Chapter 10.1, Problem 10E

In single-factor ANOVA with I treatments and J observations per treatment, let $\mu =\left(1/I\right)\sum {\mu }_i$ .

  a. Express E( $\bar{X}$..) in terms of µ. [Hint: $\bar{X}$.. = (1/I) $\sum {\bar{X}}_i]$

  b. Determine E( $E\left({\bar{X}}_{i^{.}}{}^2\right)$. [Hint: For any rv Y, E(Y²) = V(Y) ₊ [ $\left[E{\left(Y\right)}^2\right]$ ]

  c. Determine E( ${\bar{X}}^2$..).

  d. Determine E(SSTr) and then show that

     $E(MSTr)={\sigma }^2+\frac{J}{I-1}\sum {\left({\mu }_i-\mu \right)}^2$

  e. Using the result of part (d), what is E(MSTr) when H₀ is true? When H₀ is false, how does E(MSTr) compare to σ ²?

To determine

Find the expression for $E\left({\overline{X}}_{\mathrm{..}}\right)$ in terms of $\mu$.

Answer to Problem 10E

The expression for $E\left({\overline{X}}_{\mathrm{..}}\right)$ in terms of $\mu$ is $\underset{\_}{E\left({\overline{X}}_{\mathrm{..}}\right)=\mu }$.

Explanation of Solution

Given info:

A single factor ANOVA consists of I treatments and J observations per treatment, with $\mu =\frac{1}{I}{\displaystyle \sum {\mu }_i}$.

Calculation:

Denote the sample mean corresponding to the i^th treatment as ${\overline{X}}_{i\cdot }$. Then the grand mean is:

${\overline{X}}_{\mathrm{..}}=\frac{1}{I}{\displaystyle \sum _{i=1}^I{\overline{X}}_{i.}}$

Taking expectation on both sides,

$\begin{array}{c}E\left({\overline{X}}_{\mathrm{..}}\right)=E\left(\frac{1}{I}{\displaystyle \sum _{i=1}^I{\overline{X}}_{i.}}\right)\\ =\frac{1}{I}E\left({\displaystyle \sum _{i=1}^I{\overline{X}}_{i.}}\right)\left(\text{as }I\text{ is non-random}\right)\\ =\frac{1}{I}{\displaystyle \sum _{i=1}^{I}E\left({\overline{X}}_{i.}\right)}.\end{array}$

Now, ${\overline{X}}_i.$ is the mean of sample observations from i^th treatment, for $i=1,2,\mathrm{...},I$. Denote the expectation of each observation of the i^th treatment is ${\mu }_i$. For a random sample of size n taken from a distribution with mean $\mu$, it is known that the sample mean has the expected value $\mu$.

Thus,

$E\left({\overline{X}}_{i.}\right)={\mu }_i$, for $i=1,2,\mathrm{...},I$.

Hence,

$\begin{array}{c}E\left({\overline{X}}_{\mathrm{..}}\right)=\frac{1}{I}{\displaystyle \sum _{i=1}^I{\mu }_i}\\ =\mu .\end{array}$

Thus, the expression for $E\left({\overline{X}}_{\mathrm{..}}\right)$ in terms of $\mu$ is $\underset{\_}{E\left({\overline{X}}_{\mathrm{..}}\right)=\mu }$.

To determine

Find the value of $E\left({\overline{X}}_{i\text{.}}^2\right)$.

Answer to Problem 10E

The value of $E\left({\overline{X}}_{i\text{.}}^2\right)$ is $\underset{\_}{\frac{{\sigma }^2}{J}+{\mu }_i^2}$.

Explanation of Solution

Calculation:

Let ${\sigma }^2$ be the population variance for each treatment. For a random sample of size n taken from a distribution with variance ${\sigma }^2$, it is known that the sample mean has the variance $\frac{{\sigma }^2}{n}$.

Thus, the sample mean for the i^th treatment,  ${\overline{X}}_{i\cdot }$ is calculated for a total of J observations. Thus, ${\overline{X}}_{i\cdot }$ has a distribution with variance $V\left({\overline{X}}_{i\cdot }\right)=\frac{{\sigma }^2}{J}$.

Now, variance of a certain quantity is $V\left(X\right)=E\left(X^2\right)-{\left(E\left(X\right)\right)}^2$.

Thus,

$E\left(X^2\right)=V\left(X\right)+{\left(E\left(X\right)\right)}^2$.

Thus,

$\begin{array}{c}E\left({\overline{X}}_{i\text{.}}^2\right)=V\left({\overline{X}}_{i\cdot }\right)+{\left(E\left({\overline{X}}_{i\cdot }\right)\right)}^2\\ =\frac{{\sigma }^2}{J}+{\mu }_i^2.\end{array}$

Thus, value of $E\left({\overline{X}}_{i\text{.}}^2\right)$ is $\underset{\_}{\frac{{\sigma }^2}{J}+{\mu }_i^2}$.

To determine

Find the value of $E\left({\overline{X}}_{\mathrm{..}}^2\right)$.

Answer to Problem 10E

The value of $E\left({\overline{X}}_{\mathrm{..}}^2\right)$ is $\underset{\_}{\frac{{\sigma }^2}{IJ}+{\mu }^2}$.

Explanation of Solution

Calculation:

From part a, $E\left({\overline{X}}_{\mathrm{..}}\right)=\mu$.

The grand sample mean,  ${\overline{X}}_{\mathrm{..}}$ is calculated for a total of IJ observations. Thus, ${\overline{X}}_{\mathrm{..}}$ has a distribution with variance $V\left({\overline{X}}_{\mathrm{..}}\right)=\frac{{\sigma }^2}{IJ}$.

Thus,

$\begin{array}{c}E\left({\overline{X}}_{\mathrm{..}}\right)=V\left({\overline{X}}_{\mathrm{..}}\right)+{\left(E\left({\overline{X}}_{\mathrm{..}}\right)\right)}^2\\ =\frac{{\sigma }^2}{IJ}+{\mu }^2.\end{array}$

Thus, value of $E\left({\overline{X}}_{\mathrm{..}}^2\right)$ is $\underset{\_}{\frac{{\sigma }^2}{IJ}+{\mu }^2}$.

To determine

Find the expression for $E\left(\text{SSTr}\right)$.

Show that $E\left(\text{MSTr}\right)={\sigma }^2+\frac{J}{I-1}{\displaystyle \sum {\left({\mu }_i-\mu \right)}^2}$.

Answer to Problem 10E

The expression for $E\left(\text{SSTr}\right)$ is $\underset{\_}{E\left(\text{SSTr}\right)=\left(I-1\right){\sigma }^2+J{\displaystyle \sum _i{\left({\mu }_i-\mu \right)}^2}}$.

Explanation of Solution

Calculation:

For equal sample sizes, the treatment sum of squares, SSTr is:

$\begin{array}{c}\text{SSTr}={\displaystyle \sum _{i}J{\left({\overline{X}}_{i.}-{\overline{X}}_{\mathrm{..}}\right)}^2}\\ ={\displaystyle \sum _{i}J{\left({\overline{X}}_{i.}^2-2{\overline{X}}_{i.}{\overline{X}}_{\mathrm{..}}+{\overline{X}}_{\mathrm{..}}^2\right)}^2}\\ =J{\displaystyle \sum _i{\overline{X}}_{i.}{}^2}-2J{\overline{X}}_{\mathrm{..}}{\displaystyle \sum _i{\overline{X}}_{i.}}+J{\overline{X}}_{\mathrm{..}}^2{\displaystyle \sum _{i}1}\left(J\text{ and }{\overline{X}}_{\mathrm{..}}^2\text{ being independent of }I\right)\end{array}$

Now,

${\overline{X}}_{\mathrm{..}}=\frac{1}{I}{\displaystyle \sum _{i=1}^I{\overline{X}}_{i.}}$.

Thus,

${\displaystyle \sum _{i=1}^I{\overline{X}}_{i.}}=I{\overline{X}}_{\mathrm{..}}$

As a result,

$\begin{array}{c}\text{SSTr}=J{\displaystyle \sum _i{\overline{X}}_{i.}{}^2}-2J{\overline{X}}_{\mathrm{..}}\left(I{\overline{X}}_{\mathrm{..}}\right)+IJ{\overline{X}}_{\mathrm{..}}^2\\ =J{\displaystyle \sum _i{\overline{X}}_{i.}{}^2}-2IJ{\overline{X}}_{\mathrm{..}}^2+IJ{\overline{X}}_{\mathrm{..}}^2\\ =J{\displaystyle \sum _i{\overline{X}}_{i.}{}^2}-IJ{\overline{X}}_{\mathrm{..}}^2\end{array}$

Taking expectation of SSTr,

$\begin{array}{c}E\left(\text{SSTr}\right)=E\left(J{\displaystyle \sum _i{\overline{X}}_{i.}{}^2}-IJ{\overline{X}}_{\mathrm{..}}^2\right)\\ =JE\left({\displaystyle \sum _i{\overline{X}}_{i.}{}^2}\right)-E\left(IJ{\overline{X}}_{\mathrm{..}}^2\right)\left(J\text{ being non-random}\right)\\ =J{\displaystyle \sum _{i}E\left({\overline{X}}_{i.}{}^2\right)}-IJE\left({\overline{X}}_{\mathrm{..}}^2\right)\end{array}$

From part b, $E\left({\overline{X}}_{i.}{}^2\right)=\frac{{\sigma }^2}{J}+{\mu }_i^2$.

From part c, $E\left({\overline{X}}_{\mathrm{..}}^2\right)=\frac{{\sigma }^2}{IJ}+{\mu }^2$.

Substitute these values in the expression for $E\left(\text{SSTr}\right)$:

$\begin{array}{c}E\left(\text{SSTr}\right)=J{\displaystyle \sum _i\left(\frac{{\sigma }^2}{J}+{\mu }_i^2\right)}-IJE\left(\frac{{\sigma }^2}{IJ}+{\mu }^2\right)\\ ={\displaystyle \sum _i{\sigma }^2}+J{\displaystyle \sum _i{\mu }_i{}^2}-{\sigma }^2-IJ{\mu }^2\\ =I{\sigma }^2+J{\displaystyle \sum _i{\mu }_i{}^2}-{\sigma }^2-IJ{\mu }^2\\ =\left(I-1\right){\sigma }^2+J\left({\displaystyle \sum _i{\mu }_i{}^2}-I{\mu }^2\right)\end{array}$

From the expansion of SSTr, it is seen that ${\displaystyle \sum _i{\left({\overline{X}}_{i.}-{\overline{X}}_{\mathrm{..}}\right)}^2}={\displaystyle \sum _i{\overline{X}}_{i.}{}^2}-I{\overline{X}}_{\mathrm{..}}^2$. Thus,

$\begin{array}{c}E\left(\text{SSTr}\right)=\left(I-1\right){\sigma }^2+J\left({\displaystyle \sum _i{\mu }_i{}^2}-I{\mu }^2\right)\\ =\left(I-1\right){\sigma }^2+J{\displaystyle \sum _i{\left({\mu }_i-\mu \right)}^2}\end{array}$.

Thus, the expression for $E\left(\text{SSTr}\right)$ is $\underset{\_}{E\left(\text{SSTr}\right)=\left(I-1\right){\sigma }^2+J{\displaystyle \sum _i{\left({\mu }_i-\mu \right)}^2}}$.

It is known that the mean square treatment (MSTr) is, $\text{MSTr}=\frac{\text{SSTr}}{I-1}$.

Now, divide both sides of $E\left(\text{SSTr}\right)$ by $I-1$:

$\begin{array}{c}\frac{E\left(\text{SSTr}\right)}{I-1}=\frac{1}{I-1}\left[\left(I-1\right){\sigma }^2+J{\displaystyle \sum _i{\left({\mu }_i-\mu \right)}^2}\right]\\ E\left(\frac{\text{SSTr}}{I-1}\right)={\sigma }^2+\frac{J}{I-1}{\displaystyle \sum _i{\left({\mu }_i-\mu \right)}^2}\\ \underset{\_}{E\left(\text{MSTr}\right)={\sigma }^2+\frac{J}{I-1}{\displaystyle \sum _i{\left({\mu }_i-\mu \right)}^2}}\end{array}$

To determine

Find the value of $E\left(\text{MSTr}\right)$ when $H_0$  is true, using the result of part d.

Compare this value to the value of $E\left(\text{MSTr}\right)$ when $H_0$  is false.

Answer to Problem 10E

The value of $E\left(\text{MSTr}\right)$ when $H_0$  is true, using the result of part d is $\underset{\_}{{\sigma }^2}$.

The value of $E\left(\text{MSTr}\right)$ is greater when $H_0$  is false, than when $H_0$  is true.

Explanation of Solution

Calculation:

The null hypothesis $H_0$  is defined as:

$H_0:{\mu }_1={\mu }_2=\mathrm{...}={\mu }_I=0$.

When $H_0$ is true, ${\mu }_i=0$ for each $i=1,2,\mathrm{...},I$ in the expression for $H_0$.

Given that $\mu =\frac{1}{I}{\displaystyle \sum {\mu }_i}$.

Thus, when ${\mu }_i=0$ for each $i=1,2,\mathrm{...},I$, evidently,

$\begin{array}{c}\mu =\frac{1}{I}{\displaystyle \sum {\mu }_i}\\ =\frac{1}{I}{\displaystyle \sum _{i}0}\\ =0.\end{array}$

Replace ${\mu }_i=0$ for each $i=1,2,\mathrm{...},I$ and $\mu =0$ in the expression for $E\left(\text{MSTr}\right)$:

$\begin{array}{c}E\left(\text{MSTr}\right)={\sigma }^2+\frac{J}{I-1}{\displaystyle \sum _i{\left({\mu }_i-\mu \right)}^2}\\ ={\sigma }^2+\frac{J}{I-1}{\displaystyle \sum \left(0\right)}\\ ={\sigma }^2+0\\ =\underset{\_}{{\sigma }^2}.\end{array}$

When $H_0$ is false, ${\mu }_i\ne 0$ for at least one $i=1,2,\mathrm{...},I$ in the expression for $H_0$.

In such a situation, $\left({\mu }_i-\mu \right)\ne 0$ for at least one $i=1,2,\mathrm{...},I$.

As a result, ${\left({\mu }_i-\mu \right)}^2>0$ for at least one $i=1,2,\mathrm{...},I$.

Hence,

$\begin{array}{c}E\left(\text{MSTr}\right)={\sigma }^2+\frac{J}{I-1}{\displaystyle \sum _i{\left({\mu }_i-\mu \right)}^2}\\ >{\sigma }^2+0\\ E\left(\text{MSTr}\right)>{\sigma }^2.\end{array}$

Thus, when $H_0$  is false, the resultant value of $E\left(\text{MSTr}\right)$ is greater than when $H_0$  is true, leading to an overestimation of the population variance, ${\sigma }^2$.