AP Statistics Unit 5: Sampling Distributions — Free Review Games.

A sampling distribution shows how a statistic varies across repeated samples.

The expected value of x-bar is the population mean (unbiased).

\(SD(\bar{x}) = \sigma/\sqrt{n}\), which decreases as \(n\) increases.

CLT says x-bar is approximately normal for large n, regardless of population shape.

A common guideline is n >= 30 for the CLT to apply.

The standard error of \(\bar{x}\) is \(\sigma/\sqrt{n}\).

The normal approximation requires both np and n(1-p) to be at least 10.

\(SD(\hat{p}) = \sqrt{p(1-p)/n}\).

E(p-hat) = p, the true population proportion.

\(SE = 10/\sqrt{25} = 10/5 = 2\).

For skewed populations, larger samples (n >= 30-50) are needed for CLT.

\(SE = \sigma/\sqrt{n}\). Doubling \(n\) divides \(SE\) by \(\sqrt{2}\).

When sampling without replacement, n should be < 10% of N for independence.

Averaging smooths out individual variation, reducing the spread.

\(\sqrt{0.6 \times 0.4/100} = \sqrt{0.0024} = 0.049\).

The sampling distribution of a statistic shows the pattern of values that statistic would take across many samples of the same size from the same population, which is the defining mechanism behind inference. The distractor 'It is the same as the distribution of the population' is wrong because the population distribution describes individual values, while the sampling distribution describes a statistic computed from samples. Students should remember that sampling distributions are theoretical constructs built from the idea of repeated sampling, not a single observed dataset.

An unbiased statistic is one whose sampling distribution is centered at the true parameter value, meaning it neither systematically overestimates nor underestimates on average. The distractor 'It always equals the parameter exactly in every sample' is wrong because individual samples vary due to sampling variability even when the statistic is unbiased overall. The key exam idea is that bias concerns the center of the sampling distribution, while variability concerns its spread.

The sample proportion \(\hat{p}\) is calculated from categorical data and serves as the natural point estimate for the population proportion \(p\). The distractor \(\bar{x}\) is wrong because it estimates a population mean from quantitative data, not a proportion from categorical data. Students should match the type of data (categorical vs quantitative) to the correct estimator before choosing a sampling distribution formula.

The Large Counts Condition, \(np \geq 10\) and \(n(1-p) \geq 10\), ensures there are enough expected successes and failures for the binomial-based sampling distribution of \(\hat{p}\) to be well approximated by a Normal curve. The distractor '\(n \geq 30\) only' is wrong because that guideline applies to sample means under the Central Limit Theorem, not specifically to proportions, and does not account for how extreme \(p\) is. Students must always check counts of successes and failures, not just overall sample size, when working with proportions.

The Central Limit Theorem allows us to treat the sampling distribution of \(\bar{x}\) as approximately Normal for sufficiently large samples, and \(n \geq 30\) is a common rule of thumb when the population shape is unknown or moderately skewed. The distractor 'The population must be exactly Normal' is wrong because the entire point of the CLT is that Normality of \(\bar{x}\) can occur even when the population is not Normal, provided \(n\) is large enough. Students should remember that the CLT is specifically the tool used for means, while the Large Counts Condition applies to proportions.

When the underlying population is Normal, any linear combination of independent Normal variables, including the sample mean, is exactly Normal regardless of sample size. The distractor 'Approximately Normal only if \(n \geq 30\)' is wrong because the \(n \geq 30\) guideline is only needed when the population shape is unknown or non-Normal, not when the population itself is already Normal. Students should recognize that starting with a Normal population removes the need to invoke the Central Limit Theorem at all.

The standard error of \(\bar{x}\) is \(\dfrac{\sigma}{\sqrt{n}}\) because averaging reduces variability by a factor of the square root of the sample size, a direct consequence of the variance of a sum of independent random variables. The distractor \(\sigma \sqrt{n}\) is wrong because it increases with sample size rather than decreasing, which contradicts the fact that larger samples produce more precise estimates. Students should memorize this square-root relationship since it explains why quadrupling the sample size only halves the standard error.

The 10% condition requires the sample size to be no more than 10% of the population so that removing observations without replacement does not meaningfully change probabilities from one draw to the next, preserving the independence assumption used in the standard error formulas. The distractor 'That the sample size is at least 10' is wrong because that describes the Large Counts Condition for proportions, a completely different check. Students should keep these two 'rules of thumb' distinct: 10% condition is about independence, while Large Counts is about Normality of \(\hat{p}\).

The Central Limit Theorem guarantees that as sample size increases, the sampling distribution of \(\bar{x}\) becomes increasingly Normal even when the parent population is skewed. The distractor 'It becomes increasingly skewed' is wrong because averaging many observations smooths out the influence of extreme values, reducing rather than increasing skewness. This convergence toward Normality regardless of starting shape is the central idea students must apply throughout Unit 5.

Repeatedly computing \(\hat{p}\) from many samples and recording the resulting values builds up the sampling distribution of \(\hat{p}\), which describes the sample-to-sample variability of that statistic. The distractor 'The population distribution' is wrong because the population distribution describes individual defective/non-defective items, not the collection of sample proportions from repeated samples. This scenario is a good mental model for how sampling distributions are generated in practice through repeated sampling.

Checking the Large Counts Condition gives \(np = 60(0.4) = 24\) and \(n(1-p) = 60(0.6) = 36\), both of which exceed 10, so the Normal approximation is justified. The distractor 'No, because \(n\) is less than 100' is wrong because there is no requirement that \(n \geq 100\); the relevant threshold involves expected successes and failures, not an arbitrary sample size cutoff. Students should always compute both \(np\) and \(n(1-p)\) rather than relying on sample size alone.

Using \(\dfrac{\sigma}{\sqrt{n}} = \dfrac{12}{\sqrt{36}} = \dfrac{12}{6} = 2\), the standard error of the sample mean is 2. The distractor '\(12\)' is wrong because that is the population standard deviation, not the standard deviation of the sampling distribution, which is always smaller once averaging occurs. Students should always divide by \(\sqrt{n}\), not \(n\) itself, when finding the standard error of \(\bar{x}\).

Using \(SD(\hat{p}) = \sqrt{\dfrac{p(1-p)}{n}} = \sqrt{\dfrac{0.25 \times 0.75}{200}} = \sqrt{0.0009375} \approx 0.031\), this is the correct standard error. The distractor '\(0.250\)' is wrong because it simply restates \(p\) rather than computing the actual formula involving \(n\), confusing the parameter with its variability. Students should carefully apply the square root formula rather than guessing based on the value of \(p\) alone.

Since \(400\) is a tiny fraction of \(2{,}000{,}000\) (well under 10%), the 10% condition is satisfied and observations can be treated as approximately independent. The distractor 'No, because the population is too large to ever satisfy it' is wrong because a very large population actually makes the 10% condition easier to satisfy, not harder, since the sample represents an even smaller fraction of it. Students should recognize that large populations are typically friendly to independence assumptions in real surveys.

A population with strong skew and outliers requires a larger sample size before the sampling distribution of \(\bar{x}\) becomes approximately Normal, because more observations are needed to average out the extreme asymmetry. The distractor 'Sampling from a population that is already Normal' is wrong because in that case \(\bar{x}\) is exactly Normal for any \(n\), requiring no minimum sample size at all. Students should adjust their expected minimum \(n\) based on how far the population shape deviates from Normal.

The standard error is \(\dfrac{16}{\sqrt{64}} = 2\), so \(z = \dfrac{104-100}{2} = 2\), and \(P(Z > 2) \approx 0.023\) using the standard Normal table. The distractor 'Approximately \(0.159\)' corresponds to \(z = 1\), which would result from incorrectly using \(\sigma = 16\) directly as the standard error instead of dividing by \(\sqrt{n}\). Students must always compute the correct standard error before converting to a z-score for probability calculations involving \(\bar{x}\).

The standard error is \(\sqrt{\dfrac{0.65 \times 0.35}{150}} \approx 0.039\), giving \(z = \dfrac{0.70 - 0.65}{0.039} \approx 1.28\), and \(P(Z > 1.28) \approx 0.100\)–\(0.106\), matching the correct choice. The distractor 'Approximately \(0.894\)' is wrong because it represents \(P(Z < 1.28)\) rather than the upper tail probability the question asks for. Students should always double-check whether the problem asks for the area above or below a z-score before selecting a final answer.

The sampling distribution of \(\bar{x}\) is built from many computed sample means, while the population distribution is built from individual measurements, and these two distributions serve fundamentally different purposes. The distractor 'The sampling distribution always has more spread than the population distribution' is wrong because averaging reduces variability, so the sampling distribution of \(\bar{x}\) has a smaller standard deviation than the population, specifically \(\sigma/\sqrt{n}\). Recognizing this distinction is essential for correctly interpreting variability in inference problems.

Because \(SD(\hat{p}) = \sqrt{\dfrac{p(1-p)}{n}}\) depends on \(\dfrac{1}{\sqrt{n}}\), quadrupling \(n\) from 25 to 100 multiplies the standard deviation by \(\dfrac{1}{\sqrt{4}} = \dfrac{1}{2}\). The distractor 'It is multiplied by \(\dfrac{1}{4}\)' incorrectly assumes the standard deviation scales directly with \(\dfrac{1}{n}\) rather than \(\dfrac{1}{\sqrt{n}}\). Students should remember this square-root relationship whenever comparing standard errors across different sample sizes.

A single observation has standard deviation \(\sigma\), while the sample mean has standard deviation \(\sigma/\sqrt{n}\), which is smaller whenever \(n>1\), showing that averaging reduces variability. The distractor 'A sample mean is always more variable than a single observation' reverses this relationship and is incorrect because averaging cancels out some of the random fluctuation present in individual values. This comparison highlights why larger samples give more precise and reliable estimates of the population mean.

With \(p = 0.05\), we need \(np \geq 10\), so \(n \geq \dfrac{10}{0.05} = 200\), and \(n(1-p)\) will automatically be satisfied since it grows much faster; only \(n=250\) among the choices meets \(np \geq 10\). The distractor \(n = 100\) is wrong because \(np = 100(0.05) = 5\), which falls short of the required 10 expected successes. Students should remember that extreme values of \(p\) close to 0 or 1 require larger sample sizes to meet the Large Counts Condition.

The standard error of \(\bar{x}\) quantifies how much sample means typically vary from the true population mean \(\mu\) across repeated samples of the same size. The distractor 'It measures the typical distance between a single data value and the population mean' is wrong because that description applies to \(\sigma\), the population standard deviation, not the standard error of the sample mean. Correctly distinguishing between spread of individual values and spread of sample means is central to interpreting confidence intervals and hypothesis tests later in the course.

Since \(SD(\hat{p})\) is proportional to \(1/\sqrt{n}\), and \(n_2\) is 4 times \(n_1\), the standard deviation for the larger sample is \(1/\sqrt{4} = 1/2\) of the smaller sample's value. The distractor 'It is a quarter as large' incorrectly assumes the standard deviation scales with \(1/n\) instead of \(1/\sqrt{n}\). Students should always apply the square-root scaling rule when comparing standard errors across different sample sizes.

The Central Limit Theorem is powerful because it justifies using Normal probability calculations and confidence procedures for \(\bar{x}\) regardless of the population's original shape, as long as the sample size is sufficiently large. The distractor 'It guarantees the sample mean equals the population mean exactly' is wrong because the CLT describes the shape and spread of the sampling distribution, not a guarantee of a single sample's exact accuracy. This theorem is the theoretical backbone that allows z and t procedures to be applied broadly in practice.

With \(p = 0.5\) and \(n = 80\), \(np = 40\) and \(n(1-p) = 40\), both comfortably above 10, so the Large Counts Condition is satisfied and the sampling distribution of \(\hat{p}\) is approximately Normal. The distractor 'Uniform, since \(p = 0.5\)' is wrong because \(p = 0.5\) actually produces the most symmetric binomial shape, which closely resembles Normal rather than uniform. Students should recall that \(p\) near 0.5 generally requires smaller sample sizes to achieve Normality than extreme values of \(p\).

With \(n = 100 \geq 30\), the CLT applies, giving an approximately Normal sampling distribution centered at \(\mu = 40\) with standard error \(\dfrac{15}{\sqrt{100}} = 1.5\). The distractor 'It is approximately Normal with mean \(40\) and standard deviation \(15\)' incorrectly uses the population standard deviation instead of dividing by \(\sqrt{n}\) to compute the standard error. Students should always remember both parts of the CLT: the mean stays the same, but the spread shrinks by a factor of \(\sqrt{n}\).

Because the Large Counts Condition requires \(np \geq 10\) and \(n(1-p) \geq 10\), a small \(p\) like \(0.02\) requires a much larger \(n\) (at least 500) to reach 10 expected successes, compared to \(p = 0.5\) which reaches 10 successes with just \(n = 20\). The distractor 'Estimating \(p = 0.5\) requires a larger \(n\)' reverses this relationship, since values of \(p\) near 0.5 need the smallest sample sizes to satisfy Large Counts. Students should recognize that extreme proportions demand substantially larger samples for valid Normal approximations.

Since \(\dfrac{\sigma}{\sqrt{n}} = 3\) and \(\sqrt{16}=4\), solving gives \(\sigma = 3 \times 4 = 12\). The distractor '\(3\)' incorrectly assumes the population standard deviation equals the standard error, ignoring the division by \(\sqrt{n}\) that occurs when averaging. Students should practice solving the standard error formula for \(\sigma\) as well as for the standard error itself, since AP questions often work in either direction.

The correct formula for the standard error of a sample proportion is \(\sqrt{\dfrac{p(1-p)}{n}}\), which here equals \(\sqrt{\dfrac{0.55 \times 0.45}{120}} \approx 0.045\). The distractor \(\sqrt{0.55 \times 0.45} \approx 0.497\) is wrong because it omits dividing by \(n\) before taking the square root, producing a value far too large to represent sample-to-sample variability. Students must always divide \(p(1-p)\) by \(n\) before applying the square root when computing \(SD(\hat{p})\).

Increasing sample size reduces the standard error, which shrinks the typical spread of the sampling distribution, but any individual sample statistic remains a random outcome and is not guaranteed to equal the parameter exactly. The distractor 'Larger samples always produce biased statistics' is wrong because increasing \(n\) affects variability, not bias, and bias is instead determined by the sampling method and estimator used. A key big-picture idea for the exam is that larger samples make estimates more precise on average, not perfectly accurate in any single instance.

With strong skew, a small sample size like \(n = 10\) is generally insufficient for the Central Limit Theorem to smooth out the skewness, making the Normal approximation questionable. The distractor \(n = 250\) is wrong to select as risky, because such a large sample size gives ample opportunity for averaging effects to counteract the skew and produce a shape close to Normal. Students should treat the CLT's \(n \geq 30\) guideline as a loose rule that becomes even more conservative when the underlying population shows strong skew or outliers.

Regardless of sample size, the sampling distribution of \(\hat{p}\) is always centered at the true population proportion \(p\), but its spread, given by \(\sqrt{p(1-p)/n}\), decreases as \(n\) increases, so Sample B's distribution is narrower. The distractor 'Sample A's sampling distribution has less spread than Sample B's' reverses this relationship since the smaller sample actually produces greater variability. This principle, that center stays fixed while spread shrinks with larger \(n\), is one of the most tested ideas in Unit 5.

Setting \(\sqrt{\dfrac{0.5 \times 0.5}{n}} \leq 0.02\) and solving gives \(n \geq \dfrac{0.25}{0.0004} = 625\), so \(n = 625\) is the smallest choice meeting the requirement. The distractor \(n = 250\) is wrong because it yields a standard error of \(\sqrt{0.25/250} \approx 0.0316\), which exceeds the target of \(0.02\). Students should practice solving standard error inequalities for \(n\), since this type of algebraic manipulation frequently appears on the AP exam.

The Normality of \(\hat{p}\) specifically requires both \(np\) and \(n(1-p)\) to be at least 10, and when \(p\) is extreme, one of these products stays small even for moderately large \(n\), delaying convergence to Normal shape, whereas the CLT for \(\bar{x}\) depends more generally on sample size and moderate skew. The distractor 'Because \(\bar{x}\) always ignores skewness in the population' is wrong because the CLT for means does account for skewness by requiring larger \(n\) when skew is strong, it does not ignore it. Recognizing this distinction between conditions for proportions versus means helps students correctly judge Normality assumptions on multi-part free-response questions.

Checking \(np = 45 \times 0.03 = 1.35\), which is far below the required threshold of 10, shows the Large Counts Condition fails and the Normal approximation should not be trusted here. The distractor 'The Normal approximation is appropriate because \(n \geq 30\)' is wrong because the \(n \geq 30\) heuristic applies to means under the CLT, not to proportions, which instead require checking \(np\) and \(n(1-p)\) directly. This scenario illustrates why sample size alone is insufficient justification when working with proportions, especially for rare events.

The standard error is \(\dfrac{21}{\sqrt{49}} = 3\), so \(197\) and \(203\) correspond to \(z = -1\) and \(z = 1\), and the area within one standard deviation of the mean under a Normal curve is approximately \(0.683\). The distractor 'Approximately \(0.955\)' corresponds to two standard deviations rather than one, which would be the case if the bounds were \(194\) and \(206\). Students should always convert given values to z-scores using the correct standard error before applying the empirical rule or normalcdf calculations.

Because \(SD(\hat{p})\) is proportional to \(1/\sqrt{n}\), quadrupling \(n\) cuts the standard error in half, which is a larger reduction than merely doubling \(n\) (which reduces it by a factor of \(1/\sqrt{2} \approx 0.71\)). The distractor 'Changing \(p\) from \(0.5\) to \(0.4\)' is wrong because moving \(p\) away from \(0.5\) actually decreases \(p(1-p)\) only slightly and produces a much smaller reduction in standard error than a substantial increase in \(n\). Students should recognize sample size as the primary lever for controlling standard error in most practical scenarios.

The standard error is \(\sqrt{\dfrac{0.6 \times 0.4}{250}} \approx 0.031\), giving \(z = \dfrac{0.55-0.6}{0.031} \approx -1.61\), and \(P(Z < -1.61) \approx 0.054\)–\(0.056\), matching the correct answer. The distractor 'Approximately \(0.944\)' is wrong because it represents \(P(Z > -1.61)\), the complement of the requested lower-tail probability. Students should carefully identify whether a probability question asks for the tail below or above a computed z-score before finalizing their answer.

Since \(SD(\hat{p}) = \sqrt{p(1-p)/n}\), and the function \(p(1-p)\) is a downward parabola maximized at \(p = 0.5\), the standard error of \(\hat{p}\) is largest when \(p\) is at this midpoint value. The distractor 'Extreme values of \(p\) always produce more variability than moderate values' is wrong because it directly contradicts the mathematical behavior of the \(p(1-p)\) term, which actually shrinks as \(p\) moves toward 0 or 1. Understanding this parabolic relationship helps explain why polls near a 50-50 split tend to require larger samples for precise estimates.

Both samples share the same theoretical standard error, \(\dfrac{24}{\sqrt{64}} = 3\), because it depends only on \(\sigma\) and \(n\), but the two actual sample means will typically differ somewhat due to random sampling variability, even from the identical population. The distractor 'The two sample means must be identical since they come from the same population' is wrong because sampling variability guarantees that different random samples generally yield different statistic values, even under identical conditions. This distinction between a fixed theoretical standard error and the observed variation between actual samples is a subtle but important exam concept.

With \(SD(\hat{p}) = \sqrt{\dfrac{0.3 \times 0.7}{500}} \approx 0.0205\), an approximate 95% range under the empirical rule is \(p \pm 2 \times SD(\hat{p}) \approx 0.30 \pm 0.041\), giving roughly \((0.259, 0.341)\). The distractor \((0.15, 0.45)\) corresponds to using three standard deviations, which represents approximately a 99.7% range rather than 95%. Students should connect the empirical rule multipliers (1, 2, 3 standard deviations for 68%, 95%, 99.7%) directly to sampling distribution problems involving proportions and means.

For a population with only moderate skew, \(n = 40\) comfortably exceeds the common \(n \geq 30\) threshold, giving reasonable confidence that the sampling distribution of \(\bar{x}\) will be approximately Normal while still being a realistic sample size to collect. The distractor \(n = 15\) is wrong because moderate skew typically requires a sample size at or above 30 to sufficiently average out asymmetry, so \(n = 15\) would likely be too small. Students should treat \(n \geq 30\) as a flexible guideline that should be increased when skew or outliers are more pronounced.

The Central Limit Theorem addresses the shape of the sampling distribution of \(\bar{x}\), showing it approaches Normal as \(n\) grows, while the Law of Large Numbers separately addresses how the sample mean gets closer to the true population mean \(\mu\) as more data is collected. The distractor 'The CLT guarantees \(\bar{x}\) equals \(\mu\) for large samples' is wrong because the CLT concerns distributional shape, not exact convergence to a specific value, which is instead the focus of the Law of Large Numbers. Distinguishing these two related but distinct theorems helps prevent common misconceptions on the AP exam.

The standard error is \(\dfrac{12}{\sqrt{36}} = 2\), so \(z = \dfrac{82-78}{2} = 2.0\), indicating this sample mean is two standard errors above the population mean. The distractor '\(0.33\)' is wrong because it results from incorrectly dividing the difference by \(\sigma\) instead of the properly computed standard error, ignoring the effect of sample size. Students should always compute the standard error first, using \(\sigma/\sqrt{n}\), before finding a z-score for a sample mean.

With \(p = 0.05\) and \(n = 60\), \(np = 3\), far below the needed threshold of 10, meaning there are too few expected successes for the binomial-based distribution of \(\hat{p}\) to be symmetric, resulting in a right-skewed shape rather than an approximately Normal one. The distractor 'The formula for \(SD(\hat{p})\) becomes undefined when \(p\) is small' is wrong because the formula \(\sqrt{p(1-p)/n}\) remains perfectly well defined for any \(p\) strictly between 0 and 1, it simply produces a small variability value. Students should always verify both \(np\) and \(n(1-p)\) independently, since a large \(n\) alone cannot compensate for an extreme value of \(p\).