To construct the 90% and 99% confidence intervals for the population mean when the population standard deviation ($\sigma$) is known, we'll use the Z-distribution (standard normal distribution). Here’s the step-by-step process:

1. Calculate the sample mean ($\bar{x}$):

First, we add all the sample values and divide by the number of values ($n$).

$\bar{x} = \frac{\sum x_i}{n}$

Given dataTo construct the 90% and 99% confidence intervals for the population mean when the population standard deviation ($\sigma$) is known, we'll use the Z-distribution (standard normal distribution). Here’s the step-by-step process:

1. Calculate the sample mean ($\bar{x}$):

First, we add all the sample values and divide by the number of values ($n$).

$\bar{x} = \frac{\sum x_i}{n}$

Given 17.97, 17.42, 17.84, 21.62, 15.35, 22.46, 18.68, 17.34, 20.36, 16.77, 18.17, 17.84, 22.62, 22.94, 20.42, 19.54

The sample size ($n$) is 16.

$\bar{x} = \frac{17.97 + 17.42 + 17.84 + 21.62 + 15.35 + 22.46 + 18.68 + 17.34 + 20.36 + 16.77 + 18.17 + 17.84 + 22.62 + 22.94 + 20.42 + 19.54}{16} = 19.20$

2. Confidence Interval Formula:

When the population standard deviation ($\sigma$) is known, the confidence interval (CI) for the population mean ($\mu$) is given by:

$CI = \bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$

Where:

• $\bar{x}$ = sample mean
• $Z_{\alpha/2}$ = Z-score corresponding to the confidence level
• $\sigma$ = population standard deviation = 2.38
• $n$ = sample size = 16

3. 90% Confidence Interval:

For a 90% confidence level, $Z_{\alpha/2} = 1.645$.

$\text{Margin of Error} = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} = 1.645 \times \frac{2.38}{\sqrt{16}} = 1.645 \times 0.595 = 0.979$

The 90% confidence interval is:

$CI_{90\%} = 19.20 \pm 0.979$ $CI_{90\%} = (18.221, 20.179)$

4. 99% Confidence Interval:

For a 99% confidence level, $Z_{\alpha/2} = 2.576$.

$\text{Margin of Error} = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} = 2.576 \times \frac{2.38}{\sqrt{16}} = 2.576 \times 0.595 = 1.533$

The 99% confidence interval is:

$CI_{99\%} = 19.20 \pm 1.533$ $CI_{99\%} = (17.667, 20.733)$

5. Interpretation and Comparison:

• 90% Confidence Interval: We are 90% confident that the true population mean lies between 18.221 and 20.179.
• 99% Confidence Interval: We are 99% confident that the true population mean lies between 17.667 and 20.733.

Comparison: The 99% confidence interval is wider than the 90% confidence interval because increasing the confidence level increases the range of values within which the true population mean is expected to fall. This is a trade-off; a higher confidence level means more certainty but also less precision (wider interval).

Would you like more details on any step or have any further questions?

Follow-up Questions:

1. How does changing the sample size affect the width of the confidence interval?
2. Why do we use the Z-distribution instead of the t-distribution here?
3. What would happen to the confidence intervals if the population standard deviation was not known?
4. How is the confidence interval interpreted in the context of this company's stock prices?
5. What are the implications if the confidence intervals are too wide for decision-making purposes?

Tip: Always ensure the sample size is adequate when estimating population parameters to increase the precision of confidence intervals.