To calculate the margin of error (MOE) for a confidence interval, we use the following formula:

$\text{MOE} = t^* \times \frac{s}{\sqrt{n}}$

Where:

• $t^*$ is the critical value from the t-distribution for the given confidence level (90% confidence in this case).
• $s$ is the sample standard deviation.
• $n$ is the sample size.

Since the sample standard deviation $s = 43$, we need to determine the critical value $t^*$ for a 90% confidence interval. This depends on the degrees of freedom, which is $n - 1$.

Step 1: Find the critical value $t^*$

For a 90% confidence interval, the area in each tail of the t-distribution is $0.05$ (because $100\% - 90\% = 10\%$, and we split that 10% between the two tails).

We’ll need to look up the t-value for the appropriate degrees of freedom ($n-1$).

Step 2: Calculate the margin of error for each sample size

a) For $n = 13$:

• Degrees of freedom = $13 - 1 = 12$
• From the t-distribution table, the critical value $t^*$ for 12 degrees of freedom and a 90% confidence level is approximately $1.782$.

$\text{MOE} = 1.782 \times \frac{43}{\sqrt{13}} \approx 1.782 \times 11.94 \approx 21.27$

So, the margin of error for $n = 13$ is approximately 21.27.

b) For $n = 25$:

• Degrees of freedom = $25 - 1 = 24$
• From the t-distribution table, the critical value $t^*$ for 24 degrees of freedom and a 90% confidence level is approximately $1.711$.

$\text{MOE} = 1.711 \times \frac{43}{\sqrt{25}} = 1.711 \times 8.6 \approx 14.71$

So, the margin of error for $n = 25$ is approximately 14.71.

c) For $n = 46$:

• Degrees of freedom = $46 - 1 = 45$
• From the t-distribution table, the critical value $t^*$ for 45 degrees of freedom and a 90% confidence level is approximately $1.679$.

$\text{MOE} = 1.679 \times \frac{43}{\sqrt{46}} \approx 1.679 \times 6.28 \approx 10.54$

So, the margin of error for $n = 46$ is approximately 10.54.

Final Answers:

a) $n = 13$, MOE = 21.27
b) $n = 25$, MOE = 14.71
c) $n = 46$, MOE = 10.54

Would you like more details on how these values were calculated, or do you have any questions?

Here are some additional questions you might find helpful:

1. How does increasing the sample size affect the margin of error?
2. How would you calculate the margin of error for a 95% confidence level instead of 90%?
3. What is the difference between the z-distribution and t-distribution in confidence interval calculations?
4. How do you determine whether to use a t-distribution or z-distribution for margin of error calculations?
5. Can the margin of error be negative? If not, why?

Tip: When the sample size is large (typically $n > 30$), the t-distribution approaches the normal distribution, and you can use the z-value instead of the t-value.