To solve this problem, we need to determine the minimum sample size required to estimate the population proportion with a given margin of error, confidence level, and prior estimate of the population proportion $p$.

Given Information:

• Margin of error ($E$) = 0.07
• Confidence level = 95%
• Estimated proportion ($p$) = 0.92
• Complement of $p$ ($q$) = $1 - p = 1 - 0.92 = 0.08$

Formula for Minimum Sample Size:

The formula for finding the minimum sample size $n$ for estimating a population proportion is: $n = \frac{{Z^2 \cdot p \cdot q}}{{E^2}}$ where:

• $Z$ is the Z-score corresponding to the confidence level,
• $p$ is the estimated proportion,
• $q = 1 - p$,
• $E$ is the margin of error.

Step 1: Find the Z-score for a 95% Confidence Level

For a 95% confidence level, the Z-score is approximately 1.96.

Step 2: Substitute Values into the Formula

$n = \frac{{(1.96)^2 \cdot 0.92 \cdot 0.08}}{{(0.07)^2}}$

Calculating each part:

1. $Z^2 = (1.96)^2 = 3.8416$
2. $p \cdot q = 0.92 \cdot 0.08 = 0.0736$
3. $E^2 = (0.07)^2 = 0.0049$

Now, substitute these into the formula: $n = \frac{{3.8416 \cdot 0.0736}}{{0.0049}} \approx 57.7$

Since sample size must be a whole number, we round up to the next integer: $n = 58$

Answer:

The minimum sample size required is 58.

Would you like more details on the solution, or do you have any questions?

Related Questions:

1. What is the formula for calculating sample size if the population standard deviation is known?
2. How would the sample size change if the margin of error was reduced to 0.05?
3. What Z-score corresponds to a 99% confidence level?
4. Why do we round up the sample size in these calculations?
5. How does the value of $p$ affect the sample size?

Tip:

Always remember to round up when calculating sample sizes in statistics to ensure the margin of error meets the desired confidence level.