Chapter 12.1, Problem 26E

To determine

To find: The ANOVA table for the data in the Exercise 12.5.

Answer to Problem 26E

Solution: The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	10323
Error	57	325052	5702.67
Total	59	336825.33

The P-value is 0.362749.

Explanation of Solution

Given: The data are provided in the referred exercise 12.5 of three joysticks to compare the time taken to complete a navigation mission by each joystick, which is also mentioned in the table given below:

Joystick	Mean	s	n
1	279	78	20
2	245	68	20
3	258	80	20

Calculation: The ANOVA table requires the following entries, which are mentioned below in the table along with their formula of calculation:

Sources	Degree of freedom	Sum of squares	Mean sum of square	F
Groups	$I-1$	${{\displaystyle \sum {}_{\text{groups}}{n}_i\left({\overline{x}}_i-\overline{x}\right)}}^2$	$\frac{SSG}{DFG}$	MSG/MSE
Error	$N-I$	${\displaystyle \sum {}_{\text{groups}}\left(n_i-1\right)s_i{}^2}$	$\frac{SSE}{DFE}$
Total	$N-1$	${{\displaystyle \sum {}_{obs}\left(x_{ij}-\overline{x}\right)}}^2$

Sum of square of group can be calculated by the formula given below:

${{\displaystyle \sum {}_{\text{groups}}{n}_i\left({\overline{x}}_i-\overline{x}\right)}}^2$

$$ $\begin{array}{c}\overline{x}=\frac{279+245+258}{3}\\ =260.67\end{array}$

All the required calculation is provided in the table given below:

n	${\overline{x}}_i$	$\left({\overline{x}}_i-\overline{x}\right)$	${\left({\overline{x}}_i-\overline{x}\right)}^2$	$n_i\times {\left({\overline{x}}_i-\overline{x}\right)}^2$
20	279	$279-260.67=18.33$	335.99	6719.78
20	245	$245-260.67=-15.67$	245.55	4910.98
20	258	$258-260.67=-2.67$	7.13	142.58

$\begin{array}{c}SSG=6719.78+4910.98+142.58\\ =11773.3\end{array}$

Sum of square of error can be calculated by the formula given below:

${\displaystyle \sum {}_{\text{groups}}\left(n_i-1\right)s_i{}^2}$

All the required calculation is provided in the table given below:

$n_i-1$	$s_i$	$s_i{}^2$	$\left(n_i-1\right)s_i{}^2$
$20-1=19$	78	$78\times 78=6084$	$6084\times 19=115596$
$20-1=19$	68	$68\times 68=4624$	$4624\times 19=87856$
$20-1=19$	80	$80\times 80=6400$	$6400\times 19=121600$

$\begin{array}{c}SSE=115596+87856+121600\\ =325052\end{array}$

Also, the sum of squares of all the observations is calculated by adding the sum of squares of groups and sum of squares of error, which is shown below:

$\begin{array}{c}{{\displaystyle \sum {}_{obs}\left(x_{ij}-\overline{x}\right)}}^2={{\displaystyle \sum {}_{\text{groups}}{n}_i\left({\overline{x}}_i-\overline{x}\right)}}^2+{\displaystyle \sum {}_{\text{groups}}\left(n_i-1\right)s_i{}^2}\\ =11773.3+325052\\ =336825.33\end{array}$

There are total three groups, so I is equal to 3 and the total number of observations are 60 as there are 20 observations in each group. Therefore, the degree of freedom for the groups, error and total are provided in the table below:

$\begin{array}{c}\text{Groups}=I-1\\ =3-1\\ =2\end{array}$

$\begin{array}{c}\text{Error}=N-I\\ =60-3\\ =57\end{array}$

$\begin{array}{c}\text{Total}=N-1\\ =60-1\\ =59\end{array}$

Mean sum of square of groups can be calculated by the following formula:

$MSG\text{=}\frac{SSG}{DFG}$

Substitute the values of SSG and its respective degree of freedom in the above formula as follows:

$\begin{array}{c}MSG=\frac{SSG}{DFG}\\ =\frac{11773.33}{2}\\ =5886.67\end{array}$

Mean sum of square of error can be calculated by the formula mentioned below:

$MSE\text{=}\frac{SSE}{DFE}$

Substitute the values of SSE and its respective degree of freedom in the above formula as follows:

$\begin{array}{c}MSE\text{=}\frac{SSE}{DFE}\\ =\frac{325052}{57}\\ =5702.67\end{array}$

The F-statistic can be calculated by the formula mentioned below:

$F=\frac{MSG}{MSE}$

Substitute the values of MSG and MSE in the formula as follows:

$\begin{array}{c}F=\frac{MSG}{MSE}\\ =\frac{5886.67}{5702.67}\\ =1.0323\end{array}$

The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	1.0323
Error	57	325052	5702.67
Total	59	336825.33

The P-value on the basis of F statistic and degree of freedom can be calculated by using the software Microsoft excel as shown in the snapshot below:

Insert the values of x, degree of freedom1, and degree of freeom2, and press enter. The output is shown below in the snapshot:

The null hypothesis $\left(H_0\right)$ and alternative hypothesis $\left(H_{\alpha }\right)$ are shown below:

$H_0:$ ${\mu }_1={\mu }_2={\mu }_3$

$H_a:$ Means are not all equal.

The P-value 0.362749 is greater than the 0.05 level of significance, therefore the null hypothesis is not rejected.

Interpretation: Since the Null hypothesis is not rejected, so it can be concluded that the mean time of all three joysticks to complete the navigation mission is same.