(a) A 100(1- a)% confidence interval for the mean u of a normal population when the value of a is known is given by Under the same conditions as those leading to the interval, x = µ) < 1.645= 0.95. Use this to derive a one-sided interval for u that has infinite width and provides a lower confidence bound on u. x - 1.645 • (F - 100) What is this interval for the true average porosity of a certain seam of coal samples that is normally distributed with true standard deviation 0.76 if the average porosity for 20 specimens from the seam was 4.35? (If you need to use co or -0o, enter INFINITY or INFINITY, respectively. Round your answers to four decimal places.)

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Please solve both (a) & (c) and show/explain the steps, thanks!!

### Confidence Interval for the Mean

**a) A 100(1 – α)% confidence interval for the mean μ of a normal population when the value of σ is known is given by**

\[
\left( \bar{x} - z_{\alpha/2} \frac{\sigma}{\sqrt{n}}, \bar{x} + z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \right).
\]

**Under the same conditions as those leading to the interval,**

\[
P\left( \frac{\bar{x} - \mu}{\left(\frac{\sigma}{\sqrt{n}}\right)} < 1.645 \right) = 0.95.
\]

**Use this to derive a one-sided interval for μ that has infinite width and provides a lower confidence bound on μ.**

1. \((0, \bar{x} = 1.645\frac{\sigma}{\sqrt{n}})\)
2. \(\left(\bar{x} - 1.645\frac{\sigma}{\sqrt{n}}, \infty\right)\)
3. \(\left(0, \bar{x} + 1.645\frac{\sigma}{\sqrt{n}}\right)\)
4. \(\left(\bar{x} = 1.645\frac{\sigma}{\sqrt{n}}, \infty\right)\)
5. \(\left(\bar{x} + 1.645\frac{\sigma}{\sqrt{n}}, \infty\right)\)

**Correct Answer:**
- \(\left(\bar{x} - 1.645\frac{\sigma}{\sqrt{n}}, \infty\right)\) (Option 2)

### Problem

What is this interval for the true average porosity of a certain seam of coal samples that is normally distributed with true standard deviation 0.76 if the average porosity for 20 specimens from the seam was 4.35? (If you need to use ∞ or −∞, enter INFINITY or −INFINITY, respectively. Round your answers to four decimal places.)

**Answer Boxes:**

\[
( \underline{ \hspace{2em} } ,  \underline{ \hspace{10em} } )
\]
Transcribed Image Text:### Confidence Interval for the Mean **a) A 100(1 – α)% confidence interval for the mean μ of a normal population when the value of σ is known is given by** \[ \left( \bar{x} - z_{\alpha/2} \frac{\sigma}{\sqrt{n}}, \bar{x} + z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \right). \] **Under the same conditions as those leading to the interval,** \[ P\left( \frac{\bar{x} - \mu}{\left(\frac{\sigma}{\sqrt{n}}\right)} < 1.645 \right) = 0.95. \] **Use this to derive a one-sided interval for μ that has infinite width and provides a lower confidence bound on μ.** 1. \((0, \bar{x} = 1.645\frac{\sigma}{\sqrt{n}})\) 2. \(\left(\bar{x} - 1.645\frac{\sigma}{\sqrt{n}}, \infty\right)\) 3. \(\left(0, \bar{x} + 1.645\frac{\sigma}{\sqrt{n}}\right)\) 4. \(\left(\bar{x} = 1.645\frac{\sigma}{\sqrt{n}}, \infty\right)\) 5. \(\left(\bar{x} + 1.645\frac{\sigma}{\sqrt{n}}, \infty\right)\) **Correct Answer:** - \(\left(\bar{x} - 1.645\frac{\sigma}{\sqrt{n}}, \infty\right)\) (Option 2) ### Problem What is this interval for the true average porosity of a certain seam of coal samples that is normally distributed with true standard deviation 0.76 if the average porosity for 20 specimens from the seam was 4.35? (If you need to use ∞ or −∞, enter INFINITY or −INFINITY, respectively. Round your answers to four decimal places.) **Answer Boxes:** \[ ( \underline{ \hspace{2em} } , \underline{ \hspace{10em} } ) \]
### Educational Transcription

#### Problem Statement
(b) Generalize the result of part (a) to obtain a lower bound with confidence level 100(1 − α)%.

Options:
1. \( \left( -\overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}}, \infty \right) \)
2. \( \left( -\infty, \overline{x} - z_\alpha \frac{\sigma}{\sqrt{n}} \right) \)
3. \( \left( -\infty, \overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}} \right) \)
4. \( \left( \overline{x} - z_\alpha \frac{\sigma}{\sqrt{n}}, \infty \right) \) [Correct]
5. \( \left( \overline{x} - z_{\alpha/2}, \overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}} \right) \)

(c) What is an analogous interval to that of part (b) that provides an upper bound on μ?

Options:
1. \( \left( -\overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}}, \infty \right) \)
2. \( \left( -\infty, \overline{x} - z_\alpha \frac{\sigma}{\sqrt{n}} \right) \)
3. \( \left( -\infty, \overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}} \right) \) [Correct]
4. \( \left( \overline{x} - z_\alpha \frac{\sigma}{\sqrt{n}}, \infty \right) \)
5. \( \left( \overline{x} - z_{\alpha/2}, \overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}} \right) \)

#### Analysis
Compute this 99% interval for the true average stray-load loss \( \mu \) (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1,500 rpm, when \( n = 25 \) and \( \overline{x} = 58.3 \).
Transcribed Image Text:### Educational Transcription #### Problem Statement (b) Generalize the result of part (a) to obtain a lower bound with confidence level 100(1 − α)%. Options: 1. \( \left( -\overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}}, \infty \right) \) 2. \( \left( -\infty, \overline{x} - z_\alpha \frac{\sigma}{\sqrt{n}} \right) \) 3. \( \left( -\infty, \overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}} \right) \) 4. \( \left( \overline{x} - z_\alpha \frac{\sigma}{\sqrt{n}}, \infty \right) \) [Correct] 5. \( \left( \overline{x} - z_{\alpha/2}, \overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}} \right) \) (c) What is an analogous interval to that of part (b) that provides an upper bound on μ? Options: 1. \( \left( -\overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}}, \infty \right) \) 2. \( \left( -\infty, \overline{x} - z_\alpha \frac{\sigma}{\sqrt{n}} \right) \) 3. \( \left( -\infty, \overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}} \right) \) [Correct] 4. \( \left( \overline{x} - z_\alpha \frac{\sigma}{\sqrt{n}}, \infty \right) \) 5. \( \left( \overline{x} - z_{\alpha/2}, \overline{x} + z_\alpha \frac{\sigma}{\sqrt{n}} \right) \) #### Analysis Compute this 99% interval for the true average stray-load loss \( \mu \) (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1,500 rpm, when \( n = 25 \) and \( \overline{x} = 58.3 \).
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