Part CCSIR NET December 2024symmetry-is-about-the-mean-not-about-mn-over-two

Symmetry is about the mean not about mn over two

Let $X_{1}, X_{2}, \dots, X_m$ and $Y_{1}, Y_{2}, \dots, Y_{n}$ be two mutually independent random samples from populations with absolutely continuous distribution functions F_X and F_Y, respectively. For N = m + n, define $T_N = \sum(i=1..N)$ iZ $_{i}$ where $Z_{i} = 1$ if the i-th observation in the combined ordered arrangement of N observations is from F_X; and $Z_{i} = 0$ , otherwise. Then, which of the following statements are true?

A.If F_X(x) = F_Y(x) ∀x, then E(T_N) = m(N + 1)/2.✓ Correct
B.If F_X(x) = F_Y(x) ∀x, then Var(T_N) = mn(N + 1)/24.
C.If F_X(x) = F_Y(x) ∀x, then the distribution of T_N is symmetric about mn/2.
D.The minimum and maximum possible values of T_N are m(m + 1)/2 and N(N + 1)/2 − m(m + 1)/2, respectively.

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 what the inference means.

See pricing

50 are analysed free — try those first.

The trap it tests

What the inference means

p-values, size, power and error types say specific things. This was not one of them.

Drill statements like this

Symmetry is about the mean not about mn over two

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