To determine the appropriate confidence interval, we can use the t-distribution since the sample size is small (n=6).
Formula for the confidence interval of the true mean:
CI = x̄ ± (t * (s/√n))
Given:
Sample size (n) = 6
Sample mean (x̄) = 12,200 pounds
Sample standard deviation (s) = 200 pounds
To find the t-value for a 95% confidence level with 5 degrees of freedom (df = n-1 = 6-1 = 5), we can consult a t-table or use statistical software:
t = 2.571
Plugging the values into the formula:
CI = 12,200 ± (2.571 * (200/√6))
Calculating (√6):
√6 ≈ 2.449
CI = 12,200 ± (2.571 * (200/2.449))
CI = 12,200 ± 332.819
The 95% confidence interval of the true mean is (11,867.18, 12,532.82)
Which type of confidence interval should you use in this situation? A sample of 6 adult elephants had an average weight of 12,200 pounds, with a sample standard deviation of 200 pounds. Find the 95% confidence interval of the true mean.
1 answer