AP Statistics Flashcards: Sampling Distributions For Sample Proportions

What this deck covers

This deck focuses on Sampling Distributions For Sample Proportions, giving you a quick way to review the definitions, rules, and examples that matter most for AP Statistics.

How to use these flashcards

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.

Flashcard 1: Which theorem justifies the normality of the sampling distribution of the sample proportion?

Answer: Central Limit Theorem. Applies when sample size conditions are satisfied.

Flashcard 2: What is the symbol for the sample proportion?

Answer: $\bar{p}$. Standard notation for sample proportion in statistics.

Flashcard 3: What is the effect of a larger sample size on the sampling distribution?

Answer: Reduces variability of the sample proportion. Larger $n$ decreases standard error, concentrating distribution.

Flashcard 4: How does increasing the sample size affect the normality of the sampling distribution?

Answer: Increases normality approximation. Better approximation with larger sample sizes.

Flashcard 5: Calculate $\text{SD}(\bar{p})$ for $p=0.45$, $n=200$.

Answer: $\text{SD}(\bar{p}) = 0.035$. Using $\sqrt{\frac{0.45(0.55)}{200}} = \sqrt{0.0012375}$.

Flashcard 6: Determine $\text{SD}(\bar{p})$ for $p=0.15$, $n=150$.

Answer: $\text{SD}(\bar{p}) = 0.029$. Using $\sqrt{\frac{0.15(0.85)}{150}} = \sqrt{0.00085}$.

Flashcard 7: What is the central concept of the Central Limit Theorem?

Answer: Sample means approximate normality as sample size increases. Sample statistics approach normal distribution as $n$ increases.

Flashcard 8: What is the mean of the sampling distribution of the sample proportion?

Answer: The population proportion $p$. Expected value of an unbiased estimator equals the parameter.

Flashcard 9: Calculate the standard deviation of the sample proportion for $p=0.2$ and $n=100$.

Answer: $\text{SD}(\bar{p}) = 0.04$. Using $\sqrt{\frac{0.2(0.8)}{100}} = \sqrt{0.0016}$.

Flashcard 10: Identify the conditions required for the sampling distribution of the sample proportion to be approximately normal.

Answer: np ≥ 10 and n(1-p) ≥ 10. Ensures expected counts are large enough for normal approximation.

Flashcard 11: What is the relationship between population proportion $p$ and sample proportion $\bar{p}$?

Answer: $E(\bar{p}) = p$. Sample proportion is unbiased for population proportion.

Flashcard 12: Calculate the standard deviation for $p=0.6$ and $n=150$.

Answer: $\text{SD}(\bar{p}) = 0.04$. Using $\sqrt{\frac{0.6(0.4)}{150}} = \sqrt{0.0016}$.

Flashcard 13: Verify normal approximation for $n=40$, $p=0.1$.

Answer: Not valid. $np = 4 < 10$, violating normality condition.

Flashcard 14: What does the Law of Large Numbers imply for sample proportions?

Answer: Sample proportion approaches population proportion as n increases. Larger samples give more accurate estimates of population parameters.

Flashcard 15: Identify the symbol for the population proportion.

Answer: $p$. Represents the true population proportion parameter.

Flashcard 16: Calculate $\text{SD}(\bar{p})$ for $p=0.55$, $n=120$.

Answer: $\text{SD}(\bar{p}) = 0.045$. Using $\sqrt{\frac{0.55(0.45)}{120}} = \sqrt{0.00206}$.

Flashcard 17: Verify if normal approximation is valid for $n=30, p=0.4$.

Answer: Yes, valid. $np = 12 \geq 10$ and $n(1-p) = 18 \geq 10$.

Flashcard 18: What does $\text{SD}(\bar{p})$ stand for in the context of sampling distributions?

Answer: Standard deviation of the sample proportion. Measures variability of sample proportion estimates.

Flashcard 19: What is the consequence of violating normality conditions for sample proportions?

Answer: The approximation may be invalid. Normal approximation may not be accurate for inference.

Flashcard 20: Calculate $\text{SD}(\bar{p})$ for $p=0.7$, $n=90$.

Answer: $\text{SD}(\bar{p}) = 0.048$. Using $\sqrt{\frac{0.7(0.3)}{90}} = \sqrt{0.00233}$.

Flashcard 21: What does $n$ represent in the formula for $\text{SD}(\bar{p})$?

Answer: Sample size. Number of observations in the sample.

Flashcard 22: What is the impact of larger $n$ on $\text{SD}(\bar{p})$?

Answer: Decreases $\text{SD}(\bar{p})$. Standard error decreases as sample size increases.

Flashcard 23: What is the expected value of the sample proportion equal to?

Answer: The population proportion $p$. Sample proportion is an unbiased estimator of $p$.

Flashcard 24: What is the primary purpose of a sampling distribution?

Answer: To infer population parameters from sample statistics. Enables statistical inference about population parameters.

Flashcard 25: State the conditions for using the normal approximation for sample proportions.

Answer: $np \text{ and } n(1-p) \text{ both } \text{≥} 10$. Required for Central Limit Theorem to apply.

Flashcard 26: What is the shape of the sampling distribution of the sample proportion when conditions are met?

Answer: Approximately normal. Central Limit Theorem ensures normality for large samples.

Flashcard 27: What happens to the standard deviation of the sample proportion as sample size increases?

Answer: It decreases. Standard error is inversely proportional to $\sqrt{n}$.

Flashcard 28: Calculate the sample proportion's standard deviation for $p=0.5$ and $n=50$.

Answer: $\text{SD}(\bar{p}) = 0.07$. Using $\sqrt{\frac{0.5(0.5)}{50}} = \sqrt{0.005}$.

Flashcard 29: What is the typical shape of a sampling distribution as $n$ becomes large?

Answer: Normal distribution. Central Limit Theorem ensures normality for large $n$.

Flashcard 30: What is the role of a sample proportion in statistics?

Answer: Estimate the population proportion. Provides estimate of unknown population parameter.

Flashcard 31: Find the sample proportion's standard deviation for $p=0.3$ and $n=200$.

Answer: $\text{SD}(\bar{p}) = 0.032$. Using $\sqrt{\frac{0.3(0.7)}{200}} = \sqrt{0.00105}$.

Flashcard 32: Find the standard deviation of the sample proportion if $p=0.4$ and $n=25$.

Answer: $\text{SD}(\bar{p}) = 0.098$. Using $\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4(0.6)}{25}}$.

Flashcard 33: What does a sampling distribution describe?

Answer: Distribution of a statistic over many samples. Shows behavior of statistics across repeated sampling.

Flashcard 34: Determine $\text{SD}(\bar{p})$ for $p=0.25$, $n=80$.

Answer: $\text{SD}(\bar{p}) = 0.048$. Using $\sqrt{\frac{0.25(0.75)}{80}} = \sqrt{0.00234}$.

Flashcard 35: State the formula for the sample proportion $\bar{p}$.

Answer: $\bar{p} = \frac{x}{n}$. Number of successes divided by sample size.