Chapter 7.2, Problem 39E

a.

To determine

Find the distribution of $\widehat{\theta }=c_1{\overline{X}}_1+c_2{\overline{X}}_2+\mathrm{.......}+c_k{\overline{X}}_k$.

Give the reasons.

Answer to Problem 39E

The distribution of $\widehat{\theta }=c_1{\overline{X}}_1+c_2{\overline{X}}_2+\mathrm{.......}+c_k{\overline{X}}_k$ follows normal distribution with mean ${\mu }_1+{\mu }_2+\mathrm{......}+{\mu }_k$ and variance $\frac{c_1^2{\sigma }^2}{n_1}+\frac{c_2^2{\sigma }^2}{n_2}+\mathrm{....}+\frac{c_k^2{\sigma }^2}{n_k}$.

The reasons are if $Y_1,Y_2,\mathrm{......},Y_n$ has a normal distribution with mean $\mu$ and variance ${\sigma }^2$, then $\overline{Y}$ follows normal distribution with mean $\mu$, variance $\frac{{\sigma }^2}{n}$ and sum of normal variables are also normal.

Explanation of Solution

From the given information, $X_1,X_2,\mathrm{............},X_k$ follows normal distribution with mean ${\mu }_i$ and variance ${\sigma }^2$.

Let $\widehat{\theta }=c_1{\overline{X}}_1+c_2{\overline{X}}_2+\mathrm{.......}+c_k{\overline{X}}_k$.

From theorem 7.1, if $Y_1,Y_2,\mathrm{......},Y_n$ has a normal distribution with mean $\mu$ and variance ${\sigma }^2$, then $\overline{Y}$ follows normal distribution with mean $\mu$ and variance $\frac{{\sigma }^2}{n}$.

By using theorem 7.1, as ${\overline{X}}_1\sim N\left({\mu }_1,\frac{{\sigma }^2}{n_1}\right)$, ${\overline{X}}_2\sim N\left({\mu }_2,\frac{{\sigma }^2}{n_2}\right)$,…….., ${\overline{X}}_k\sim N\left({\mu }_k,\frac{{\sigma }^2}{n_2}\right)$, thus for constants $c_1,c_2,\mathrm{...},c_k$ it can be said that $c_1{\overline{X}}_1\sim N\left(c_1{\mu }_1,\frac{c_1^2{\sigma }^2}{n_1}\right)$, $c_2{\overline{X}}_2\sim N\left(c_2{\mu }_2,\frac{c_2^2{\sigma }^2}{n_2}\right)$,…………….., $c_k{\overline{X}}_k\sim N\left(c_k{\mu }_k,\frac{c_k^2{\sigma }^2}{n_k}\right)$.

Then,

$\begin{array}{c}\widehat{\theta }=c_1{\overline{X}}_1+c_2{\overline{X}}_2+\mathrm{.......}+c_k{\overline{X}}_k\\ \sim N\left({\mu }_1+{\mu }_2+\mathrm{......}+{\mu }_k,\frac{c_1^2{\sigma }^2}{n_1}+\frac{c_2^2{\sigma }^2}{n_2}+\mathrm{....}+\frac{c_k^2{\sigma }^2}{n_k}\right)\end{array}$

Thus, the distribution of $\widehat{\theta }=c_1{\overline{X}}_1+c_2{\overline{X}}_2+\mathrm{.......}+c_k{\overline{X}}_k$ follows normal distribution with mean ${\mu }_1+{\mu }_2+\mathrm{......}+{\mu }_k$ and variance $\frac{c_1^2{\sigma }^2}{n_1}+\frac{c_2^2{\sigma }^2}{n_2}+\mathrm{....}+\frac{c_k^2{\sigma }^2}{n_k}$.

The reasons are if $Y_1,Y_2,\mathrm{......},Y_n$ has a normal distribution with mean $\mu$ and variance ${\sigma }^2$, then $\overline{Y}$ follows normal distribution with mean $\mu$, variance $\frac{{\sigma }^2}{n}$ and sum of normal variables are also normal.

b.

To determine

Find the distribution of $\frac{\text{SSE}}{{\sigma }^2}$.

Give the reasons.

Answer to Problem 39E

The distribution of $\frac{\text{SSE}}{{\sigma }^2}$ is chi-square with parameter $\left(n_1+n_2+\mathrm{.......}+n_k\right)-k$ df.

The reasons are if $Y_1,Y_2,\mathrm{......},Y_n$ has a normal distribution with mean $\mu$ and variance ${\sigma }^2$, then $\frac{\left(n-1\right)S^2}{{\sigma }^2}=\frac{1}{{\sigma }^2}{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}$ follows chi-square distribution with $\left(n-1\right)$ df and sum of chi-square distributions is also chi-square.

Explanation of Solution

From the given information, $\text{SSE}={\displaystyle \sum _{i=1}^k\left(n_i-1\right)S_i^2}$.

From the theorem 7.3, if $Y_1,Y_2,\mathrm{......},Y_n$ has a normal distribution with mean $\mu$ and variance ${\sigma }^2$, then $\frac{\left(n-1\right)S^2}{{\sigma }^2}=\frac{1}{{\sigma }^2}{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}$ follows chi-square distribution with $\left(n-1\right)$ df.

Then, by using theorem 7.3, $\frac{\left(n_1-1\right)S_1^2}{{\sigma }^2}$ follows chi-square distribution with $\left(n_1-1\right)$ df, $\frac{\left(n_2-1\right)S_2^2}{{\sigma }^2}$ follows chi-square distribution with $\left(n_2-1\right)$ df,…….and $\frac{\left(n_k-1\right)S_k^2}{{\sigma }^2}$ follows chi-square distribution with $\left(n_k-1\right)$ df.

Then, $\frac{\text{SSE}}{{\sigma }^2}=\frac{{\displaystyle \sum _{i=1}^k\left(n_i-1\right)S_i^2}}{{\sigma }^2}$ follows chi-square distribution with $\left(n_1-1\right)+\left(n_2-1\right)+\mathrm{.......}+\left(n_k-1\right)=\left(n_1+n_2+\mathrm{.......}+n_k\right)-k$ df.

Thus, the distribution of $\frac{\text{SSE}}{{\sigma }^2}$ is chi-square with parameter $\left(n_1+n_2+\mathrm{.......}+n_k\right)-k$ df.

The reasons are if $Y_1,Y_2,\mathrm{......},Y_n$ has a normal distribution with mean $\mu$ and variance ${\sigma }^2$, then $\frac{\left(n-1\right)S^2}{{\sigma }^2}=\frac{1}{{\sigma }^2}{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}$ follows chi-square distribution with $\left(n-1\right)$ df and sum of chi-square distributions is also chi-square.

c.

To determine

Find the distribution of $\frac{\widehat{\theta }-\theta }{\sqrt{\left(\frac{c_1^2}{n_1}+\frac{c_2^2}{n_2}+\mathrm{......}+\frac{c_k^2}{n_k}\right)MSE}}$.

Give the reasons.

Answer to Problem 39E

The distribution of $\frac{\widehat{\theta }-\theta }{\sqrt{\left(\frac{c_1^2}{n_1}+\frac{c_2^2}{n_2}+\mathrm{......}+\frac{c_k^2}{n_k}\right)MSE}}$ has a t distribution with $n_1+n_2+\mathrm{.......}+n_k-k$.

The reason is that if $Y_1,Y_2,\mathrm{........},Y_n$ follow normal distribution with mean $\mu$ and variance ${\sigma }^2$, then $T=\frac{\overline{Y}-\mu }{S/\sqrt{n}}$ has a t distribution with $\left(n-1\right)$ df.

Explanation of Solution

From the given information, $\text{MSE}=\frac{\text{SSE}}{n_1+n_2+\mathrm{.......}+n_k-k}$.

From the definition 7.2, if $Y_1,Y_2,\mathrm{........},Y_n$ follow normal distribution with mean $\mu$ and variance ${\sigma }^2$, then $T=\frac{\overline{Y}-\mu }{S/\sqrt{n}}$ has a t distribution with $\left(n-1\right)$ df.

Then, by using the definition 7.2, $\frac{\widehat{\theta }-\theta }{\sqrt{\left(\frac{c_1^2}{n_1}+\frac{c_2^2}{n_2}+\mathrm{......}+\frac{c_k^2}{n_k}\right)MSE}}$ has a t distribution with $n_1+n_2+\mathrm{.......}+n_k-k$.

Thus, the distribution of $\frac{\widehat{\theta }-\theta }{\sqrt{\left(\frac{c_1^2}{n_1}+\frac{c_2^2}{n_2}+\mathrm{......}+\frac{c_k^2}{n_k}\right)MSE}}$ has a t distribution with $n_1+n_2+\mathrm{.......}+n_k-k$.

The reason is that if $Y_1,Y_2,\mathrm{........},Y_n$ follow normal distribution with mean $\mu$ and variance ${\sigma }^2$, then $T=\frac{\overline{Y}-\mu }{S/\sqrt{n}}$ has a t distribution with $\left(n-1\right)$ df.