Topic 7.2 Constructing a Confidence Interval for a Population Mean: check the conditions and construct a one-sample t-interval for a population mean, using the t critical value, the standard error, and the correct degrees of freedom.

What this topic is asking

The College Board (Topic 7.2) wants you to construct and interpret a one-sample t-interval for a population mean $\mu$: check the conditions, compute $\bar{x}$, find the $t$ critical value with $n - 1$ degrees of freedom, the standard error $s/\sqrt{n}$, and the margin of error, then state and interpret the interval.

The interval and why it uses $t$

We use the $t$-distribution, not the normal $z$, because the population standard deviation $\sigma$ is unknown and replaced by the sample standard deviation $s$. Estimating $s$ from the data adds extra uncertainty, so $t$ is wider (heavier-tailed) than the normal, more so for small samples. As $n$ grows, $s$ estimates $\sigma$ better and $t$ approaches $z$. The degrees of freedom $df = n - 1$ index which $t$-distribution to use.

Checking the conditions

The shape condition has the most nuance. For $n \ge 30$, the central limit theorem covers you. For a small sample, you must inspect a graph of the data: a roughly symmetric shape with no outliers supports the procedure, while strong skew or an outlier undermines it. On the exam, state which justification applies (large $n$, or a described plot).

Reading the result and the parts

Interpret the interval as: "We are $C\%$ confident the true mean of [context] is between [low] and [high]," and, if asked, interpret the confidence level separately ("in repeated sampling, about $C\%$ of intervals built this way would capture the true mean"). The margin of error $t^{*} s/\sqrt{n}$ shows the same three levers as proportions: higher confidence raises $t^{*}$ and widens the interval; larger $n$ shrinks $s/\sqrt{n}$ and narrows it (precision improving with $\sqrt{n}$); more variable data (larger $s$) widens it. Sample-size questions ("how large for a margin of error of at most $m$?") are solved by rearranging $t^{*} s/\sqrt{n} \le m$, often using $z^{*}$ as an approximation when $df$ is unknown in advance.

Try this

Q1. A sample of $n = 10$ gives $s = 5$. Find the standard error and the degrees of freedom. [2 points]

• Cue. $SE = \dfrac{5}{\sqrt{10}} \approx 1.58$; $df = n - 1 = 9$.

Q2. Why does a mean interval use $t$ instead of $z$? [1 point]

• Cue. Because $\sigma$ is unknown and estimated by $s$; that extra uncertainty makes the $t$-distribution (wider than normal) the correct model.