To solve this problem and find the margin of error (MOE) for a 95% confidence interval to estimate the population proportion, let's proceed step-by-step:

Given Information:

• Sample size ($n$) = 165
• Positive outcomes ($x$) = 76
• Confidence level = 95%

Step 1: Calculate the sample proportion ($\hat{p}$):

$\hat{p} = \frac{x}{n} = \frac{76}{165} \approx 0.4606$

Step 2: Find the critical value ($z^*$):

For a 95% confidence level, the critical value $z^*$ corresponds to 1.96 (from the standard normal distribution table).

Step 3: Calculate the standard error (SE):

The formula for standard error is: $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ Substituting the values: $SE = \sqrt{\frac{0.4606(1 - 0.4606)}{165}} = \sqrt{\frac{0.4606 \times 0.5394}{165}} = \sqrt{\frac{0.2483}{165}} \approx \sqrt{0.001505} \approx 0.0388$

Step 4: Compute the margin of error (MOE):

The formula for MOE is: $MOE = z^* \times SE$ Substituting the values: $MOE = 1.96 \times 0.0388 \approx 0.0760$

Final Answer:

The margin of error for the 95% confidence interval is approximately 0.076 (or 7.6%).

Would you like further explanation or assistance with confidence intervals?

Related Questions:

1. How does the critical value $z^*$ change for a 90% confidence level?
2. What would happen to the MOE if the sample size were doubled?
3. Can the population proportion ($\hat{p}$) be used to predict future outcomes?
4. How is the MOE affected by a change in the confidence level?
5. What assumptions must be met to use this formula for confidence intervals?

Tip:

Always verify that the sample size is large enough ($n \hat{p} \geq 5$ and $n(1 - \hat{p}) \geq 5$) to ensure the normal approximation is valid!