AP Statistics / Unit 4: Inference for Quantitative Data: Means / Topic 4.2
NUM8ERS study notes · Topic 4.2

Constructing a Confidence Interval for a Population Mean or Population Mean Difference

Turn a sample average into an interval estimate. Learn when to use tt, how to calculate a margin of error, and how paired observations become 11 sample of differences.

2026–27 curriculumOne-sample tt intervalsMatched pairs4 worked examples8 practice questions
Estimatexˉ\bar{x}, or dˉ\bar d for paired differences
Add uncertaintyMargin of error=t∗SE⁡\text{Margin of error}=t^*\operatorname{SE}
Check the methodRandomization, independence and shape

By the end of this lesson, you should be able to:

  • Choose a one-sample tt-interval and identify its population parameter.
  • Describe tt-distributions and select t∗t^* using confidence level and degrees of freedom.
  • Check the relevant conditions for single measurements or paired differences.
  • Calculate standard error, margin of error and interval endpoints.

Before you start: Review sample means, sample SD⁡\operatorname{SD}, square roots, random sampling and Topic 4.1’s sampling distributions.

First time learning this? Follow the bottle example, then work through the paired-data example.

Here to revise? Review the key formulas, then try the practice questions.

The concept in 60 seconds

A sample mean is an estimate, not the whole answer. If a sample of bottles averages 501.2 mL501.2 \,\mathrm{mL}, we can use that value to estimate the population mean fill volume. A confidence interval adds a margin of error to show the uncertainty created by random sampling.

The construction: estimate±margin of error\text{estimate}\pm\text{margin of error}.

For a population mean with unknown population SD⁡\operatorname{SD}, use xˉ±t∗sn\bar x\pm t^*\frac{s}{\sqrt n}, after checking the conditions. The critical value t∗t^* comes from a tt-distribution with df=n−1\mathrm{df}=n-1.

For paired data, first calculate 11 difference for each pair. Then construct the same kind of one-sample interval using the mean and SD⁡\operatorname{SD} of those differences.

This lesson focuses on choosing and constructing the interval. We also practise a clear interpretation; evaluating claims from an interval is developed in Topic 4.3.

A bottle-filling investigation

Fictional teaching investigation: A quality-control team wants to estimate the mean fill volume of a lot of 10,00010{,}000 bottles. It selects an SRS of 2525 bottles without replacement. The sample mean is 501.2 mL501.2 \,\mathrm{mL}, and the sample standard deviation is 7.5 mL7.5 \,\mathrm{mL}. Previous process information supports an approximately normal population model. The population mean μ\mu and population SD⁡\operatorname{SD} σ\sigma are unknown.

The team wants a 95%95\% confidence interval for μ\mu, the mean fill volume of all bottles in this lot. It is estimating a mean quantitative measurement, not the proportion of bottles above a cutoff.

Predict: Would xˉ±7.5\bar{x} \pm 7.5 describe the uncertainty of an average properly? Which quantity measures individual variation, and which estimates variation in sample averages?

Visual guide 1: identify the pieces before calculating
Estimatexˉ=501.2 mL\bar{x}=501.2 \,\mathrm{mL}

The sample average estimates the unknown population average.

Estimated sampling spreadSE⁡=7.525=1.5 mL\operatorname{SE}=\frac{7.5}{\sqrt{25}}=1.5 \,\mathrm{mL}

The SD⁡\operatorname{SD} of individual values is 7.5 mL7.5 \,\mathrm{mL}; the estimated SD⁡\operatorname{SD} of the sample mean is 1.5 mL1.5 \,\mathrm{mL}.

Allow for uncertaintyt∗≈2.0639t^*\approx 2.0639

At 95%95\% confidence with df=24\mathrm{df}=24, multiply SE⁡\operatorname{SE} by this critical value.

The confidence level and degrees of freedom determine t∗t^*. The sample statistics determine the center and estimated standard error. Check the sampling design and shape before interpreting the interval.

Topic 4.1 supplied population parameters to study sampling distributions. Here the population parameters are unknown, so we estimate sampling spread with sn\frac{s}{\sqrt{n}} and use tt rather than substituting 1.961.96 automatically.

Key ideas and notation

Keep the population target and the sample inputs distinct.
Symbol / termMeaningIn the bottle investigation
μ\muThe population mean, a fixed unknown parameter.Mean fill volume of all 10,00010{,}000 bottles in this lot.
xˉ\bar{x}The sample mean, our point estimate of μ\mu.501.2 mL501.2 \,\mathrm{mL}.
σ\sigma and ssPopulation SD⁡\operatorname{SD} and sample SD⁡\operatorname{SD}; σ\sigma is unknown here.Use s=7.5 mLs=7.5 \,\mathrm{mL} to estimate individual population variability.
nn and NNNumber sampled and number in the population.n=25n=25 bottles; N=10,000N=10{,}000 bottles.
SE⁡xˉ\operatorname{SE}_{\bar x}Estimated standard deviation of the sample mean: sn\frac{s}{\sqrt{n}}.1.5 mL1.5 \,\mathrm{mL}.
t∗t^*Positive critical value for the central confidence area of the relevant tt-distribution.About 2.06392.0639 for 95%95\% and df=24\mathrm{df}=24.
df\mathrm{df}Degrees of freedom for a one-sample tt procedure.n−1=24n-1=24.
ME⁡\operatorname{ME}Margin of error: t∗×SE⁡t^* \times \operatorname{SE}.About 3.096 mL3.096 \,\mathrm{mL}.
μd\mu_dPopulation mean of the paired differences, in a stated order.For before−after\text{before}-\text{after} completion times, the mean time reduction in the population.
dˉ\bar d and sds_dMean and sample SD⁡\operatorname{SD} of the observed paired differences.Compute them from the difference list, not from either original list alone.

Why n−1n-1? Once the sample mean is calculated, the nn deviations from that mean must sum to 00. Only n−1n-1 of those deviations can vary freely. That is the degrees of freedom used in the one-sample tt model.

Understand tt and t∗t^*

A tt-distribution is symmetric and bell-shaped around 00. Compared with the standard normal curve, it has heavier tails, especially at small degrees of freedom. It allows for the extra uncertainty caused by estimating σ\sigma with ss.

The tt family contains different curves for different df\mathrm{df} values. As df\mathrm{df} increases, the curves approach the standard normal curve. Do not claim that every tt curve has SD⁡\operatorname{SD} 11. The standardized tt model is not an ordinary normal distribution.

For a two-sided 95%95\% interval, the central area is 0.950.95 and each tail has area 0.0250.025. The positive cutoff t∗t^* therefore has cumulative area 0.9750.975 to its left.

Visual guide 2: connect heavier tails with a confidence cutoff
Student t distributions and a central confidence areaLeft panel compares the standard normal density with t distributions of degrees of freedom 5 and 24 on shared axes. Right panel shows the t distribution with df 24, central area 95 percent between minus 2.0639 and plus 2.0639, with 2.5 percent in each tail. −4 −2 0 2 4 Standardized t or z value 0.0 0.1 0.2 0.3 0.4 Probability density t curves have heavier tails Standard normal t, df = 5 t, df = 24 −4 −2 0 2 4 Standardized t or z
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