If n=30, 𝑥¯(x-bar)=35, and s=7, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population.Give your answers to one decimal place. < 𝜇 <

To construct a confidence interval for the population mean (μ) at a 99% confidence level, given that the sample size (n) is 30, the sample mean (x̄) is 35, and the sample standard deviation (s) is 7, follow these steps:

1. Identify the critical value (z) for a 99% confidence level:* Since the sample size is less than 30, we use the t-distribution. For a 99% confidence level and degrees of freedom (df) = n - 1 = 30 - 1 = 29, we look up the t-value in the t-distribution table. The critical value (t*) for 29 degrees of freedom at a 99% confidence level is approximately 2.756.

2. Calculate the standard error (SE) of the mean: $SE = \frac{s}{\sqrt{n}} = \frac{7}{\sqrt{30}} \approx 1.278$

3. Calculate the margin of error (ME): $ME = t^* \times SE = 2.756 \times 1.278 \approx 3.5$

4. Construct the confidence interval: $\text{Lower limit} = x̄ - ME = 35 - 3.5 = 31.5$ $\text{Upper limit} = x̄ + ME = 35 + 3.5 = 38.5$

Therefore, the 99% confidence interval for the population mean (μ) is: $31.5 < \mu < 38.5$