Question

What is the confidence interval for the population mean assuming the population has a normal distribution and n=12, x=29.5, s=4.1, and 99% confidence

Answer 1

Conditions

What is the confidence interval for the population mean assuming the population has a normal distribution and $n=12$, $x=29.5$, $s=4.1$, and 99% confidence

Solution

$$\mathbb{P} \left( \bar{X} - z_{\frac{1 - \alpha}{2}} \frac{\sigma}{\sqrt{n}} \leq \mu \leq \bar{X} + z_{\frac{1 - \alpha}{2}} \frac{\sigma}{\sqrt{n}} \right) = \alpha$$ $$z_{0.99} = 2.326$$ $$29.5 - 2.326 \cdot \frac{4.1}{\sqrt{12}} \leq \mu \leq 29.5 + 2.326 \cdot \frac{4.1}{\sqrt{12}}$$ $$24.747 \leq \mu \leq 32.253$$

Answer: $24.747 \leq \mu \leq 32.253$