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Q 8.105.

Expert-verified
Found in: Page 335

### Elementary Statistics

Book edition 9th
Author(s) Weiss, Neil
Pages 590 pages
ISBN 9780321989505

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# One-Sided One-Mean z-Intervals. Presuming that the assumptions for a one-mean z-interval are satisfied, we have the following formulas for $\left(1-\alpha \right)$-level confidence bounds for a population mean $\mu$ :- Lower confidence bound: $\stackrel{~}{x}-{z}_{\sigma }·\sigma /\sqrt{n}$- Upper confidence bound: $\overline{x}+{z}_{\alpha }·\sigma /\sqrt{n}$Interpret the preceding formulas for lower and upper confidence bounds in words.

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

See the step by step solution

## Step 1: Concept

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

## Step 2: Calculation

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

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

$100\left(1-\alpha \right)%$ level upper confidence bound for population means $\mu$ is $\overline{x}+{z}_{\alpha }×\frac{\sigma }{\sqrt{n}}$

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

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