Chapter 13.4, Problem 6E

Suppose that independent samples of sizes n₁, n₂,…, n_k are taken from each of k normally distributed populations with means μ₁, μ₂,…,μ_k and common variances, all equal to σ². Let Y_ij denote the jth observation from population i, for j = 1, 2, …, n_i and i = 1, 2,…, k, and let n = n₁ + n₂ + …+ n_k

1. a. Recall that

   $SSE={\displaystyle \underset{i=1}{\overset{k}{\sum }}\left(n_i-1\right)}{s}_i^2\text{ where }{s}_i^2=\frac{1}{n_i-1}{\displaystyle \underset{j=1}{\overset{n_i}{\sum }}{\left(Y_{ij}-{\bar{Y}}_{i·}\right)}^2}$

   Argue that SSE/σ² has a χ²distribution with (n₁ − 1) + (n₂ − l) + … + (n_k − 1) = n − k df.

   1. Exercises proceed by an asterisk are optional.

2. b. Argue that under the null hypothesis. H₀:μ₁ = μ₂ = …= μ_k all the Y_ij’s  are independent. normally distributed random variables with the same mean and variance. Use Theorem 7.3 to argue further that, under the null hypothesis.

   $Total\text{ SS}={\displaystyle \underset{i=1}{\overset{k}{\sum }}{\displaystyle \underset{j=1}{\overset{n_i}{\sum }}{\left(Y_{ij}-\bar{Y}\right)}^2}}$

   is such that (Total SS)/σ² has a χ² distribution with n – 1 df.

3. c. In Section 13.3. we argued that SST is a function of only the sample means and that SSE is a function of only the sample variances. Hence. SST and SSE are independent. Recall that Total SS = SST + SSE. Use the results of Exercise 13.5 and parts (a) and (b) to show that, under the hypothesis H₀:μ₁ = μ₂ = …= μ_k SST/σ²: has χ² distribution with k − 1 df.
4. d. Use the results of part(a)–(c) to argue that, under the hypothesis H₀: μ₁ = μ₂ = …= μ_k,  F = MST/MSE has an F distribution with k − 1 and n − k numerator and denominator degrees of freedom, respectively.

To determine

Show that $\frac{\text{SSE}}{{\sigma }^2}$ has a chi-square distribution with the $\left(n-k\right)$ degrees of freedom.

Explanation of Solution

Result:

• The function $\frac{\left(n_i-1\right)S_i^2}{{\sigma }^2}$ follows a chi-square distribution with $\left(n_i-1\right)$.
• If X and Y are an independent random variable and have a chi-square distribution with ‘a’ and ‘b’ the degrees of freedom; then, $X+Y$ follows a chi-square distribution with $a+b$ the degrees of freedom.

The formula for the sum of squares for error is as follows:

$\begin{array}{l}\text{SSE = }{\displaystyle \sum _{i=1}^k\left(n_i-1\right)S_i^2}\\ \text{Where }{S}_i^2=\frac{1}{\left(n_i-1\right)}{\displaystyle \sum _{i=1}^k{\left(Y_{ij}-{\overline{Y}}_{i.}\right)}^2}\end{array}$

Thus, the $\frac{\text{SSE}}{{\sigma }^2}$ can be expressed as follows:

$\begin{array}{c}\frac{\text{SSE}}{{\sigma }^2}=\frac{{\displaystyle \sum _{i=1}^k\left(n_i-1\right)S_i^2}}{{\sigma }^2}\\ ={\displaystyle \sum _{i=1}^k\frac{\left(n_i-1\right)S_i^2}{{\sigma }^2}}\\ =\frac{\left(n_1-1\right)S_1^2}{{\sigma }^2}+\frac{\left(n_2-1\right)S_2^2}{{\sigma }^2}+\mathrm{....}+\frac{\left(n_k-1\right)S_k^2}{{\sigma }^2}\end{array}$

That is,

$\frac{\text{SSE}}{{\sigma }^2}$ is a sum of k chi-square distributions. Therefore, $\frac{\text{SSE}}{{\sigma }^2}$ has a chi-square distribution.

The degree of freedom is calculated as follows:

$\begin{array}{c}\text{Degrees of freedom}=\left(n_1-1\right)+\left(n_2-1\right)+\mathrm{...}+\left(n_k-1\right)\\ ={\displaystyle \sum _{i=1}^k{n}_i}-k\\ =n-k\end{array}$

Thus, $\frac{\text{SSE}}{{\sigma }^2}$ has a chi-square distribution with the $\left(n-k\right)$ degrees of freedom.

To determine

Prove that $\frac{\text{Total SS}}{{\sigma }^2}$ has a chi-square distribution with the $\left(n-1\right)$ degrees of freedom.

Explanation of Solution

The formula for the sum of squares for total is as follows:

$\text{Total SS = }{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^{n_i}{\left(Y_{ij}-\overline{Y}\right)}^2}}$

If the null hypothesis is true; then, $Y_{ij}$ is independent and normally distributed with the same mean and same variance.

Thus, the $\frac{\text{Total SS}}{{\sigma }^2}$ can be expressed as follows:

$\begin{array}{c}\frac{\text{Total SS }}{{\sigma }^2}\text{= }\frac{1}{{\sigma }^2}{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^{n_i}{\left(Y_{ij}-\overline{Y}\right)}^2}}\\ ={\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^{n_i}\frac{{\left(Y_{ij}-\overline{Y}\right)}^2}{{\sigma }^2}}}\end{array}$

Use Theorem7.3 $\frac{\text{Total SS}}{{\sigma }^2}$ that has a chi-square distribution with the $\left(n-1\right)$ degrees of freedom.

To determine

Show that $\frac{\text{SST}}{{\sigma }^2}$ has a chi-square distribution with the $\left(k-1\right)$ degrees of freedom.

Explanation of Solution

From Part (a), $\frac{\text{SSE}}{{\sigma }^2}$ has a chi-square distribution with the $\left(n-k\right)$ degrees of freedom.

From Part (b), $\frac{\text{Total SS}}{{\sigma }^2}$ has a chi-square distribution with the $\left(n-1\right)$ degrees of freedom.

Formula for the total sum of square is as follows:

$\begin{array}{c}\text{Total SS}=\text{SST+SSE}\\ \frac{\text{Total SS}}{{\sigma }^2}=\frac{\text{SST}}{{\sigma }^2}\text{+}\frac{\text{SSE}}{{\sigma }^2}\\ \frac{\text{SST}}{{\sigma }^2}=\frac{\text{Total SS}}{{\sigma }^2}-\frac{\text{SSE}}{{\sigma }^2}\end{array}$

From Exercise 13.5, it is clear that $\frac{\text{SST}}{{\sigma }^2}$ has a chi-square distribution.

The degrees of freedom associated with $\frac{\text{SST}}{{\sigma }^2}$ is calculated as follows:

$\begin{array}{c}\text{Degrees of freedom}=\left(n-1\right)-\left(n-k\right)\\ =\left(k-1\right)\end{array}$

Thus, $\frac{\text{SST}}{{\sigma }^2}$ has a chi-square distribution with the $\left(k-1\right)$ degrees of freedom.

To determine

Prove that $F=\frac{MST}{MSE}$ has an F distribution with the $\left(k-1\right)$ and $\left(n-k\right)$ numerator and denominator degrees of freedom.

Explanation of Solution

From Part (a), $\frac{\text{SSE}}{{\sigma }^2}$ has a chi-square distribution with the $\left(n-k\right)$ degrees of freedom.

From Part (c), $\frac{\text{SST}}{{\sigma }^2}$ has a chi-square distribution with the $\left(k-1\right)$ degrees of freedom.

Formula for the F statistic is as follows:

$\begin{array}{c}F=\frac{MST}{MSE}\\ =\frac{\frac{SST}{\left(k-1\right)}}{\frac{SSE}{\left(n-k\right)}}\end{array}$

That is, F statistic is a ratio of two chi-square distributions divided by their respective degrees of freedom. Thus, $F=\frac{MST}{MSE}$ has an F distribution with the $\left(k-1\right)$ and $\left(n-k\right)$ numerator and denominator degrees of freedom, respectively.