Confidence Interval for a Population Proportion - AP Statistics

1. Calculate the 95% CI for $\bar{p} = 0.45$, $n = 200$.

   $(0.385, 0.515)$. Using $0.45 \pm 1.96\sqrt{\frac{0.45(0.55)}{200}} = (0.385, 0.515)$.

2. What is the sample proportion for 80 successes in 200 trials?

   $\bar{p} = 0.4$. Calculated as $\frac{80}{200} = 0.4$.

3. Calculate the 90% CI for $\bar{p} = 0.75$, $n = 150$.

   $(0.693, 0.807)$. Using $0.75 \pm 1.645\sqrt{\frac{0.75(0.25)}{150}} = (0.693, 0.807)$.

4. Identify the margin of error when $z^* = 1.96$ and SE = 0.05.

   0.098. Margin of error equals $z^* \times SE = 1.96 \times 0.05 = 0.098$.

5. State the margin of error formula for a population proportion CI.

   $z^* \, \sqrt{\frac{\bar{p}(1-\bar{p})}{n}}$. Half-width of the confidence interval around $\bar{p}$.

6. What is the impact of a small sample size on confidence intervals?

   Results in wider confidence intervals. Small $n$ increases standard error, reducing precision.

7. What is the formula for the confidence interval of a population proportion?

   $\bar{p} \, \pm \, z^* \, \sqrt{\frac{\bar{p}(1-\bar{p})}{n}}$. Standard formula using sample proportion, critical value, and standard error.

8. What is the minimum sample size needed for a valid confidence interval?

   Both $np \, \geq \, 10$ and $n(1-p) \, \geq \, 10$. Ensures normal approximation is valid for both successes and failures.

9. What is the standard error formula for a population proportion?

   $\sqrt{\frac{\bar{p}(1-\bar{p})}{n}}$. Measures the sampling variability of the sample proportion.

10. Find the standard error for $\bar{p} = 0.5$, $n = 250$.

    $0.0316$. Using $\sqrt{\frac{0.5(0.5)}{250}} = 0.0316$.

11. Identify the sample proportion if 30 out of 100 are successful.

    $\bar{p} = 0.3$. Sample proportion equals number of successes divided by sample size.

12. Identify the effect of a narrower confidence interval on precision.

    Increases precision. Narrower intervals provide more precise estimates of the parameter.

13. State the z-score associated with a 90% confidence interval.

    $z^* = 1.645$. Captures the middle 90% of the standard normal distribution.

14. What is the relationship between CI width and variability?

    Higher variability increases CI width. More variability requires wider intervals to maintain confidence level.

15. What is the effect of a larger sample size on the standard error?

    Decreases the standard error. Standard error is inversely proportional to $\sqrt{n}$.

16. What is the interpretation of a 95% confidence interval?

    95% confident the true proportion is within the interval. Describes the long-run capture rate of the true parameter.

17. What is the effect of increasing sample size on CI width?

    Decreases CI width. Larger $n$ reduces standard error, making the interval narrower.

18. Which statistical distribution is used for the critical value in proportion CI?

    Standard normal distribution. Used because sample proportions are approximately normally distributed.

19. Find and correct the error: $CI = \bar{p} \, \pm \, z^* \, \sqrt{\frac{1-\bar{p}}{n}}$.

    $CI = \bar{p} \, \pm \, z^* \, \sqrt{\frac{\bar{p}(1-\bar{p})}{n}}$. Missing $\bar{p}$ term in the numerator of the standard error formula.

20. State the effect of a higher confidence level on CI width.

    Increases the width of the CI. Higher confidence requires larger critical value and margin of error.

21. How does variability in data affect the CI width?

    Greater variability increases CI width. Higher variability increases uncertainty, requiring wider intervals.

22. What does $z^*$ represent in the confidence interval formula?

    The critical value from the standard normal distribution. Corresponds to the desired confidence level (e.g., 1.96 for 95%).

23. What happens to the CI if the confidence level is increased?

    The CI becomes wider. Higher confidence requires larger critical value, increasing margin of error.

24. What is the critical value for an 80% confidence level?

    $z^* = 1.282$. Captures the middle 80% of the standard normal distribution.

25. Define the term 'population proportion'.

    The fraction or percentage of the population with a particular characteristic. Represents the parameter $p$ we're trying to estimate with confidence intervals.

26. Calculate the margin of error for $\bar{p} = 0.6$, $n = 100$, $z^* = 1.96$.

    0.096. Using formula: $1.96 \times \sqrt{\frac{0.6(0.4)}{100}} = 0.096$.

27. What mathematical operation is used to adjust the sample proportion in CI?

    Addition and subtraction. The $\pm$ creates the interval bounds around the point estimate.

28. What is the typical confidence level used in constructing confidence intervals?

    95%. Most commonly used standard in statistical practice.

29. What assumptions must be met to construct a CI for a proportion?

    Random sample and $np, n(1-p) \, \geq \, 10$. Required for valid normal approximation to the sampling distribution.

30. What does $\bar{p}$ represent in a confidence interval formula?

    Sample proportion. The point estimator for the population proportion $p$.

31. How does the confidence level affect the critical value?

    Higher confidence level increases critical value. More confidence requires larger $z^*$ to capture more area.

32. What is the critical value for a 99% confidence level?

    $z^* = 2.576$. Captures the middle 99% of the standard normal distribution.

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