Q 8.146. - Elementary Statistics

Question

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

• Lower confidence bound: $\overline{x}-t_{\alpha }-s/\sqrt{\pi }$
• Upper confidence bound: $\hat{\mathrm{x}}+t_{\alpha }·s/\sqrt{n}$

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

Step-by-Step Solution

Answer

$\mu \ge \overline{x}-t_{\alpha }\times \frac{s}{\sqrt{n}}$or $\mu \le \overline{x}+t_{\alpha }\times \frac{s}{\sqrt{n}}$

Part (a) Step 1: Given information

One-Sided One-Mean $t-$Intervals.

Part (a) Step 2: Concept

Formula used:  Lower confidence bound $\overline{x}-t_{\alpha }-s/\sqrt{\pi }$ and upper confidence bound $\hat{\mathrm{x}}+t_{\alpha }·s/\sqrt{n}$

Part (a) Step 3: Calculation

When the population standard deviation, $\sigma$, is unknown, the lower confidence bound $100(1-\alpha )\%$ for the population mean $\mu$ is given by $\overline{x}-t\alpha \frac{s}{\sqrt{n}}$

$\therefore$ We are $100(1-\alpha )\%$confident that the population mean is greater than or equal to the value $\overline{x}-t_{\alpha }\times \frac{s}{\sqrt{n}}$ i.e, $\mu \ge \overline{x}-t_{\alpha }\times \frac{s}{\sqrt{n}}$

When the population standard deviation, $\sigma$, is unknown, the upper confidence bound $100(1-\alpha )\%$ for the population mean $\mu$ is given by $\overline{x}+t_{\alpha }\times \frac{s}{\sqrt{n}}$

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