Sampling Distributions for Sample Proportions - AP Statistics
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What does a sampling distribution describe?
What does a sampling distribution describe?
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Distribution of a statistic over many samples. Shows behavior of statistics across repeated sampling.
Distribution of a statistic over many samples. Shows behavior of statistics across repeated sampling.
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What is the effect of a larger sample size on the sampling distribution?
What is the effect of a larger sample size on the sampling distribution?
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Reduces variability of the sample proportion. Larger $n$ decreases standard error, concentrating distribution.
Reduces variability of the sample proportion. Larger $n$ decreases standard error, concentrating distribution.
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Calculate the standard deviation for $p=0.6$ and $n=150$.
Calculate the standard deviation for $p=0.6$ and $n=150$.
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$\text{SD}(\bar{p}) = 0.04$. Using $\sqrt{\frac{0.6(0.4)}{150}} = \sqrt{0.0016}$.
$\text{SD}(\bar{p}) = 0.04$. Using $\sqrt{\frac{0.6(0.4)}{150}} = \sqrt{0.0016}$.
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What is the shape of the sampling distribution of the sample proportion when conditions are met?
What is the shape of the sampling distribution of the sample proportion when conditions are met?
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Approximately normal. Central Limit Theorem ensures normality for large samples.
Approximately normal. Central Limit Theorem ensures normality for large samples.
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What is the central concept of the Central Limit Theorem?
What is the central concept of the Central Limit Theorem?
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Sample means approximate normality as sample size increases. Sample statistics approach normal distribution as $n$ increases.
Sample means approximate normality as sample size increases. Sample statistics approach normal distribution as $n$ increases.
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Identify the symbol for the population proportion.
Identify the symbol for the population proportion.
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$p$. Represents the true population proportion parameter.
$p$. Represents the true population proportion parameter.
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What does the Law of Large Numbers imply for sample proportions?
What does the Law of Large Numbers imply for sample proportions?
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Sample proportion approaches population proportion as n increases. Larger samples give more accurate estimates of population parameters.
Sample proportion approaches population proportion as n increases. Larger samples give more accurate estimates of population parameters.
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Which theorem justifies the normality of the sampling distribution of the sample proportion?
Which theorem justifies the normality of the sampling distribution of the sample proportion?
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Central Limit Theorem. Applies when sample size conditions are satisfied.
Central Limit Theorem. Applies when sample size conditions are satisfied.
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Calculate the standard deviation of the sample proportion for $p=0.2$ and $n=100$.
Calculate the standard deviation of the sample proportion for $p=0.2$ and $n=100$.
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$\text{SD}(\bar{p}) = 0.04$. Using $\sqrt{\frac{0.2(0.8)}{100}} = \sqrt{0.0016}$.
$\text{SD}(\bar{p}) = 0.04$. Using $\sqrt{\frac{0.2(0.8)}{100}} = \sqrt{0.0016}$.
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Determine $\text{SD}(\bar{p})$ for $p=0.15$, $n=150$.
Determine $\text{SD}(\bar{p})$ for $p=0.15$, $n=150$.
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$\text{SD}(\bar{p}) = 0.029$. Using $\sqrt{\frac{0.15(0.85)}{150}} = \sqrt{0.00085}$.
$\text{SD}(\bar{p}) = 0.029$. Using $\sqrt{\frac{0.15(0.85)}{150}} = \sqrt{0.00085}$.
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Calculate $\text{SD}(\bar{p})$ for $p=0.7$, $n=90$.
Calculate $\text{SD}(\bar{p})$ for $p=0.7$, $n=90$.
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$\text{SD}(\bar{p}) = 0.048$. Using $\sqrt{\frac{0.7(0.3)}{90}} = \sqrt{0.00233}$.
$\text{SD}(\bar{p}) = 0.048$. Using $\sqrt{\frac{0.7(0.3)}{90}} = \sqrt{0.00233}$.
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What is the typical shape of a sampling distribution as $n$ becomes large?
What is the typical shape of a sampling distribution as $n$ becomes large?
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Normal distribution. Central Limit Theorem ensures normality for large $n$.
Normal distribution. Central Limit Theorem ensures normality for large $n$.
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Verify normal approximation for $n=40$, $p=0.1$.
Verify normal approximation for $n=40$, $p=0.1$.
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Not valid. $np = 4 < 10$, violating normality condition.
Not valid. $np = 4 < 10$, violating normality condition.
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Calculate $\text{SD}(\bar{p})$ for $p=0.55$, $n=120$.
Calculate $\text{SD}(\bar{p})$ for $p=0.55$, $n=120$.
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$\text{SD}(\bar{p}) = 0.045$. Using $\sqrt{\frac{0.55(0.45)}{120}} = \sqrt{0.00206}$.
$\text{SD}(\bar{p}) = 0.045$. Using $\sqrt{\frac{0.55(0.45)}{120}} = \sqrt{0.00206}$.
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What does $n$ represent in the formula for $\text{SD}(\bar{p})$?
What does $n$ represent in the formula for $\text{SD}(\bar{p})$?
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Sample size. Number of observations in the sample.
Sample size. Number of observations in the sample.
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What is the role of a sample proportion in statistics?
What is the role of a sample proportion in statistics?
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Estimate the population proportion. Provides estimate of unknown population parameter.
Estimate the population proportion. Provides estimate of unknown population parameter.
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What is the consequence of violating normality conditions for sample proportions?
What is the consequence of violating normality conditions for sample proportions?
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The approximation may be invalid. Normal approximation may not be accurate for inference.
The approximation may be invalid. Normal approximation may not be accurate for inference.
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Calculate $\text{SD}(\bar{p})$ for $p=0.45$, $n=200$.
Calculate $\text{SD}(\bar{p})$ for $p=0.45$, $n=200$.
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$\text{SD}(\bar{p}) = 0.035$. Using $\sqrt{\frac{0.45(0.55)}{200}} = \sqrt{0.0012375}$.
$\text{SD}(\bar{p}) = 0.035$. Using $\sqrt{\frac{0.45(0.55)}{200}} = \sqrt{0.0012375}$.
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What is the primary purpose of a sampling distribution?
What is the primary purpose of a sampling distribution?
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To infer population parameters from sample statistics. Enables statistical inference about population parameters.
To infer population parameters from sample statistics. Enables statistical inference about population parameters.
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What is the mean of the sampling distribution of the sample proportion?
What is the mean of the sampling distribution of the sample proportion?
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The population proportion $p$. Expected value of an unbiased estimator equals the parameter.
The population proportion $p$. Expected value of an unbiased estimator equals the parameter.
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Calculate the sample proportion's standard deviation for $p=0.5$ and $n=50$.
Calculate the sample proportion's standard deviation for $p=0.5$ and $n=50$.
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$\text{SD}(\bar{p}) = 0.07$. Using $\sqrt{\frac{0.5(0.5)}{50}} = \sqrt{0.005}$.
$\text{SD}(\bar{p}) = 0.07$. Using $\sqrt{\frac{0.5(0.5)}{50}} = \sqrt{0.005}$.
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What happens to the standard deviation of the sample proportion as sample size increases?
What happens to the standard deviation of the sample proportion as sample size increases?
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It decreases. Standard error is inversely proportional to $\sqrt{n}$.
It decreases. Standard error is inversely proportional to $\sqrt{n}$.
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State the conditions for using the normal approximation for sample proportions.
State the conditions for using the normal approximation for sample proportions.
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$np \text{ and } n(1-p) \text{ both } \text{≥} 10$. Required for Central Limit Theorem to apply.
$np \text{ and } n(1-p) \text{ both } \text{≥} 10$. Required for Central Limit Theorem to apply.
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Find the sample proportion's standard deviation for $p=0.3$ and $n=200$.
Find the sample proportion's standard deviation for $p=0.3$ and $n=200$.
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$\text{SD}(\bar{p}) = 0.032$. Using $\sqrt{\frac{0.3(0.7)}{200}} = \sqrt{0.00105}$.
$\text{SD}(\bar{p}) = 0.032$. Using $\sqrt{\frac{0.3(0.7)}{200}} = \sqrt{0.00105}$.
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How does increasing the sample size affect the normality of the sampling distribution?
How does increasing the sample size affect the normality of the sampling distribution?
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Increases normality approximation. Better approximation with larger sample sizes.
Increases normality approximation. Better approximation with larger sample sizes.
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What is the relationship between population proportion $p$ and sample proportion $\bar{p}$?
What is the relationship between population proportion $p$ and sample proportion $\bar{p}$?
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$E(\bar{p}) = p$. Sample proportion is unbiased for population proportion.
$E(\bar{p}) = p$. Sample proportion is unbiased for population proportion.
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Verify if normal approximation is valid for $n=30, p=0.4$.
Verify if normal approximation is valid for $n=30, p=0.4$.
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Yes, valid. $np = 12 \geq 10$ and $n(1-p) = 18 \geq 10$.
Yes, valid. $np = 12 \geq 10$ and $n(1-p) = 18 \geq 10$.
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Find the standard deviation of the sample proportion if $p=0.4$ and $n=25$.
Find the standard deviation of the sample proportion if $p=0.4$ and $n=25$.
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$\text{SD}(\bar{p}) = 0.098$. Using $\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4(0.6)}{25}}$.
$\text{SD}(\bar{p}) = 0.098$. Using $\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4(0.6)}{25}}$.
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What is the symbol for the sample proportion?
What is the symbol for the sample proportion?
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$\bar{p}$. Standard notation for sample proportion in statistics.
$\bar{p}$. Standard notation for sample proportion in statistics.
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Identify the conditions required for the sampling distribution of the sample proportion to be approximately normal.
Identify the conditions required for the sampling distribution of the sample proportion to be approximately normal.
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np ≥ 10 and n(1-p) ≥ 10. Ensures expected counts are large enough for normal approximation.
np ≥ 10 and n(1-p) ≥ 10. Ensures expected counts are large enough for normal approximation.
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