Sampling Distributions

What you'll learn

This revision guide covers sampling distributions, a fundamental concept in AP Statistics that bridges descriptive and inferential statistics. You'll understand how sample statistics vary from sample to sample, learn to describe sampling distributions of sample means and proportions, and apply the Central Limit Theorem to solve problems involving probability and inference.

Key terms and definitions

Sampling distribution — the probability distribution of a statistic (such as a sample mean or sample proportion) obtained from all possible samples of a fixed size from a population.

Standard error — the standard deviation of a sampling distribution, measuring the typical distance between a sample statistic and the population parameter.

Central Limit Theorem (CLT) — states that for a sufficiently large sample size, the sampling distribution of the sample mean is approximately normal, regardless of the population's distribution.

Sample mean (x̄) — the arithmetic average of observations in a sample, used as an estimate of the population mean μ.

Sample proportion (p̂) — the fraction of individuals in a sample with a particular characteristic, used as an estimate of the population proportion p.

Sampling variability — the natural variation in a statistic from sample to sample due to random chance.

Unbiased estimator — a statistic whose sampling distribution has a mean equal to the parameter being estimated.

Law of Large Numbers — as sample size increases, sample statistics tend to get closer to the population parameter.

Core concepts

Sampling distributions of the sample mean

When taking repeated random samples of size n from a population with mean μ and standard deviation σ, the sampling distribution of x̄ has specific properties:

Mean of the sampling distribution:

• E(x̄) = μ
• The sample mean is an unbiased estimator of the population mean

Standard deviation of the sampling distribution (standard error):

• σₓ̄ = σ/√n
• Standard error decreases as sample size increases
• This formula assumes sampling with replacement or sampling from an infinite population
• For finite populations with sampling without replacement, use the finite population correction: σₓ̄ = (σ/√n)√[(N-n)/(N-1)], where N is the population size

Shape of the sampling distribution:

• If the population is normally distributed, the sampling distribution of x̄ is normal for any sample size
• If the population is not normal, the Central Limit Theorem applies when n ≥ 30 (general guideline)
• For moderately skewed populations, smaller samples may suffice; for heavily skewed populations, larger samples may be needed

The key insight is that even when sampling from non-normal populations, the sampling distribution of x̄ becomes approximately normal as sample size increases.

The Central Limit Theorem in practice

The Central Limit Theorem is arguably the most important theorem in statistics. It allows us to make probability statements about sample means even when we don't know the population distribution.

Conditions for applying the CLT:

• Random sampling from the population
• Independence: individual observations must be independent
  • For sampling without replacement, the 10% condition: n ≤ 0.10N
• Sample size: n ≥ 30 is a common threshold, though this depends on the population shape

Applications:

• Calculate probabilities involving sample means using normal distribution methods
• Standardize using z = (x̄ - μ)/(σ/√n)
• Make inferences about population parameters

Example scenario: A coffee shop in London serves beverages with a mean volume of 250 mL and standard deviation of 8 mL. For samples of 40 beverages, the sampling distribution of x̄ is approximately normal with mean 250 mL and standard error 8/√40 ≈ 1.26 mL, even if individual beverage volumes aren't normally distributed.

Sampling distributions of the sample proportion

When sampling from a population where the proportion with a characteristic is p, the sampling distribution of p̂ has the following properties:

Mean of the sampling distribution:

• E(p̂) = p
• The sample proportion is an unbiased estimator of the population proportion

Standard deviation of the sampling distribution (standard error):

• σp̂ = √[p(1-p)/n]
• Standard error decreases as n increases
• Standard error is largest when p = 0.5

Shape of the sampling distribution:

• Approximately normal when both np ≥ 10 and n(1-p) ≥ 10
• These conditions ensure the sampling distribution is sufficiently symmetric and not too discrete

Key differences from the sample mean:

• The standard error formula includes the parameter p itself
• The normality condition depends on both n and p
• The sampling distribution is discrete but approximated by a continuous normal distribution

Combining random variables

Understanding how to combine random variables is essential for more complex sampling distribution problems.

For independent random variables X and Y:

• E(X ± Y) = E(X) ± E(Y)
• Var(X ± Y) = Var(X) + Var(Y) (note: always addition, even for differences)
• SD(X ± Y) = √[Var(X) + Var(Y)]

For a constant a and random variable X:

• E(aX + b) = aE(X) + b
• Var(aX + b) = a²Var(X)
• SD(aX + b) = |a|SD(X)

Application to sampling distributions: When comparing two independent samples, such as x̄₁ - x̄₂:

• Mean: E(x̄₁ - x̄₂) = μ₁ - μ₂
• Standard error: σₓ̄₁₋ₓ̄₂ = √(σ₁²/n₁ + σ₂²/n₂)
• Shape: approximately normal if both individual sampling distributions are approximately normal

Bias and variability in estimators

An effective estimator should be both unbiased and have low variability.

Bias:

• Bias = E(statistic) - parameter
• An unbiased estimator has bias = 0
• Both x̄ and p̂ are unbiased estimators
• Sample standard deviation s is approximately unbiased for σ (exactly unbiased when using n-1 in the denominator)

Variability:

• Lower variability means more precise estimates
• Increasing sample size always reduces variability
• Trade-off: larger samples cost more time and resources

Comparing estimators:

• If two estimators are both unbiased, prefer the one with smaller standard error
• An unbiased estimator with high variability may perform worse than a slightly biased estimator with very low variability in practice
• The sampling distribution provides a complete picture of an estimator's behavior

The 10% condition and independence

Independence of observations is crucial for sampling distribution formulas to be valid.

The 10% condition:

• When sampling without replacement, observations are technically dependent
• If n ≤ 0.10N, the dependence is negligible
• Standard error formulas can be used without the finite population correction
• Example: sampling 50 students from a school of 800 satisfies the condition (50 < 80)

Why independence matters:

• Standard error formulas assume independence
• Violation leads to incorrect probability calculations
• Common violations: cluster sampling, systematic sampling, or convenience sampling
• Proper random sampling ensures independence

Worked examples

Example 1: Sampling distribution of the sample mean

Question: The masses of mangoes exported from Jamaica are normally distributed with mean 350 g and standard deviation 45 g. A random sample of 9 mangoes is selected.

(a) Describe the sampling distribution of the sample mean mass. [3 marks]

(b) Find the probability that the sample mean mass exceeds 370 g. [3 marks]

Solution:

(a) Since the population is normally distributed:

• Shape: The sampling distribution of x̄ is exactly normal [1 mark]
• Mean: E(x̄) = μ = 350 g [1 mark]
• Standard error: σₓ̄ = σ/√n = 45/√9 = 15 g [1 mark]

(b) We need P(x̄ > 370):

• Standardize: z = (370 - 350)/15 = 20/15 = 1.33 [1 mark]
• P(Z > 1.33) = 1 - 0.9082 = 0.0918 [1 mark]
• The probability is 0.092 (or 9.2%) [1 mark]

Example 2: Sampling distribution of the sample proportion

Question: In a large Caribbean university, 35% of students commute to campus by public transport. A random sample of 120 students is selected.

(a) Verify that the sampling distribution of p̂ is approximately normal. [2 marks]

(b) Calculate the probability that between 30% and 40% of the sample use public transport. [4 marks]

Solution:

(a) Check the conditions:

• np = 120(0.35) = 42 ≥ 10 ✓ [1 mark]
• n(1-p) = 120(0.65) = 78 ≥ 10 ✓ [1 mark]
• Both conditions satisfied, so sampling distribution is approximately normal

(b) First, find the standard error:

• σp̂ = √[p(1-p)/n] = √[0.35(0.65)/120] = √0.0019 = 0.0435 [1 mark]

For p̂ = 0.30: z = (0.30 - 0.35)/0.0435 = -1.15 [1 mark] For p̂ = 0.40: z = (0.40 - 0.35)/0.0435 = 1.15 [1 mark]

P(-1.15 < Z < 1.15) = 0.8749 - 0.1251 = 0.7498 ≈ 0.75 [1 mark]

Example 3: Comparing two independent samples

Question: The times taken to complete a mathematics exam in two schools are independent and normally distributed. School A has mean 95 minutes and standard deviation 12 minutes. School B has mean 88 minutes and standard deviation 15 minutes. Random samples of 30 students from School A and 40 students from School B are taken.

Find the probability that the mean time for School A exceeds the mean time for School B by more than 10 minutes. [5 marks]

Solution:

Let x̄ₐ and x̄ᵦ represent the sample means.

• E(x̄ₐ - x̄ᵦ) = 95 - 88 = 7 minutes [1 mark]
• σₓ̄ₐ₋ₓ̄ᵦ = √(12²/30 + 15²/40) = √(4.8 + 5.625) = √10.425 = 3.23 minutes [2 marks]
• Since both populations are normal, x̄ₐ - x̄ᵦ is exactly normal [1 mark]

We need P(x̄ₐ - x̄ᵦ > 10):

• z = (10 - 7)/3.23 = 0.93
• P(Z > 0.93) = 1 - 0.8238 = 0.1762 ≈ 0.18 [1 mark]

Common mistakes and how to avoid them

• Confusing population standard deviation σ with standard error σ/√n — Always divide by √n when working with sampling distributions of means. The standard error measures variability of the statistic, not individual observations.

• Using the Central Limit Theorem when n < 30 without justification — If n < 30, you must state that the population is approximately normal to use normal calculations. Don't automatically apply the CLT for small samples.

• Forgetting to check the 10% condition — When sampling without replacement, verify n ≤ 0.10N before using standard formulas. This is especially important in finite population contexts.

• Not verifying conditions for normal approximation of p̂ — Always check both np ≥ 10 and n(1-p) ≥ 10 before treating the sampling distribution of a proportion as normal. State these checks explicitly in exam answers.

• Adding standard deviations instead of variances — When combining independent random variables, remember SD(X ± Y) = √[Var(X) + Var(Y)], not SD(X) + SD(Y).

• Misidentifying what is random — The sample statistic (x̄ or p̂) is the random variable with a sampling distribution, not the individual observations or the parameter itself.

Exam technique for "Sampling Distributions"

• Command word "describe" — When asked to describe a sampling distribution, provide three components: shape (normal, approximately normal, or exact distribution), center (mean/expected value), and spread (standard error). This typically earns 3 marks.

• Show your working for probability calculations — Always write the standardization formula z = (value - mean)/standard error, then the probability statement P(Z < z), even if using a calculator. This allows partial credit if you make an arithmetic error.

• Verify conditions explicitly — When using the CLT or normal approximation for proportions, write out the conditions and verify them. Examiners award marks for checking assumptions, not just performing calculations.

• Use context in conclusions — Don't just give a numerical answer. Relate probabilities back to the context: "There is a 9.2% probability that the mean mass of 9 randomly selected mangoes exceeds 370 g." This demonstrates statistical thinking and may earn communication marks.

Quick revision summary

Sampling distributions describe how statistics vary across repeated samples. The sample mean x̄ has mean μ and standard error σ/√n. The sample proportion p̂ has mean p and standard error √[p(1-p)/n]. The Central Limit Theorem states that sampling distributions become approximately normal for large n (typically n ≥ 30), enabling probability calculations. Always verify conditions: randomness, independence (10% condition), and sample size requirements. Standard error decreases with larger samples, following the √n pattern.