Questions: Chi-Square Distribution: Theory and Tests

Question 1

Z₁, Z₂, Z₃, Z₄ are independent standard normal random variables. Which expression follows a chi-square distribution with 4 degrees of freedom?

• A. Z₁ + Z₂ + Z₃ + Z₄
• B. (Z₁ + Z₂ + Z₃ + Z₄)²
• C. |Z₁| + |Z₂| + |Z₃| + |Z₄|
• D. Z₁² + Z₂² + Z₃² + Z₄²

Answer: (D)

χ²(k) is defined as the sum of k *squared* independent standard normals. Option A is a sum of standard normals, which is N(0, 4). Option B is the square of a single sum, not the sum of individual squares. Option C sums absolute values (giving a half-normal-based distribution), not squares. Only option D matches: each Zᵢ² contributes 1 degree of freedom, and four independent squared normals give χ²(4).

Question 2

A goodness-of-fit test assigns survey responses to 6 categories, with n = 300 total responses. What are the degrees of freedom for the chi-square test statistic?

• A. 300 (the sample size)
• B. 6 (the number of categories)
• C. 5 (categories minus one)
• D. 299 (sample size minus one)

Answer: (C)

Degrees of freedom in a goodness-of-fit test equal k − 1, where k is the number of categories. With 6 categories, df = 5. The −1 arises because observed counts must sum to n, imposing one constraint and removing one degree of freedom. The sample size n affects expected counts and power but does not determine the degrees of freedom. Option D (n − 1 = 299) would apply to a t-test on a single mean.

Question 3

A chi-square test statistic can take negative values, so unusually negative values provide evidence against the null hypothesis.

• T. True
• F. False

Answer: False

A chi-square statistic is a sum of squared terms — either squared standard normals (by definition) or terms of the form (O_i − E_i)²/E_i — which are always non-negative. The distribution is bounded below at 0 and is right-skewed. Negative values are impossible. Chi-square tests are always one-tailed, rejecting H₀ only when the statistic is *large*. A value near 0 indicates the observed counts closely match the null model.

Question 4

The expected value of a chi-square distribution with k degrees of freedom equals k.

• T. True
• F. False

Answer: True

Since χ²(k) = Z₁² + Z₂² + ⋯ + Z_k² and each Zᵢ is standard normal, E[Zᵢ²] = Var(Zᵢ) = 1. By linearity of expectation, E[χ²(k)] = k × 1 = k. Each independent squared standard normal contributes exactly 1 to the expected value, and k independent contributions sum to k. The variance is 2k, following from Var(Zᵢ²) = E[Zᵢ⁴] − (E[Zᵢ²])² = 3 − 1 = 2.

Question 5

Explain what the goodness-of-fit test statistic Σ(O_i − E_i)²/E_i measures and why a large value provides evidence against the null hypothesis.

Model answer: The statistic measures the overall discrepancy between observed counts (O_i) and what the null hypothesis predicts (E_i), with each squared difference scaled by 1/E_i to make it unit-free. Scaling matters: a deviation of 5 in a category with expected count 10 is more alarming than the same deviation in a category with expected count 500. Under H₀, observed counts should scatter randomly close to expected counts. A large statistic means the data deviate far more than chance would typically produce. Comparing to the χ²(k−1) distribution quantifies exactly how unlikely this much deviation is under H₀.

The connection to the definition of χ²(k) is direct: when H₀ is true, each term (O_i − E_i)/√E_i is approximately a standard normal, so the sum of their squares is approximately chi-square distributed. This is why the test statistic has the form it does — it is constructed to follow a known distribution under H₀.