To construct a confidence interval for the population mean $\mu$ at an 80% confidence level with the given data, we use the formula:

$\text{Confidence Interval} = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right)$

Given values:

• Sample mean, $\bar{x} = 42$
• Sample standard deviation, $s = 6$
• Sample size, $n = 12$
• Confidence level = 80%

Since the population is assumed to be normally distributed, we can use the standard normal distribution for the $z$-score.

Step-by-Step Calculation:

1. Determine the $z$-score for an 80% confidence level:

   • For an 80% confidence level, there is 10% in each tail of the distribution.
   • The corresponding $z$-score for 80% confidence is approximately $z = 1.28$.

2. Calculate the Standard Error (SE): $\text{SE} = \frac{s}{\sqrt{n}} = \frac{6}{\sqrt{12}} \approx 1.73$

3. Construct the Confidence Interval: $\bar{x} \pm z \cdot \text{SE} = 42 \pm 1.28 \cdot 1.73$ $= 42 \pm 2.2$

4. Resulting Interval: $39.8 < \mu < 44.2$

Final Answer:

$39.8 < \mu < 44.2$

Would you like further details or have any questions?

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1. How do confidence levels affect interval width?
2. Why is the z-score for 80% confidence level approximately 1.28?
3. What changes if we increase the confidence level to 90%?
4. How is standard error calculated for different sample sizes?
5. Why do we assume normal distribution for confidence intervals?

Tip: Higher confidence levels result in wider intervals, capturing the true mean with greater certainty.