To construct a 99% confidence interval for the mean lifetime of the batteries, we use the formula for the confidence interval when the population standard deviation ($\sigma$) is known:

$\text{Confidence Interval} = \bar{x} \pm z \frac{\sigma}{\sqrt{n}}$

Given:

• Sample mean ($\bar{x}$) = 120.1 hours
• Population standard deviation ($\sigma$) = 20 hours
• Sample size ($n$) = 50
• Confidence level = 99%

Step 1: Determine the $z$-value for a 99% confidence level

For a 99% confidence level, the critical value ($z$) is approximately 2.576 (from the standard normal distribution table).

Step 2: Calculate the standard error (SE)

The standard error of the mean is given by: $SE = \frac{\sigma}{\sqrt{n}}$ $SE = \frac{20}{\sqrt{50}} \approx \frac{20}{7.071} \approx 2.828$

Step 3: Compute the margin of error (ME)

The margin of error is calculated as: $ME = z \cdot SE$ $ME = 2.576 \cdot 2.828 \approx 7.284$

Step 4: Determine the confidence interval

The confidence interval is: $\text{Confidence Interval} = \bar{x} \pm ME$ $\text{Confidence Interval} = 120.1 \pm 7.284$ $\text{Confidence Interval} = [120.1 - 7.284, 120.1 + 7.284]$ $\text{Confidence Interval} = [112.82, 127.38]$

Final Answer:

The 99% confidence interval for the mean lifetime of the batteries is: $[112.82 \, \text{hours}, \, 127.38 \, \text{hours}]$

Let me know if you’d like more details or further clarification.

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Related Questions:

1. How do you interpret a confidence interval in the context of this problem?
2. How would the interval change if the sample size were larger?
3. What happens to the margin of error if the confidence level is reduced to 95%?
4. How can we compute the required sample size for a specific margin of error?
5. Why is it appropriate to use the $z$-distribution in this case instead of the $t$-distribution?

Tip:

Always confirm the assumptions of normality and known standard deviation before applying the $z$-interval formula.