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Q 8.105.

Expert-verifiedFound in: Page 335

Book edition
9th

Author(s)
Weiss, Neil

Pages
590 pages

ISBN
9780321989505

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: $\stackrel{~}{x}-{z}_{\sigma}\xb7\sigma /\sqrt{n}$

- Upper confidence bound: $\overline{x}+{z}_{\alpha}\xb7\sigma /\sqrt{n}$

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

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. $\mu \ge \overline{x}+{z}_{\alpha}\times \frac{\sigma}{\sqrt{n}}$

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

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. $\mu \ge \overline{x}+{z}_{\alpha}\times \frac{\sigma}{\sqrt{n}}$

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