Q 8.105. - Elementary Statistics

Question

One-Sided One-Mean z-Intervals. Presuming that the assumptions for a one-mean z-interval are satisfied, we have the following formulas for $(1-\alpha )$-level confidence bounds for a population mean \(\mu\) :

- Lower confidence bound: $\tilde{x}-z_{\sigma }·\sigma /\sqrt{n}$

- Upper confidence bound: $\overline{x}+z_{\alpha }·\sigma /\sqrt{n}$

Interpret the preceding formulas for lower and upper confidence bounds in words.

Step-by-Step Solution

Answer

We are $100(1-\alpha )\%$ certain that the population mean is less than or equal to $\overline{x}+z_{\alpha }\times \frac{\sigma }{\sqrt{n}}$. i.e. $\mu \ge \overline{x}+z_{\alpha }\times \frac{\sigma }{\sqrt{n}}$

Step 1: Concept

The formula used: One mean z- interval procedure $\overline{x}-z_{\alpha }\times \frac{\sigma }{\sqrt{n}}$

Step 2: Calculation

The lower confidence bound for the population mean $\mu$ is $\overline{x}-z_{\alpha }\times \frac{\sigma }{\sqrt{n}}$, which is $100(1-\alpha )\%$

As a result, we are 100 percent certain that the population mean is larger than or equal to $\overline{x}-z_a\times \frac{\sigma }{\sqrt{n}}$i.e. $\mu \ge \overline{x}-z_{\alpha }\times \frac{\sigma }{\sqrt{n}}$

$100(1-\alpha )\%$ level upper confidence bound for population means $\mu$ is $\overline{x}+z_{\alpha }\times \frac{\sigma }{\sqrt{n}}$

As a result, we are $100(1-\alpha )\%$ certain that the population mean is less than or equal to $\overline{x}+z_{\alpha }\times \frac{\sigma }{\sqrt{n}}$. i.e. $\mu \ge \overline{x}+z_{\alpha }\times \frac{\sigma }{\sqrt{n}}$