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Let's break down the questions one by one.

### i) **What is the point estimate of the proportion of the population of business travelers who believe a frequent flyer program is the most important factor when choosing an airline carrier?**

The point estimate of a population proportion is given by the ratio of the number of successes to the total sample size.

From the problem:
- 618 business travelers listed the frequent flyer program as the most important factor.
- The total number of business travelers surveyed was 1993.

Thus, the point estimate (\(\hat{p}\)) is:
\[
\hat{p} = \frac{618}{1993} \approx 0.3101
\]
So, the point estimate is approximately **0.3101** or **31.01%**.

---

### ii) **Develop a 95% confidence interval estimate of the population proportion.**

To calculate the confidence interval for a population proportion, we use the formula:
\[
\hat{p} \pm Z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
Where:
- \(\hat{p}\) is the point estimate (0.3101 from part i),
- \(Z\) is the Z-score corresponding to the 95% confidence level (which is 1.96),
- \(n\) is the sample size (1993).

First, calculate the standard error (SE):
\[
SE = \sqrt{\frac{0.3101(1 - 0.3101)}{1993}} \approx 0.0105
\]

Now, compute the margin of error (ME):
\[
ME = 1.96 \times 0.0105 \approx 0.0206
\]

Finally, the confidence interval is:
\[
0.3101 \pm 0.0206
\]

Thus, the 95% confidence interval is approximately:
\[
(0.2895, 0.3307)
\]

So, the interval is **(0.2895, 0.3307)** or **28.95% to 33.07%**.

---

### iii) **How large a sample would be required to report the margin of error of 0.01 at 95% confidence?**

To determine the required sample size (\(n\)) for a specific margin of error (ME), the formula is:
\[
n = \left( \frac{Z^2 \cdot \hat{p} \cdot (1 - \hat{p})}{ME^2} \right)
\]

Given:
- \(Z = 1.96\),
- \(\hat{p} = 0.3101\),
- Desired margin of error (ME) = 0.01.

Substitute into the formula:
\[
n = \frac{(1.96)^2 \cdot 0.3101 \cdot (1 - 0.3101)}{(0.01)^2}
\]
\[
n = \frac{3.8416 \cdot 0.3101 \cdot 0.6899}{0.0001}
\]
\[
n \approx 8246.3
\]

Thus, a sample size of approximately **8247** would be required to achieve a margin of error of 0.01 at 95% confidence.

---

### Recommendation

It would require a much larger sample (about 8247 respondents) to achieve a margin of error of 0.01, which is significantly larger than the current sample size of 1993. Whether USA Today should attempt to provide this level of precision depends on factors such as available resources and the importance of that precision to their conclusions.

---

Would you like more details or have any questions about this?

Here are 5 related questions that build on this:
1. What happens to the confidence interval if we increase the confidence level to 99%?
2. How does increasing the sample size affect the margin of error in confidence intervals?
3. What would be the impact on the margin of error if we chose a different proportion estimate, say 0.5?
4. Why is the Z-score 1.96 for a 95% confidence level, and how can it be determined for other confidence levels?
5. Could this point estimate be biased if the survey is not representative of all business travelers?

**Tip:** Increasing the sample size can decrease the margin of error, improving the precision of your estimates, but it also increases cost and time needed for data collection.

What is the point estimate of the proportion of the population of business travelers who believe a frequent flyer program is the most important factor when choosing an airline carrier? Develop a 95% confidence interval estimate of the population proportion. How large a sample would be required to report the margin of error of .01 at 95% confidence?

Calculate the point estimate, 95% confidence interval, and required sample size for determining how many business travelers prioritize frequent flyer programs when choosing an airline. Learn key statistical methods including population proportion, margin of error, and sample size estimation.

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