Assuming the population has an approximate normal distribution, if a sample size n=26  has a sample mean x=46
 
 with a sample standard deviation s=3 , find the margin of error at a 90% confidence level. Round the answer to two decimal places.

To calculate the margin of error (ME) for a sample, we use the following formula for a confidence interval with a t-distribution (since the sample size is small, \(n = 26\)):

\[
ME = t_{\alpha/2} \times \frac{s}{\sqrt{n}}
\]

Where:
- \(t_{\alpha/2}\) is the critical value from the t-distribution for a 90% confidence level and \(n-1 = 25\) degrees of freedom.
- \(s = 3\) is the sample standard deviation.
- \(n = 26\) is the sample size.

### Steps:
1. **Find the t-critical value** for a 90% confidence level with 25 degrees of freedom.
   Using t-tables or a calculator, the t-critical value for \(25\) degrees of freedom at 90% confidence is approximately **1.708**.

2. **Calculate the standard error (SE)**:
\[
SE = \frac{s}{\sqrt{n}} = \frac{3}{\sqrt{26}} \approx 0.5883
\]

3. **Calculate the margin of error (ME)**:
\[
ME = 1.708 \times 0.5883 \approx 1.00
\]

### Final answer:
The margin of error at a 90% confidence level is approximately **1.00**.

Would you like more details or have any questions? Here are five related questions to explore this topic further:

1. How does the confidence level affect the margin of error?
2. What would the margin of error be for a 95% confidence level?
3. What happens to the margin of error if the sample size increases?
4. How do you interpret the margin of error in the context of this problem?
5. What is the role of the t-distribution in calculating confidence intervals for small samples?

**Tip:** The t-distribution is used instead of the z-distribution for smaller samples (typically \(n < 30\)) because it accounts for the increased variability in estimates.

Assuming the population has an approximate normal distribution, if a sample size n=26 has a sample mean x=46 with a sample standard deviation s=3, find the margin of error at a 90% confidence level. Round the answer to two decimal places.

Learn how to calculate the margin of error for a sample with n=26, sample mean x=46, and sample standard deviation s=3, using a 90% confidence level. This problem demonstrates the use of the t-distribution in small-sample confidence intervals, with a step-by-step explanation of key statistical concepts.

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