To construct a 91% confidence interval for the mean amount of household garbage discarded in one week, we follow these steps:

1. Given Data:

• Sample size ($n$): 58 households
• Sample mean ($\bar{x}$): 25.863 lbs
• Population standard deviation ($\sigma$): 10.531 lbs
• Confidence level: 91%

2. Determine the Z-score for 91% confidence level:

For a 91% confidence level, the critical value corresponds to the middle 91% of the normal distribution, leaving 4.5% in each tail. Using a Z-table, the Z-score for 91% confidence level is approximately 1.695.

3. Calculate the Margin of Error (E):

The margin of error is given by the formula:

$E = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$

Substituting the known values:

$E = 1.695 \times \frac{10.531}{\sqrt{58}}$ $E \approx 1.695 \times 1.381 = 2.341 \text{ lbs}$

4. Construct the Confidence Interval:

The confidence interval is given by:

$\text{CI} = \left( \bar{x} - E, \bar{x} + E \right)$

Substituting the values:

$\text{CI} = \left( 25.863 - 2.341, 25.863 + 2.341 \right)$ $\text{CI} = (23.522, 28.204) \text{ lbs}$

Final Answer:

The 91% confidence interval for the mean amount of household garbage discarded in one week is approximately (23.52 lbs, 28.20 lbs).

Would you like more details on any step or a clarification on confidence intervals?

Here are 5 related questions to deepen your understanding:

1. What would happen to the confidence interval if the sample size increased?
2. How does the confidence level affect the width of the confidence interval?
3. What is the importance of using the Z-score in constructing confidence intervals?
4. How would the confidence interval change if the standard deviation was unknown?
5. What would the confidence interval look like with a 95% confidence level instead?

Tip: Increasing the sample size decreases the margin of error, making your confidence interval narrower and more precise.