Construct the indicated confidence interval for the population mean $\mu$ using the $t$-distribution. Assume the population is normally distributed.
\[
c=0.99, \bar{x}=14.6, s=4.0, n=6
\]
(Round to one decimal place as needed.)

Step 1 ：The problem is asking for a confidence interval for the population mean using the t-distribution. The confidence level is 0.99, the sample mean is 14.6, the sample standard deviation is 4.0, and the sample size is 6.

Step 2 ：The formula for a confidence interval using the t-distribution is: \(\bar{x} \pm t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(t_{\alpha/2, n-1}\) is the t-score for the desired level of confidence with \(n-1\) degrees of freedom, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：First, we need to find the t-score for a 0.99 confidence level with 5 degrees of freedom. The t-score is 4.032142983557535.

Step 4 ：Then we can substitute the given values into the formula to find the confidence interval. The margin of error is 6.584461919772896.

Step 5 ：The 99% confidence interval for the population mean \(\mu\) using the t-distribution is \(\boxed{(8.0, 21.2)}\).