Chapter 14.2, Problem 63E

(a)

To determine

Compute the differences that exist between the burning times of the four brands of flares.

Explanation of Solution

The null and alternative hypothesis can be written as follows:

$\begin{array}{l}{\text{H}}_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4\\ {\text{H}}_1:\text{At least two means differ}\end{array}$

$\begin{array}{c}\overline{\text{x}}=\frac{10\left(61.6\right)+10\left(57.3\right)+10\left(61.8\right)+10\left(51.8\right)}{10+10+10+10}\\ =\frac{616+573+618+518}{40}\\ =\frac{2,325}{40}\\ =58.125\end{array}$

SST can be calculated as follows:

$\begin{array}{c}\text{SST}=\sum {\text{n}}_{\text{j}}\left(\overline{{\text{x}}_{\text{j}}}-\overline{\text{x}}\right)\\ =10{\left(61.6-58.13\right)}^2+10{\left(57.3-58.13\right)}^2+10\left(61.8-58.13\right)+10{\left(51.8-58.13\right)}^2\\ =662.7\end{array}$

SSE can be calculated as follows:

$\begin{array}{c}\text{SSE}=\sum \left({\text{n}}_{\text{j}}-1\right){\text{s}}_{\text{j}}^2\\ =\left(10-1\right)80.49+\left(10-1\right)70.46+\left(10-1\right)22.18+\left(10-1\right)75.29\\ =2,236\end{array}$

Thus, the value of SSE is 2,236.

Table-1 shows the ANOVA Table as follows:

Table-2
Source	Degree of freedom	Sum of square	Mean square	F
Treatments	k-1 = 3	SST = 662.7	220.9	3.56
Error	n- k = 36	SSE = 2,236	62.11

Thus, the estimated F-value is 3.56 and p-value is 0.0236. Here, there is an adequate evidence to accomplish that differences exist between the flares with respect to burning times.

(b)

To determine

Compute theLSD with Bonferroni adjustment.

Explanation of Solution

t-value can be written as follows:

$\begin{array}{c}\text{C}=\frac{4\left(3\right)}{2}\\ =6\\ {\alpha }_{\text{E}}=0.05\\ \alpha =\frac{{\alpha }_{\text{E}}}{\text{C}}\\ {\text{t}}_{\left(\frac{\alpha }{2,\text{n}-\text{k}}\right)}=2.794\end{array}$

LSD can be calculated as follows:

$\begin{array}{c}\text{LSD}={\text{t}}_{\left(\frac{\alpha }{2,\text{n}-\text{k}}\right)}\sqrt{\text{MSE}\left(\frac{1}{{\text{n}}_{\text{i}}}+\frac{1}{{\text{n}}_{\text{j}}}\right)}\\ =2.794\sqrt{62.11\left(\frac{1}{\text{10}}+\frac{1}{\text{10}}\right)}\\ =9.45\end{array}$

Table-2 shows the means and difference as follows:

Table-2
Treatment	Means			Difference
i = 1, j =2	61.6	57.3		4.3
i = 1, j =3	61.6	61.8		-2
i = 1, j =4	61.6	51.8		9.8
i =2, j =3	57.3	61.8		-4.5
i =2, j =4	57.3	51.8		5.5
i = 3, j =4	61.8	51.8		10

Therefore, from the above calculation, the mean of flares C and D differs.

(c)

To determine

Compute the Tukey’s method.

Explanation of Solution

$\begin{array}{c}{\text{q}}_{\alpha }\left(\text{k,v}\right)={\text{q}}_{0.05}\left(4,136\right)\approx 3.79\\ \varpi ={\text{q}}_{\alpha }\left(\text{k,v}\right)\sqrt{\frac{\text{MSE}}{{\text{n}}_{\text{g}}}}\\ =3.79\sqrt{\frac{62.11}{10}}\\ =9.45\end{array}$

Table-2 shows the means and difference as follows:

Table-2
Treatment	Means		Difference
i = 1, j =2	61.6	57.3	4.3
i = 1, j =3	61.6	61.8	-2
i = 1, j =4	61.6	51.8	9.8
i =2, j =3	57.3	61.8	-4.5
i =2, j =4	57.3	51.8	5.5
i = 3, j =4	61.8	51.8	10

Therefore, from the above calculation, the mean of flares A and D, C and D differs.