numpy as np from scipy.stats import t

Given data

data = [49.1, 83.5, 56.6, 55.5, 51.8, 67.5, 29.4, 17.9, 61.8, 60.7, 38.8, 75.3]

Sample mean and standard deviation

mean = np.mean(data) std_dev = np.std(data, ddof=1) # Sample standard deviation

Sample size

n = len(data)

Degrees of freedom

df = n - 1

T critical value for 99% confidence interval

confidence_level = 0.99 t_critical = t.ppf((1 + confidence_level) / 2, df)

Margin of error

margin_of_error = t_critical * (std_dev / np.sqrt(n))

Confidence interval

lower_bound = mean - margin_of_error upper_bound = mean + margin_of_error

(lower_bound, upper_bound)The 99% confidence interval for the population mean is approximately (37.35, 70.63).

Would you like further details on how this calculation was done? Here are five related questions for further exploration:

1. How is the t-distribution critical value calculated for different confidence levels?
2. What happens to the confidence interval when the sample size increases?
3. How does the standard deviation affect the width of the confidence interval?
4. Why is the sample mean used instead of the population mean in this calculation?
5. What is the difference between a 95% and a 99% confidence interval in terms of reliability?

Tip: Always check if your data meets the normality assumption when using t-distributions for confidence intervals.