Final answer:
To construct a 99% confidence interval for the population mean with n=14, a sample mean of 11.9, and a sample standard deviation of 2.1, we use the t-distribution and find that the population mean μ lies between 10.206 and 13.594.
Step-by-step explanation:
To construct a 99% confidence interval for the population mean μ with a sample of size n=14, a sample mean × of 11.9, and a sample standard deviation s of 2.1, we first need to recognize that the sample size is small (n<30) and the population standard deviation is unknown, so we use the t-distribution.
We will use the formula for the confidence interval:
Confidence Interval = × ± (t* × (s/√n))
To find the critical value t*, we need to look up the t-value corresponding to a 99% confidence level and degrees of freedom df=n-1, which in this case is 14-1=13.
Using a t-distribution table, we find that t* is approximately 3.012 for a two-tailed test at a 99% confidence level with 13 degrees of freedom. The margin of error (ME) will be:
ME = t* × (s/√n) = 3.012 × (2.1/√14) ≈ 1.694
Thus, the 99% confidence interval is:
(× - ME, × + ME) = (11.9 - 1.694, 11.9 + 1.694) = (10.206, 13.594)
So, with 99% confidence, the population mean μ lies between 10.206 and 13.594.