Cake slicing hypothesis problem

Question

A cake weighing one kilogram is cut into two pieces, and each piece is weighed separately. Denote the measured weights of the two pieces by $X$ and $Y$ . Assume that the errors in obtaining $X$ and $Y$ are independent and normally distributed with mean zero and the same (unknown) variance. Devise a test for the hypothesis that the true weights of the two pieces are equal.

Now, if I consider the quantity $X+Y-1$ it gives a measure of the error in measurement. Can I write it as $e_1+e_2$ where $e_1,e_2$ are the errors committed during measurement of the first slice and the second slice respectively? Then as $e_i \sim N(0,\sigma^2)$, so, $e_1+e_2 \sim N(0,2\sigma^2)$, so I get $X+Y \sim N(1,2 \sigma^2)$ Our null hypothesis is $H_0: \mu-\nu=0$ against $H_1: \mu-\nu\neq 0$ Now, I just cannot construct the test from here . Can anyone help?

Answer 1

Let $\mu$ and $\nu$ be the true weights of the two slices. Your hypotheses are $H_0 : \mu = \nu$ and $H_1 : \mu \ne \nu$.

To design a test, come up with a statistic (involving $X$ and $Y$) whose distribution is known under the null.

Answer 2

I would suggest writing the model as $X=\mu_1+\varepsilon_1$ and $Y=\mu_2+\varepsilon_2$ where $\varepsilon_1,\varepsilon_2$ are i.i.d $N(0,\sigma^2)$.

This gives $X\sim N(\mu_1,\sigma^2)$, independent of $Y\sim N(\mu_2,\sigma^2)$, where $\sigma^2$ is unknown.

Since $X,Y$ denote observed weights in the sample, $\mu_1,\mu_2$ are the population (true) mean weights.

(In this setup, $\mu_1+\mu_2=1$.) Now there are several options for testing $H_0:\mu_1-\mu_2=0$.