The image you uploaded shows a question about constructing a confidence interval for the mean noise level at six airports. The data given are:

137, 132, 145, 152, 209, and 169 decibels, and the task is to construct an 80% confidence interval for the mean noise level, assuming the population is approximately normal.

Steps to Solve:

1. Find the sample mean (x̄):

   • Add all the values together: $\text{Sum} = 137 + 132 + 145 + 152 + 209 + 169 = 944$
   • Divide by the number of data points (6): $\text{Sample mean} = \frac{944}{6} = 157.33$

2. Find the sample standard deviation (s):

   • Calculate each deviation from the mean: $(137 - 157.33)^2 = 406.99, \quad (132 - 157.33)^2 = 636.49, \quad (145 - 157.33)^2 = 153.39$ $(152 - 157.33)^2 = 28.49, \quad (209 - 157.33)^2 = 2676.39, \quad (169 - 157.33)^2 = 135.47$
   • Sum the squared deviations: $406.99 + 636.49 + 153.39 + 28.49 + 2676.39 + 135.47 = 4037.22$
   • Divide by the sample size minus 1 (n-1 = 5): $\text{Variance} = \frac{4037.22}{5} = 807.44$
   • Take the square root to find the standard deviation: $\text{Standard deviation} (s) = \sqrt{807.44} = 28.43$

3. Find the t-critical value for a 80% confidence interval:

   • For a confidence level of 80% and a sample size of 6, use a t-distribution with degrees of freedom (df) = 5. From a t-table or calculator, the critical value (t*) for 80% confidence and df = 5 is approximately 1.476.

4. Calculate the margin of error (ME):

   • Use the formula: $\text{ME} = t^* \times \frac{s}{\sqrt{n}} = 1.476 \times \frac{28.43}{\sqrt{6}} \approx 1.476 \times 11.61 = 17.13$

5. Construct the confidence interval:

   • The lower and upper bounds of the confidence interval are: $\text{Lower endpoint} = x̄ - \text{ME} = 157.33 - 17.13 = 140.20$ $\text{Upper endpoint} = x̄ + \text{ME} = 157.33 + 17.13 = 174.46$

Final Answer:

• Lower endpoint: 140.2
• Upper endpoint: 174.5

Would you like further clarification on any step, or have additional questions related to this?

Further questions to explore:

1. What happens if we use a different confidence level, such as 90% or 95%?
2. How does increasing the sample size affect the confidence interval?
3. How would you calculate a confidence interval for a population variance instead of the mean?
4. What is the difference between a t-distribution and a normal distribution?
5. How do you interpret a confidence interval in a real-world context?

Tip: When dealing with small sample sizes, always use the t-distribution instead of the normal distribution for better accuracy in confidence interval calculations.