Chapter 14.2, Problem 44E

(a)

To determine

Compute the LSD method.

Explanation of Solution

t-value can be written as follows:

$\begin{array}{c}\alpha =0.05\\ {\text{t}}_{\left(\frac{\alpha }{2,\text{n}-\text{k}}\right)}={\text{t}}_{\left(0.025,20\right)}\\ =2.086\end{array}$

LSD can be calculated as follows:

$\begin{array}{c}\text{LSD}={\text{t}}_{\left(\frac{\alpha }{2,\text{n}-\text{k}}\right)}\sqrt{\text{MSE}\left(\frac{1}{{\text{n}}_{\text{i}}}+\frac{1}{{\text{n}}_{\text{j}}}\right)}\\ =2.086\sqrt{\text{125}\left(\frac{1}{\text{5}}+\frac{1}{\text{10}}\right)}\\ =14.75\end{array}$

Table-1 shows the mean and difference as follows:

Table-1

Treatment	Means		Difference
i = 1, j =2	227	205	22
i = 1, j =3	227	219	8
i = 1, j =4	227	248	-21
i = 1, j =5	227	202	25
i = 2, j =3	205	219	-14
i = 2, j =4	205	248	-43
i = 2, j =5	205	202	3
i = 3, j =4	219	248	-29
i = 3, j =5	219	202	17
i = 4, j =5	248	202	46

Therefore, from the above calculation, the pairs of mean differ, ${\mu }_1\text{ and }{\mu }_2$, ${\mu }_1\text{ and }{\mu }_4$, ${\mu }_1\text{ and }{\mu }_5$, ${\mu }_1\text{ and }{\mu }_4$, ${\mu }_3\text{ and }{\mu }_5$, and ${\mu }_1\text{ and }{\mu }_2$.

(b)

To determine

Compute the Bonferroni adjustment.

Explanation of Solution

t-value can be written as follows:

$\begin{array}{c}\text{C}=\frac{5\left(4\right)}{2}\\ =10\\ {\alpha }_{\text{E}}=0.05\\ \alpha =\frac{{\alpha }_{\text{E}}}{\text{C}}\\ {\text{t}}_{\left(\frac{\alpha }{2,\text{n}-\text{k}}\right)}={\text{t}}_{\left(0.0025,20\right)}\\ =3.153\end{array}$

LSD can be calculated as follows:

$\begin{array}{c}\text{LSD}={\text{t}}_{\left(\frac{\alpha }{2,\text{n}-\text{k}}\right)}\sqrt{\text{MSE}\left(\frac{1}{{\text{n}}_{\text{i}}}+\frac{1}{{\text{n}}_{\text{j}}}\right)}\\ =3.153\sqrt{125\left(\frac{1}{5}+\frac{1}{5}\right)}\\ =22.3\end{array}$

Table-1 shows the mean and difference as follows:

Table-1

Treatment	Means		Difference
i = 1, j =2	227	205	22
i = 1, j =3	227	219	8
i = 1, j =4	227	248	-21
i = 1, j =5	227	202	25
i = 2, j =3	205	219	-14
i = 2, j =4	205	248	-43
i = 2, j =5	205	202	3
i = 3, j =4	219	248	-29
i = 3, j =5	219	202	17
i = 4, j =5	248	202	46

Therefore, from the above calculation, the following pairs differ. ${\mu }_2$ and ${\mu }_5$, ${\mu }_2$ and ${\mu }_4$, ${\mu }_3$ and ${\mu }_4$,and ${\mu }_4$ and ${\mu }_5$.

(c)

To determine

Compute the Tukey’s multiple comparison.

Explanation of Solution

$\begin{array}{c}{\text{q}}_{\alpha }\left(\text{k,v}\right)={\text{q}}_{0.05}\left(5,20\right)=4.23\\ \varpi ={\text{q}}_{\alpha }\left(\text{k,v}\right)\sqrt{\frac{\text{MSE}}{{\text{n}}_{\text{g}}}}\\ =4.23\sqrt{\frac{125}{5}}\\ =21.15\end{array}$

Table-1 shows the mean and difference as follows:

Table-1

Treatment	Means		Difference
i = 1, j =2	227	205	22
i = 1, j =3	227	219	8
i = 1, j =4	227	248	-21
i = 1, j =5	227	202	25
i = 2, j =3	205	219	-14
i = 2, j =4	205	248	-43
i = 2, j =5	205	202	3
i = 3, j =4	219	248	-29
i = 3, j =5	219	202	17
i = 4, j =5	248	202	46

Therefore, from the above calculation, the following pairs differ. ${\mu }_2$ and ${\mu }_5$, ${\mu }_2$ and ${\mu }_4$, ${\mu }_3$ and ${\mu }_4$, and ${\mu }_4$ and ${\mu }_5$.