To find the sample size needed to achieve a margin of error of 8 or less with a 95% confidence level, given that the standard deviation is 11, we can use the formula for the margin of error (E) in a confidence interval for a population mean:

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

Where:

• $E$ is the margin of error.
• $Z$ is the Z-score corresponding to the desired confidence level.
• $\sigma$ is the population standard deviation.
• $n$ is the sample size.

Given:

• $E = 8$
• $\sigma = 11$
• Confidence level = 0.95, so the Z-score $Z = 1.96$ (for a two-tailed test).

We need to solve for $n$:

$8 = 1.96 \times \frac{11}{\sqrt{n}}$

Now, let's solve for $n$.

$\sqrt{n} = \frac{1.96 \times 11}{8}$

$n = \left(\frac{1.96 \times 11}{8}\right)^2$

Let me calculate that for you.The calculated sample size $n$ is approximately 7.26. Since the sample size must be a whole number, you would need to round up to the next whole number. Therefore, a sample size of 8 is required to achieve a margin of error of 8 or less with a 95% confidence level when the standard deviation is 11.

Would you like any further details or have any other questions?

Here are some related questions you might consider:

1. How does changing the confidence level affect the required sample size?
2. What happens to the sample size if the margin of error is reduced?
3. How does the standard deviation influence the sample size calculation?
4. What is the impact of a smaller sample size on the margin of error?
5. Can you calculate the sample size for a different margin of error?

Tip: Always round up when calculating sample size because fractional samples aren't possible. This ensures the margin of error requirement is met.