To find the margin of error (ME) for a confidence interval when the sample standard deviation ($s$) is used, we apply the formula:

$ME = t^* \cdot \frac{s}{\sqrt{n}}$

Where:

• $t^*$ is the critical $t$-value corresponding to the given confidence level ($c$) and degrees of freedom ($df = n-1$).
• $s$ is the sample standard deviation.
• $n$ is the sample size.

Step-by-Step Solution:

1. Determine Degrees of Freedom ($df$):
   $df = n - 1 = 15 - 1 = 14$

2. Find the Critical $t^*$-Value:
   For a confidence level of $c = 0.90$, the area in each tail is $(1 - c)/2 = (1 - 0.90)/2 = 0.05$. Using a $t$-distribution table or calculator, the critical $t^*$ value for $df = 14$ is approximately: $t^* = 1.761$

3. Calculate Standard Error ($\text{SE}$):
   $\text{SE} = \frac{s}{\sqrt{n}} = \frac{5.4}{\sqrt{15}} = \frac{5.4}{3.873} \approx 1.394$

4. Compute Margin of Error ($ME$):
   $ME = t^* \cdot \text{SE} = 1.761 \cdot 1.394 \approx 2.452$

Final Answer:

The margin of error is approximately 2.45.

Would you like a detailed explanation on how to find $t^*$ or work through additional examples?

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Related Questions:

1. How would the margin of error change if $c = 0.95$?
2. How is the formula adjusted if the population standard deviation ($\sigma$) is known?
3. What are the implications of increasing the sample size ($n$) on the margin of error?
4. Can you explain why the $t$-distribution is used instead of the normal distribution for small sample sizes?
5. What is the effect of a larger standard deviation ($s$) on the margin of error?

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Tip: Always use the $t$-distribution for margin of error calculations when the sample size is small ($n < 30$) and the population standard deviation is unknown.