the margin of error and explain how to make it smaller.
• Common mistakes:
  • Interpreting the interval as a probability statement about $\mu$.
  • Using $\sigma$ formulas or z critical values when only $s$ is available.
  • Forgetting to include context and the population/parameter in the interpretation.

Confidence Interval for a Difference of Two Means

The goal: comparing two population means

Often you’re not just estimating one mean—you’re comparing two groups. For example:

• Do students who sleep at least 8 hours have a higher mean test score than those who sleep less?
• Is the mean recovery time different for two medical treatments?

Here, you want the difference between two population means:

$\mu_1 - \mu_2$

A confidence interval for a difference of two means gives a plausible range of values for that parameter.

Two-sample setting: what data structure you need

To use a two-sample t interval (for independent groups), you need:

• Two independent samples (or two independent randomized groups).
• A quantitative variable measured for both groups.

Each group has its own sample statistics:

• Group 1: $\bar{x}_1$, $s_1$, $n_1$
• Group 2: $\bar{x}_2$, $s_2$, $n_2$

Your point estimate of $\mu_1 - \mu_2$ is

$\bar{x}_1 - \bar{x}_2$

Why the standard error is different for two means

A sample mean varies from sample to sample. When you subtract two sample means, the variability adds (in a variance sense), so the standard error for the difference is

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

This captures two sources of sampling variability—one from each group.

The two-sample t confidence interval (independent samples)

A two-sample t interval for $\mu_1 - \mu_2$ is

$\left(\bar{x}_1 - \bar{x}_2\right) \pm t^*\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

The main new practical issue is degrees of freedom. Many technologies compute an approximate $df$ automatically (often associated with Welch’s method). On AP Statistics, using technology for $t^*$ and $df$ is acceptable; if you must approximate by hand, a common conservative choice sometimes taught is

$df = \min(n_1 - 1,\ n_2 - 1)$

If your course emphasizes calculator-based inference, you’ll typically report the interval produced by the two-sample t-interval procedure.

Conditions (what you must justify)

You typically justify conditions for each group plus independence between groups.

1. Random: Each sample is random, or treatments were randomly assigned.
2. Independence within each group: If sampling without replacement, check the 10% condition separately:

$n_1 \le 0.10N_1$

$n_2 \le 0.10N_2$

3. Independent groups: The two samples/groups do not influence each other (no pairing, no repeated measures on the same individuals).
4. Normal (within each group): Each group’s data are approximately normal, or each sample size is large enough for the sampling distribution of each mean to be approximately normal.

A major warning sign is outliers or strong skew in either group when sample sizes are small.

Paired data is a different procedure (important distinction)

Students often confuse “two groups” with “two-sample.” If measurements are naturally matched (same subjects before/after, twins, matched pairs), you do not use the two-sample t interval above. Instead, you compute differences for each pair and run a one-sample t interval on the differences. In this section, the focus is the independent two-sample interval, but you should always ask: “Are the observations paired?”

Interpreting a two-mean interval (including the ‘zero check’)

A correct interpretation names $\mu_1 - \mu_2$ and uses context:

“We are $C$% confident that the true difference in mean (context) between (population 1) and (population 2) is between (lower) and (upper), where the difference is defined as $\mu_1 - \mu_2$.”

A powerful insight comes from checking whether the interval contains 0:

• If 0 is inside the interval, then a difference of 0 is plausible, so there is not clear evidence of a difference (at a level consistent with that confidence).
• If 0 is not inside the interval, then 0 is not plausible, suggesting a real difference in means.

Be careful: this is a relationship to hypothesis testing ideas, but the interpretation is still about plausible values for $\mu_1 - \mu_2$.

Worked example: two-sample t interval for $\mu_1 - \mu_2$

A school compares mean weekly study time for students in two programs.

• Program 1: $n_1 = 40$, $\bar{x}_1 = 12.4$ hours, $s_1 = 3.6$ hours
• Program 2: $n_2 = 35$, $\bar{x}_2 = 10.8$ hours, $s_2 = 4.1$ hours

Construct a 95% confidence interval for $\mu_1 - \mu_2$, where $\mu_1$ is the mean study time for Program 1 and $\mu_2$ is the mean study time for Program 2.

Step 1: Choose procedure and check conditions

• We are estimating $\mu_1 - \mu_2$ with $\sigma_1$ and $\sigma_2$ unknown, so use a two-sample t interval.
• Random: assume each group is a random sample from its program (or students were randomly sampled).
• Independence within groups: reasonable if each sample is less than 10% of its program population.
• Independent groups: two different sets of students.
• Normal: both sample sizes are fairly large (35 and 40), so inference is typically reasonable.

Step 2: Compute the point estimate

$\bar{x}_1 - \bar{x}_2 = 12.4 - 10.8 = 1.6$

Step 3: Compute the standard error

$SE = \sqrt{\frac{3.6^2}{40} + \frac{4.1^2}{35}}$

Compute pieces:

$\frac{3.6^2}{40} = \frac{12.96}{40} = 0.324$

$\frac{4.1^2}{35} = \frac{16.81}{35} \approx 0.4803$

So

$SE = \sqrt{0.324 + 0.4803} = \sqrt{0.8043} \approx 0.8968$

Step 4: Find $t^*$
Using technology (or a conservative df approximation), $df$ will be around the smaller of 39 and 34 if using $\min(n_1-1,n_2-1)$, giving $df = 34$. For 95% confidence, $t^*$ is a bit above 2.

Step 5: Build the interval

$\left(\bar{x}_1 - \bar{x}_2\right) \pm t^*SE = 1.6 \pm t^*(0.8968)$

If $t^*$ is approximately 2.03, then

$ME = 2.03(0.8968) \approx 1.82$

So the interval is

$\left(1.6 - 1.82,\ 1.6 + 1.82\right) = \left(-0.22,\ 3.42\right)$

Interpretation
“We are 95% confident that the true mean difference in weekly study time (Program 1 minus Program 2) is between about -0.22 and 3.42 hours.”

Because 0 is inside the interval, a true difference of 0 hours is plausible based on these data.

What goes wrong most often in two-mean intervals

• Mixing up the order of subtraction: You must define the parameter as $\mu_1 - \mu_2$ and then compute $\bar{x}_1 - \bar{x}_2$ in the same order. If you swap, your interval changes sign.
• Treating paired data as independent: If the same individuals are measured twice, independence is violated; use a one-sample interval on differences instead.
• Forgetting the square root or squaring incorrectly in the standard error:

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

Students sometimes compute $\frac{s_1}{n_1}$ instead of $\frac{s_1^2}{n_1}$, which can drastically shrink the interval incorrectly.

Exam Focus

• Typical question patterns:
  • “Construct and interpret a confidence interval for $\mu_1 - \mu_2$” and state what it suggests about a difference.
  • Determine whether the situation is independent two-sample or paired, and justify the correct method.
  • Explain how changing $n_1$ or $n_2$ affects the margin of error.
• Common mistakes:
  • Reporting an interval but interpreting it as “most sample means will fall here” instead of interpreting it as plausible values for $\mu_1 - \mu_2$.
  • Using $df = n_1 + n_2 - 2$ automatically (that df corresponds to a pooled approach under equal-variance assumptions, which is not the default in many AP Stats courses).
  • Failing to address conditions for both groups (especially checking for skew/outliers separately).