To find the standard error (Part 3 of 6) in this scenario, we use the formula:

$\text{Standard Error} = \frac{\sigma}{\sqrt{n}}$

where:

• $\sigma = 10$ (the population standard deviation),
• $n = 76$ (the sample size).

Calculation:

$\text{Standard Error} = \frac{10}{\sqrt{76}} \approx \frac{10}{8.7178} \approx 1.1456$

So, the standard error, rounded to four decimal places, is 1.1456.

Would you like further details or have any questions?

Related Questions:

1. What is the formula for constructing a confidence interval for the mean?
2. Why is a critical value of 1.645 used for a 90% confidence level?
3. How would the standard error change with a larger sample size?
4. How does knowing the population standard deviation (as opposed to the sample) impact the calculation?
5. What would be the steps if you were calculating a 95% confidence interval instead?

Tip:

Remember, the standard error decreases as the sample size increases, improving the precision of your confidence interval.