5.2.1 Single Sample: Estimating 𝜇

What do we know

Case 1: Confidence Interval on $\mu$ with known $\sigma^2$

Note: One-tailed and Two-tailed

Given $\sigma^2$, and $n$ of a random sample with mean $\bar x$, (the population is not required to be normal since CLT can be applied) a $100(1-\alpha)\%$ confidence interval for $\mu$ is given by

$$ \bar X-z_{\alpha / 2}\frac{\sigma}{\sqrt n } \lt \mu \lt\bar X+z_{\alpha / 2}\frac{\sigma}{\sqrt n } $$

Note: Derivation

Case 2: One-sided confidence bounds on $\mu$ with known $\sigma^2$

Given $\sigma^2$, and $n$ of a random sample with mean $\bar x$, (the population is not required to be normal since CLT can be applied) a one-sided $100(1-\alpha)\%$ confidence bound for $\mu$ is given by

$$ \begin{align*} \mu \lt \bar X + z_\alpha\frac{\sigma}{\sqrt n},\ &upper \\ \mu \gt \bar X - z_\alpha\frac{\sigma}{\sqrt n}, \ &lower \end{align*} $$

Note: the “+” and “-” are different.

“upper”: greater than $\mu$, so $\mu$ is at the left of upper bound
Note: Derivation

Case 3: Confidence Interval on $\mu$, with unknown $\sigma^2$

If we don’t have $\sigma$, we can’t use the $Z = \frac{\bar X - \mu}{\sigma / \sqrt n}$ (standard normal distribution) stuff. However, if we get $S$ instead, we cant use the $T=\frac{\bar X - \mu}{S / \sqrt n}$ stuff. However, it requires a normal population.

Given $n$ and $s$ of a random sample from a normal population with mean $\bar x$, where $n < 30$, a one-sided $100(1-\alpha)\%$ confidence bound for $\mu$ is given by

$$ \bar X -t_{\alpha / 2}\frac{S}{\sqrt n}\lt \mu \lt \bar X +t_{\alpha / 2}\frac{S}{\sqrt n} $$