To determine the margin of error for a confidence interval to estimate the population mean, we use the following formula for the margin of error (ME):

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

Where:

• $t_{\alpha/2}$ is the critical value of the t-distribution for a given confidence level.
• $s$ is the sample standard deviation.
• $n$ is the sample size.
• $\alpha$ is the level of significance ($\alpha = 1 - \text{confidence level}$).

Given:

• $n = 16$
• $s = 14.1$

Now, let's calculate the margin of error for each of the three confidence levels.

a) 80% Confidence Level

For an 80% confidence level, $\alpha = 1 - 0.80 = 0.20$, so $\alpha/2 = 0.10$.

We need the critical value $t_{0.10}$ for $n - 1 = 16 - 1 = 15$ degrees of freedom. Using a t-table or calculator, $t_{0.10} \approx 1.341$.

Now calculate the margin of error:

$ME_{80\%} = 1.341 \times \frac{14.1}{\sqrt{16}} = 1.341 \times \frac{14.1}{4} = 1.341 \times 3.525 = 4.72$

So, the margin of error for an 80% confidence level is approximately 4.72.

b) 90% Confidence Level

For a 90% confidence level, $\alpha = 1 - 0.90 = 0.10$, so $\alpha/2 = 0.05$.

We need the critical value $t_{0.05}$ for $15$ degrees of freedom. From the t-table or calculator, $t_{0.05} \approx 1.753$.

Now calculate the margin of error:

$ME_{90\%} = 1.753 \times \frac{14.1}{\sqrt{16}} = 1.753 \times 3.525 = 6.17$

So, the margin of error for a 90% confidence level is approximately 6.17.

c) 99% Confidence Level

For a 99% confidence level, $\alpha = 1 - 0.99 = 0.01$, so $\alpha/2 = 0.005$.

We need the critical value $t_{0.005}$ for $15$ degrees of freedom. From the t-table or calculator, $t_{0.005} \approx 2.947$.

Now calculate the margin of error:

$ME_{99\%} = 2.947 \times \frac{14.1}{\sqrt{16}} = 2.947 \times 3.525 = 10.39$

So, the margin of error for a 99% confidence level is approximately 10.39.

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Summary of Results:

• 80% Confidence Level: Margin of Error = 4.72
• 90% Confidence Level: Margin of Error = 6.17
• 99% Confidence Level: Margin of Error = 10.39

Would you like further explanation or have any questions on this?

Here are some related questions for you to explore:

1. How does the sample size affect the margin of error for confidence intervals?
2. What happens to the margin of error if the sample standard deviation is larger?
3. How would you calculate a confidence interval once the margin of error is determined?
4. What is the difference between the t-distribution and the z-distribution in confidence intervals?
5. How do you interpret a confidence interval in terms of population parameters?

Tip: A larger sample size generally reduces the margin of error, providing a more precise estimate of the population parameter.