Find the margin of error for the given values of $c, s$, and $n$.
\[
c=0.95, s=5, n=14
\]
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The margin of error is (Round to three decimal places as needed.)

Step 1 ：We are given that the confidence level $c=0.95$, the standard deviation $s=5$, and the sample size $n=14$.

Step 2 ：We first need to find the t-score for a 95% confidence level and 13 degrees of freedom. Using a t-distribution table, we find that the t-score is approximately 2.160.

Step 3 ：We can now calculate the margin of error $E$ using the formula $E = t \cdot \frac{s}{\sqrt{n}}$.

Step 4 ：Substituting the given values into the formula, we get $E = 2.160 \cdot \frac{5}{\sqrt{14}}$.

Step 5 ：Solving this expression, we find that the margin of error $E$ is approximately 2.887.

Step 6 ：Final Answer: The margin of error for the given values of $c, s$, and $n$ is \(\boxed{2.887}\).