 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/28k/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The given identities imply

\frac{1}{n} \sum_{i = 1}^{n} (\overline{X}_{n - 1, i}^{2} - \overline{X}_{n}^{2}) = \frac{1}{n ( n - 1 ) ^{2}} \sum_{i = 1}^{n} (X_{i} - \overline{X}_{n})^{2} .

s_{n}^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} (X_{i} - \overline{X}_{n})^{2},

we have the exact identity

T_{JACK} = T_{n} - \frac{s _{n}^{2}}{n} .

s_{n}^{2} P 1

, so

n (T_{JACK} - T_{n}) = - \frac{s _{n}^{2}}{n} P 0.

The Slutsky theorem and part (c) now give

n (T_{JACK} - μ^{2}) d N (0, 4 μ^{2}) .

 New to topics? Read the docs here!