Step 1 :Given that the level of confidence $c=0.90$, the standard deviation $s=3.4$, and the sample size $n=8$, we need to find the margin of error.
Step 2 :First, we need to find the degrees of freedom, which is calculated as $n-1$. So, the degrees of freedom is $8-1=7$.
Step 3 :Next, we look up the t-score in the t-distribution table for a level of confidence of $0.90$ and degrees of freedom of $7$. The t-score is $1.895$.
Step 4 :We then substitute these values into the formula for the margin of error: $E = t \cdot \frac{s}{\sqrt{n}}$
Step 5 :Substituting the given values, we get $E = 1.895 \cdot \frac{3.4}{\sqrt{8}}$
Step 6 :Calculating the above expression, we find that the margin of error $E$ is approximately $2.278$.
Step 7 :Final Answer: The margin of error for the given values of $c, s$, and $n$ is \(\boxed{2.278}\).