Difficulty: Intermediate | Prerequisites: Basic probability, normal distribution properties, Chapter 1 notes.
These notes cover the mechanics of confidence interval width, the chi-square distribution as a sum of squared standardised normals, and the sampling distribution of the sample variance S². Together they form the inferential backbone you need before tackling regression in earnest.
Confidence interval width shrinks in proportion to 1/√n, so halving the width requires roughly four times the sample size. When you standardise independent normal observations and square them, the sum follows a chi-square distribution with degrees of freedom equal to the number of observations. The sample variance S² is an unbiased estimator of σ², and its own variance depends on σ⁴ and the sample size through the chi-square relationship.
Confidence interval (CI)
A range of values, computed from sample data, that is expected to contain the true population parameter with a stated probability (e.g. 95%). Think of it as: a net you cast around your estimate, wide enough that you are fairly sure the true value is inside.
Confidence interval length (width)
The distance from the lower bound to the upper bound of a CI. For a normal population with known variance, the length equals 2 × z*(σ/√n). In simple terms, this means: a bigger sample gives a tighter interval.
Chi-square distribution (χ²)
The distribution of the sum of k independent squared standard normal random variables. Its single parameter is the degrees of freedom, k. Think of it as: the natural distribution that shows up whenever you add up squared "z-scores."
Degrees of freedom (df)
The number of independent pieces of information that go into an estimate. For a chi-square formed by summing n squared standardised normals, df = n. When estimating the mean first, df drops to n – 1. In simple terms, this means: how many values are genuinely free to vary once your constraints are in place.
Sample variance (S²)
The unbiased estimator of the population variance σ², defined as S² = Σ(Xᵢ – X̄)² / (n – 1). Think of it as: the average squared distance of each observation from the sample mean, corrected so it does not systematically underestimate.
Standard error (SE)
The standard deviation of a sampling distribution of an estimator. For S², the standard error equals σ²√(2/(n – 1)). In simple terms, this means: how much your estimate of the variance would bounce around if you repeated the study many times.
For a 95% CI for the mean of a normal population with known σ, the interval length is L = 2 × z₀.₀₂₅(σ/√n).
Length is inversely proportional to √n. If you want the interval k times shorter, you need k² times the observations.
Worked example: a sample of n = 100 gave L = 4. To get L = 1 (four times shorter), you need roughly n = 100 × 4² = 1,600 observations.
The z* value cancels when you take the ratio of the two lengths, so you do not need to know σ to compute the required sample size from an existing CI.
If X₁, ..., Xₙ are i.i.d. N(μ, σ²), then each Zᵢ = (Xᵢ – μ)/σ is standard normal.
The sum ΣZᵢ² = Σ((Xᵢ – μ)/σ)² follows a χ²(n) distribution (chi-square with n degrees of freedom).
This holds because the definition of the chi-square distribution is precisely a sum of independent squared standard normals.
When μ is replaced by X̄ (estimated from the data), the degrees of freedom drop by one: Σ((Xᵢ – X̄)/σ)² ~ χ²(n – 1). That lost degree of freedom reflects the constraint that the deviations from X̄ must sum to zero.
Since (n – 1)S²/σ² ~ χ²(n – 1), and the variance of a χ²(k) random variable is 2k, we can derive Var(S²) directly.
Var((n – 1)S²/σ²) = 2(n – 1). Rearranging: Var(S²) = 2σ⁴/(n – 1).
This tells you how much the sample variance bounces around the true variance across repeated samples.
The result depends on σ⁴ (the fourth power of the population standard deviation), which is why estimating variability is inherently harder than estimating the mean.
SE(S²) = √Var(S²) = σ²√(2/(n – 1)).
In practice σ² is unknown, so you plug in S² to get the estimated standard error: Ŝ̂E(S²) = S²√(2/(n – 1)).
As n grows, the standard error shrinks, meaning your estimate of the variance becomes more precise.
Confidence interval length for the mean (known σ):
L = 2 z_{\alpha/2} \frac{\sigma}{\sqrt{n}}Sample size ratio for a target length:
\frac{n_2}{n_1} = \left(\frac{L_1}{L_2}\right)^2Chi-square from standardised normals (known μ):
\sum_{i=1}^{n} \left(\frac{X_i - \mu}{\sigma}\right)^2 \sim \chi^2(n)Chi-square from deviations around X̄:
\sum_{i=1}^{n} \left(\frac{X_i - \bar{X}}{\sigma}\right)^2 \sim \chi^2(n-1)Variance of S²:
\text{Var}(S^2) = \frac{2\sigma^4}{n-1}Standard error of S²:
\text{SE}(S^2) = \sigma^2 \sqrt{\frac{2}{n-1}}Students often think that halving the CI width requires doubling the sample size. It does not. Because width scales with 1/√n, you need four times the sample size to halve the width.
Students often confuse χ²(n) with χ²(n – 1). The distinction hinges on whether you use the true mean μ or the sample mean X̄. Using X̄ costs one degree of freedom.
Students sometimes assume Var(S²) depends on σ². It depends on σ⁴ (sigma to the fourth power), which is why variance estimation is less precise than mean estimation for the same sample size.
Students occasionally forget that the chi-square result requires normality. If the population is not normal, Σ((Xᵢ – μ)/σ)² does not follow a chi-square distribution exactly, though the CLT helps for large n.
⚠️ Sample size calculations appear on nearly every introductory inference exam. Know the ratio method: n₂/n₁ = (L₁/L₂)².
⚠️ Expect a question asking you to identify the distribution of Σ((Xᵢ – μ)/σ)² or Σ((Xᵢ – X̄)/σ)² and state the correct degrees of freedom.
⚠️ Deriving Var(S²) from the chi-square relationship is a classic derivation question. The chain of logic is: (n – 1)S²/σ² ~ χ²(n – 1) → Var of chi-square is 2k → rearrange.
⚠️ The standard error of S² is often tested as a follow-up: take the square root of the variance you just derived.
True or False: To reduce CI width by a factor of 3, you need 3 times as many observations. Answer: False. You need 9 times as many (3²).
Fill in the blank: Σ((Xᵢ – μ)/σ)² for n = 25 follows a ___ distribution. Answer: χ²(25).
True or False: Replacing μ with X̄ in the sum of squared standardised values does not change the degrees of freedom. Answer: False. It reduces df by 1.
Fill in the blank: Var(S²) = ___ / (n – 1). Answer: 2σ⁴.
True or False: The chi-square distribution with k degrees of freedom has mean k and variance k. Answer: False. The mean is k but the variance is 2k.
Q: A random sample of size 100 from a normal population produced a 95% CI of length 4. How many observations are needed for a CI of length 1?
A: Use the ratio n₂/n₁ = (L₁/L₂)² = (4/1)² = 16. So n₂ = 100 × 16 = 1,600.
Q: Let X₁, ..., X₂₅ be i.i.d. N(μ, σ²). What is the distribution of Σ((Xᵢ – μ)/σ)²?
A: Each (Xᵢ – μ)/σ is standard normal, and the terms are independent. The sum of 25 independent squared standard normals is χ²(25).
Q: Now suppose we replace μ with X̄ in the expression above. What distribution does the sum follow, and why does the answer change?
A: It follows χ²(24). Estimating μ from the data imposes the constraint Σ(Xᵢ – X̄) = 0, removing one degree of freedom.
Q: Derive Var(S²) when the underlying population is normal.
A: Start from (n – 1)S²/σ² ~ χ²(n – 1). The variance of a χ²(k) variable is 2k, so Var((n – 1)S²/σ²) = 2(n – 1). Pull the constant out: (n – 1)²/σ⁴ × Var(S²) = 2(n – 1). Solve: Var(S²) = 2σ⁴/(n – 1).
Q: What is the standard error of S²?
A: SE(S²) = √Var(S²) = σ²√(2/(n – 1)). In practice, replace σ² with S² to obtain the estimated standard error.
This connects to simple linear regression because the residual sum of squares in regression, divided by σ², also follows a chi-square distribution (with n – 2 degrees of freedom for simple linear regression). Understanding how degrees of freedom are "spent" here prepares you for the F-test and t-tests on regression coefficients.
The variance-of-S² result reappears when you construct confidence intervals for σ² itself, which uses chi-square quantiles directly.
confidence interval width, sample size determination, power of a confidence interval, chi-square distribution, chi-squared, sum of squared standard normals, degrees of freedom, df, sample variance, S-squared, S², variance of variance estimator, Var(S²), standard error of S², SE(S²), normal population inference, STAT 5302, applied linear regression, Weisberg Chapter 2