Question

Problem 2. Consider the one-way layout ANOVA model, where we assume that Yij = μί-cij,に1, . . . , I and J 1, . . . ,J, where μί's are fixed unknown with zero mean treatment means and eiy's are random errors , al such that Σ-lai -0 and E[Yj-μ + ai,1- Show that there exists unique numbers μ, ai, a. b. Show that the null hypothesis Ho : μ,-...- μι is equivalent to Ho : 01 ,-. . .-a1-0

Answer 1

a.

Let

μί

Now, consider:

$\begin{align*} \sum_{i=1}^{I} \alpha_i &= \sum_{i=1}^{I} (\mu_i - \mu) \\ &= \sum_{i=1}^I \mu _i - \sum_{i=1}^I \mu \\ &= \sum_{i=1}^I \mu _ i - \mu*I \\ &= \sum_{i=1}^I \mu_i - \frac{\sum_{i=1}^I \mu_i}{I} *I \\ &= \sum_{i=1}^I \mu _i - \sum_{i=1}^I \mu_i \\ &= 0 \end{align*}$

Also, consider :

$\begin{align*} E[Y_{ij}] &= E[\mu_i + e_{ij}] \\ &= E[\mu_i] + E[e_{ij}] \\ &= E[\mu_i] \ \ \ \ \ \ (e_{ij}\text{'s are error terms with zero mean)} \\ &= \mu_i \ \ \ \ \ \ \ (\mu_i \text{ is a constant}) \\ &= \mu + \alpha_i \ \ \ \ ; i = 1,2,...,I \end{align*}$

To prove uniqueness, let us consider another set of numbers $\begin{align*} \mu' , \beta_1,...,\beta_I \end{align*}$ which satisfy:
$\begin{align*} \sum_{i=1}^I \beta_i = 0 \ \ \ \ and \ \ \ \ E[Y_{ij}] = \mu' + \beta_i \ \ \ ; i = 1,...,I \end{align*}$

$\begin{align*} E[Y_{ij}] &= \mu' + \beta_i \ \ \ ; i = 1,...,I \\ \Rightarrow \mu + \alpha_i &= \mu' + \beta_i \ \ \ ; i = 1,...,I \\ \Rightarrow \alpha_i - \beta_i &= \mu' - \mu \ \ \ ; i = 1,...,I \end{align*}$

Now,

$\begin{align*} \sum_{i=1}^{I} \alpha_i - \sum_{i=1}^{I} \beta_i &= 0 - 0 \\ \Rightarrow \sum_{i=1}^{I} (\alpha_i - \beta_i) &= 0 \\ \Rightarrow \sum_{i=1}^{I} (\mu' - \mu) &= 0 \\ \Rightarrow \mu' - \mu &= 0 \\ \Rightarrow \mu' &= \mu \end{align*}$

Moreover,

$\begin{align*} \alpha_i - \beta_i &= \mu' - \mu \ \ \ \ \forall i = 1,...,I \\ \Rightarrow \alpha_i - \beta_i &= 0\ \ \ \ \ \ \ \ \ \ \forall i = 1,...,I \\ \Rightarrow \alpha_i &= \beta_i \ \ \ \ \ \ \ \ \ \forall i = 1,...,I \end{align*}$

Thus, $\mu, \alpha_1,...,\alpha_I$ are unique.

b.

The hypothesis

$\begin{align*} \mu_1 = ... = \mu_I &\Leftrightarrow \mu + \alpha_1 = ... =\mu + \alpha_I \\ & \Leftrightarrow \alpha_1 = ... = \alpha_I \end{align*}$

Now since, $\begin{align*} \alpha_1 = ... = \alpha_I \end{align*}$

and $\begin{align*} \sum_{i=1}^I \alpha_i &= 0 \end{align*}$

Thus, we get:

$\begin{align*} \sum_{i=1}^I \alpha_1 &= 0 \\ \Rightarrow \alpha_1*I &=0 \\ \Rightarrow \alpha_1 &= 0 \\ \Rightarrow \alpha_1 &= \alpha_2 = ... = \alpha_I = 0 \end{align*}$

Thus:
$\begin{align*} \mu_1 = ... = \mu_I & \Leftrightarrow \alpha_1 = ... = \alpha_I = 0 \end{align*}$

Hence Shown

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