To calculate the margin of error $E$, you can use the formula:

$E = z_c \times \frac{\sigma}{\sqrt{n}}$

where:

• $z_c$ is the critical value associated with the confidence level (in this case, $z_c = 1.645$ for 90% confidence),
• $\sigma = 2.6$ is the population standard deviation,
• $n = 100$ is the sample size.

Step-by-step calculation:

1. Calculate the square root of the sample size:

$\sqrt{n} = \sqrt{100} = 10$

2. Divide the standard deviation by the square root of the sample size:

$\frac{\sigma}{\sqrt{n}} = \frac{2.6}{10} = 0.26$

3. Multiply by the critical value $z_c$:

$E = 1.645 \times 0.26 = 0.4277$

Final Answer:

$E \approx 0.428$

So, the margin of error is $\boxed{0.428}$ (rounded to three decimal places).

Would you like more details on the process or have any questions?

Here are five related questions:

1. How is the critical value $z_c$ determined for other confidence levels?
2. How does increasing the sample size affect the margin of error?
3. What happens to the margin of error if the population standard deviation increases?
4. Can you use this formula for non-normal distributions? If not, what adjustments are needed?
5. What is the relationship between confidence level and margin of error?

Tip: Increasing the sample size reduces the margin of error, providing a more precise estimate.