Estimating sigma is what t charges for

For a simple random sample of size $n$ from a population with mean $\mu$ and standard deviation $\sigma$,

$$\mu_{\bar{x}} = \mu, \qquad \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}.$$

The first says $\bar{x}$ is unbiased. The second is the one that gets rebuilt wrong, in three ways:

1. Dividing by $n$ instead of $\sqrt{n}$. With $\sigma = 9.2$ and $n = 25$, the spread is $\frac{9.2}{5} = 1.84$, not $\frac{9.2}{25} = 0.368$.
2. Reporting the data's spread as the mean's spread. The individual measurements still vary by about 9.2; it is the average of 25 of them that varies by 1.84.
3. Expecting the sample's own $s$ to shrink. It does not. $s$ estimates $\sigma$, so a larger sample pins it down more precisely around 9.2 rather than driving it toward zero.

The square root also sets the price of precision: quadrupling $n$ halves $\sigma_{\bar{x}}$, and cutting it to a third takes nine times the data. The 10% condition applies as always when sampling without replacement.

Two routes give the sampling distribution of $\bar{x}$ an approximately normal shape:

1. The population is normal. Then $\bar{x}$ is exactly normal for every $n$, including $n = 2$.
2. The sample is large enough. The Central Limit Theorem makes $\bar{x}$ approximately normal whatever the population's shape, with $n \ge 30$ the usual working guideline and more needed when the population is strongly skewed.

Neither route makes the population normal or the data normal: a large sample from a skewed population produces a visibly skewed histogram, and only the distribution of the statistic straightens out. That was Topic 2.12's point, and it is the assumption every procedure in this unit rests on.

In a real study $\sigma$ is not available, so the sample's own standard deviation $s$ replaces it. The resulting quantity

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

is no longer a z-score. Its denominator is itself an estimate that varies from sample to sample, so the whole ratio has more variability than a z-score does, and it follows a t distribution rather than a normal one.

The quantity $\frac{s}{\sqrt{n}}$ is called the standard error of $\bar{x}$: the estimated standard deviation of the sampling distribution. For 25 batteries with $s = 9.2$ hours, $SE = \frac{9.2}{\sqrt{25}} = 1.84$ hours.

The rule for the rest of the unit is short: if $\sigma$ is known, use z; if only $s$ is available, use t. In practice that means t essentially always, and a z procedure for a mean built from the sample's $s$ is the standard error of this unit.

Each t distribution is identified by its degrees of freedom, $df = n - 1$ for a one-sample procedure. Four properties matter:

1. Symmetric and centered at 0, like the standard normal.
2. Heavier tails, so more area sits far from the center and critical values are larger than the corresponding z values.
3. Shape depends on df. Small df means very heavy tails; $t^{*}$ at 95% confidence is 4.303 for $df = 2$, 2.571 for $df = 5$, 2.064 for $df = 24$, and 2.042 for $df = 30$.
4. It approaches the standard normal as df grows, but never quite arrives: at $df = 100$, $t^{*} = 1.984$ against $z^{*} = 1.960$.

Three consequences. Reading a critical value from a z table when df is small understates the interval's width badly. Using $n$ instead of $n - 1$ for df picks the wrong member of the family. And "t and z are the same at large n" is a reasonable approximation at $n = 200$ and simply false at $n = 12$, where $t^{*} = 2.201$ is 12% larger than 1.960.