Questions: Distribution of the Sample Mean

Question 1

A population has mean μ = 50 and standard deviation σ = 10. A random sample of n = 25 is drawn. What is the standard error of the sample mean?

• A. 10 — the standard error equals the population standard deviation
• B. 2 — SE = σ/√n = 10/5
• C. 4 — SE = σ²/n = 100/25
• D. 0.4 — SE = σ/n = 10/25

Answer: (B)

The standard error is SE(X̄) = σ/√n = 10/√25 = 10/5 = 2. Option A confuses SE with σ itself (the point of SE is that averages are less variable than individual observations). Option C confuses Var(X̄) = σ²/n = 4 with the standard error — you must take the square root to get SE. Option D divides σ by n directly, skipping the square root.

Question 2

A researcher wants to cut the standard error of their sample mean in half. They currently have n = 100 observations. How many total observations do they need?

• A. 150 — adding half again is enough
• B. 200 — doubling the sample size halves the standard error
• C. 400 — SE shrinks as 1/√n, so halving SE requires quadrupling n
• D. 10,000 — SE shrinks as 1/n, so halving SE requires squaring n

Answer: (C)

Because SE = σ/√n, halving SE means √n must double, which requires n to quadruple: from 100 to 400. This is the square root law of diminishing returns. Option B is the most common misconception — doubling n multiplies SE by 1/√2 ≈ 0.71, a reduction of only 29%, not 50%.

Question 3

The sample mean X̄ is an unbiased estimator of the population mean μ, meaning E[X̄] = μ for any sample size n.

• T. True
• F. False

Answer: True

Unbiasedness follows directly from linearity of expectation: E[X̄] = E[(X₁ + ··· + Xₙ)/n] = (E[X₁] + ··· + E[Xₙ])/n = nμ/n = μ. This holds regardless of population shape and for any n ≥ 1. Unbiasedness means the sample mean is 'on target' on average — not that any individual sample will equal μ.

Question 4

If the population is not normally distributed, the sample mean X̄ cannot be used in statistical inference because its distribution is unknown.

• T. True
• F. False

Answer: False

This is false — the Central Limit Theorem guarantees that for any population with finite mean and variance, the standardized sample mean converges to a standard normal distribution as n grows. In practice, the normal approximation is reliable for n ≥ 30 for moderately skewed populations. Inference doesn't require the population to be normal; it only requires n to be large enough for the CLT to apply.

Question 5

Why does the variance of the sample mean decrease as sample size increases, and what does the square root in SE = σ/√n imply about the returns to collecting more data?

Model answer: When you average n independent observations, extreme values in different observations partially cancel each other out. The more observations, the more cancellation, so the average is less variable — specifically, Var(X̄) = σ²/n, shrinking proportionally to 1/n. But standard deviation (the relevant scale measure) is the square root of variance, giving SE = σ/√n. This square root creates diminishing returns: doubling n reduces SE by only a factor of √2 ≈ 1.41, not 2. To halve SE you must quadruple n. Collecting data gets increasingly expensive per unit of precision gained.

The variance formula Var(X̄) = σ²/n follows from the independence of observations and the variance addition rule: Var(X₁ + ··· + Xₙ) = nσ², then dividing by n² gives σ²/n. The square root law is why large observational studies are expensive and why there are practical limits to how precisely we can estimate population parameters from samples.