Part CCSIR NET June 2025the-intercept-is-a-parameter-so-five-of-them-leave-n-minus-5-not-n-minus-4

The intercept is a parameter so five of them leave n minus 5 not n minus 4

An analyst fits a multiple linear regression model $Y_{i} = β_{0} + β_{1} X_{1 i} + \dots + β_{4} X_{4 i} + ε_{i}, i = 1, 2, \dots, n$ , using the method of least squares. However, his coordinator insists that the intercept and two regressors $Z_{1} = X_{1} + X_{3}$ and $Z_{2} = X_{2} - X_{4}$ are enough to represent the model. Suppose the coordinator's claim can be tested in the form of a general linear hypothesis, viz., $H_{0} : L^{T} β = 0$ against $H_{A} : H_{0}$ is not true, where $β^{T} = (β_{0}, β_{1}, \dots, β_{4})$ . Suppose we have n observations on the response Y and each regressor. Further, assume that the errors with or without restrictions are independent $N (0, σ^{2})$ variables. Then, which of the following statements are true?

A.A possible choice of L is the 5 × 2 matrix with rows (0, 0), (1, 0), (0, 1), (−1, 0), (0, 1)✓
B.The test statistic for testing $H_{0}$ against H_A follows an F-distribution with (2, n − 4) degrees of freedom under $H_{0}$
C.Sum of squares residuals under the restrictions $L^{T} β = 0$ follows $σ^{2} χ_{n - 3}^{2}$ distribution✓
D.Sum of squares residuals without the restrictions $L^{T} β = 0$ follows a non-central $σ^{2} χ_{n - 4}^{2}$ distribution

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: OLS is the BLUE in every linear model

The intercept is a parameter so five of them leave n minus 5 not n minus 4

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