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This deck focuses on Confidence Interval For A Population Proportion, giving you a quick way to review the definitions, rules, and examples that matter most for AP Statistics.
Study Confidence Interval For A Population Proportion in AP Statistics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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What is the effect of increasing sample size on CI width?
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Decreases CI width. Larger n reduces standard error, making the interval narrower.
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This deck focuses on Confidence Interval For A Population Proportion, giving you a quick way to review the definitions, rules, and examples that matter most for AP Statistics.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: Decreases CI width. Larger n reduces standard error, making the interval narrower.
Answer: 0.096. Using formula: 1.96×1000.6(0.4)=0.096.
Answer: (0.385,0.515). Using 0.45±1.962000.45(0.55)=(0.385,0.515).
Answer: 95%. Most commonly used standard in statistical practice.
Answer: pˉ=0.4. Calculated as 20080=0.4.
Answer: The CI becomes wider. Higher confidence requires larger critical value, increasing margin of error.
Answer: (0.693,0.807). Using 0.75±1.6451500.75(0.25)=(0.693,0.807).
Answer: Decreases the standard error. Standard error is inversely proportional to n.
Answer: 0.098. Margin of error equals z∗×SE=1.96×0.05=0.098.
Answer: z∗=1.282. Captures the middle 80% of the standard normal distribution.
Answer: z∗npˉ(1−pˉ). Half-width of the confidence interval around pˉ.
Answer: 0.0316. Using 2500.5(0.5)=0.0316.
Answer: CI=pˉ±z∗npˉ(1−pˉ). Missing pˉ term in the numerator of the standard error formula.
Answer: Results in wider confidence intervals. Small n increases standard error, reducing precision.
Answer: pˉ±z∗npˉ(1−pˉ). Standard formula using sample proportion, critical value, and standard error.
Answer: Both np≥10 and n(1−p)≥10. Ensures normal approximation is valid for both successes and failures.
Answer: npˉ(1−pˉ). Measures the sampling variability of the sample proportion.
Answer: 0.0316. Using 2500.5(0.5)=0.0316.
Answer: pˉ=0.3. Sample proportion equals number of successes divided by sample size.
Answer: Increases precision. Narrower intervals provide more precise estimates of the parameter.
Answer: pˉ=0.4. Calculated as 20080=0.4.
Answer: z∗=1.645. Captures the middle 90% of the standard normal distribution.
Answer: z∗=1.645. Captures the middle 90% of the standard normal distribution.
Answer: Higher variability increases CI width. More variability requires wider intervals to maintain confidence level.
Answer: Decreases the standard error. Standard error is inversely proportional to n.
Answer: Higher variability increases CI width. More variability requires wider intervals to maintain confidence level.
Answer: Increases precision. Narrower intervals provide more precise estimates of the parameter.
Answer: 95% confident the true proportion is within the interval. Describes the long-run capture rate of the true parameter.
Answer: z∗npˉ(1−pˉ). Half-width of the confidence interval around pˉ.
Answer: Addition and subtraction. The ± creates the interval bounds around the point estimate.
Answer: Decreases CI width. Larger n reduces standard error, making the interval narrower.
Answer: Greater variability increases CI width. Higher variability increases uncertainty, requiring wider intervals.
Answer: Standard normal distribution. Used because sample proportions are approximately normally distributed.
Answer: 95% confident the true proportion is within the interval. Describes the long-run capture rate of the true parameter.
Answer: Increases the width of the CI. Higher confidence requires larger critical value and margin of error.
Answer: z∗=2.576. Captures the middle 99% of the standard normal distribution.
Answer: CI=pˉ±z∗npˉ(1−pˉ). Missing pˉ term in the numerator of the standard error formula.
Answer: (0.385,0.515). Using 0.45±1.962000.45(0.55)=(0.385,0.515).
Answer: Increases the width of the CI. Higher confidence requires larger critical value and margin of error.
Answer: Results in wider confidence intervals. Small n increases standard error, reducing precision.
Answer: Random sample and np,n(1−p)≥10. Required for valid normal approximation to the sampling distribution.
Answer: Greater variability increases CI width. Higher variability increases uncertainty, requiring wider intervals.
Answer: The critical value from the standard normal distribution. Corresponds to the desired confidence level (e.g., 1.96 for 95%).
Answer: The critical value from the standard normal distribution. Corresponds to the desired confidence level (e.g., 1.96 for 95%).
Answer: The CI becomes wider. Higher confidence requires larger critical value, increasing margin of error.
Answer: z∗=1.282. Captures the middle 80% of the standard normal distribution.
Answer: The fraction or percentage of the population with a particular characteristic. Represents the parameter p we're trying to estimate with confidence intervals.
Answer: npˉ(1−pˉ). Measures the sampling variability of the sample proportion.
Answer: 0.096. Using formula: 1.96×1000.6(0.4)=0.096.
Answer: Addition and subtraction. The ± creates the interval bounds around the point estimate.
Answer: Sample proportion. The point estimator for the population proportion p.
Answer: pˉ=0.3. Sample proportion equals number of successes divided by sample size.
Answer: pˉ±z∗npˉ(1−pˉ). Standard formula using sample proportion, critical value, and standard error.
Answer: 95%. Most commonly used standard in statistical practice.
Answer: Random sample and np,n(1−p)≥10. Required for valid normal approximation to the sampling distribution.
Answer: (0.693,0.807). Using 0.75±1.6451500.75(0.25)=(0.693,0.807).
Answer: 0.098. Margin of error equals z∗×SE=1.96×0.05=0.098.
Answer: Sample proportion. The point estimator for the population proportion p.
Answer: The fraction or percentage of the population with a particular characteristic. Represents the parameter p we're trying to estimate with confidence intervals.
Answer: Standard normal distribution. Used because sample proportions are approximately normally distributed.
Answer: Higher confidence level increases critical value. More confidence requires larger z∗ to capture more area.
Answer: z∗=2.576. Captures the middle 99% of the standard normal distribution.
Answer: Higher confidence level increases critical value. More confidence requires larger z∗ to capture more area.