To determine the minimum sample size needed for a 90% confidence interval for a population proportion with no prior estimate of the proportion, we can use the following formula for sample size $n$:

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

where:

• $Z$ is the z-score corresponding to the desired confidence level (for a 90% confidence level, $Z \approx 1.645$),
• $p$ is the estimated proportion of the population. When no prior estimate is available, we use $p = 0.5$ to maximize the sample size, since $p(1 - p)$ is largest when $p = 0.5$,
• $E$ is the margin of error, which is given as 0.09 in this case.

Substitute the values:

1. Set $Z = 1.645$,
2. Use $p = 0.5$,
3. Set $E = 0.09$.

The formula becomes:

$n = \frac{{(1.645)^2 \cdot 0.5 \cdot (1 - 0.5)}}{{(0.09)^2}}$

Step-by-Step Calculation:

1. Calculate $(1.645)^2 = 2.705025$,
2. Calculate $0.5 \cdot (1 - 0.5) = 0.25$,
3. Multiply: $2.705025 \times 0.25 = 0.67625625$,
4. Divide by $(0.09)^2 = 0.0081$.

$n = \frac{0.67625625}{0.0081} \approx 83.12$

Since sample size must be a whole number, round up to ensure the desired margin of error:

$n = 84$

Conclusion:

You need a sample size of 84 to ensure the sample proportion does not deviate from the population proportion by more than 0.09 at a 90% confidence level.

Would you like further details on the calculations, or do you have any questions?

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1. What is the impact of increasing the confidence level on sample size?
2. How would a prior estimate for the population proportion affect sample size?
3. Why is 0.5 used as an estimate for $p$ when no prior estimate is available?
4. How does changing the margin of error affect the required sample size?
5. How is the z-score determined for different confidence levels?

Tip: Increasing the desired confidence level will increase the required sample size, as it narrows the confidence interval.