To address the questions, let's break them down step by step:

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1. Determining the minimum sample size for 90% confidence:

To calculate the required sample size for estimating a population proportion, we use the formula for margin of error:

$n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$

Where:

• $n$ = minimum sample size
• $Z$ = critical value for the desired confidence level (90% confidence corresponds to $Z = 1.645$)
• $p$ = estimated population proportion (if unknown, use $p = 0.5$ for maximum variability)
• $E$ = margin of error (in this case, 0.05)

If $p = 0.5$ (maximum variability assumption):

$n = \frac{(1.645)^2 \cdot 0.5 \cdot (1 - 0.5)}{0.05^2}$

$n = \frac{2.706 \cdot 0.25}{0.0025}$

$n = \frac{0.6765}{0.0025} = 270.6$

Since the sample size must be an integer, round up to 271.

Thus, you would need to survey at least 271 people.

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2. Effect of increasing confidence level on sample size:

If the confidence level is increased (e.g., from 90% to 95%), the critical value $Z$ becomes larger (for 95%, $Z = 1.96$). This increases the numerator of the formula, which directly increases the sample size.

The reason for this is:

• A higher confidence level requires a larger margin of safety to ensure the estimate captures the true population proportion.
• Consequently, more data is needed to reduce variability in the estimate.

For instance, using 95% confidence, the calculation would be:

$n = \frac{(1.96)^2 \cdot 0.5 \cdot (1 - 0.5)}{0.05^2}$

$n = \frac{3.8416 \cdot 0.25}{0.0025}$

$n = \frac{0.9604}{0.0025} = 384.16$

Rounded up to 385, the sample size increases significantly compared to 90% confidence.

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Summary:

1. For 90% confidence, you need at least 271 respondents.
2. Increasing the confidence level would result in a larger sample size because more confidence requires reducing uncertainty, which means gathering more data.

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Do you want details on why $p = 0.5$ is chosen when $p$ is unknown or any step above?
Here are 5 related questions to deepen understanding:

1. What factors influence the choice of confidence level in real-world surveys?
2. Why does increasing confidence lead to higher sample sizes in statistical estimation?
3. How does changing the margin of error affect the required sample size?
4. Can we calculate the sample size for a specific known $p$ instead of $p = 0.5$?
5. What is the tradeoff between confidence levels and practicality in survey design?

Tip: Always choose $p = 0.5$ when no prior estimate is available, as it maximizes the required sample size, ensuring adequate precision.