Chapter 14, Problem 1RE

To determine

To find: The AVOVA table.

Answer to Problem 1RE

The AVOVA table is shown in Table-1.

Explanation of Solution

Given information:

The given data is,

Plant	Concentration
A	438	619	732	638
B	857	1014	1153	883	1053
C	925	786	1179	786
D	893	891	917	695	675	595

Calculation:

The sample size are $4$ and $n_1=3,n_2=3,n_3=3,n_4=3$ .

The total number in all the samples combined is,

  $\begin{array}{c}N=3+3+3+3\\ =12\end{array}$

From the given data the sample means are,

  $\begin{array}{c}\overline{x_1}=\frac{7.88+7.72+7.68}{3}\\ =7.76\\ \overline{x_2}=\frac{7.81+7.64+7.85}{3}\\ =7.7667\end{array}$

Further solve,

  $\begin{array}{c}\overline{x_3}=\frac{7.80+7.63+7.87}{3}\\ =7.78\\ \overline{x_4}=\frac{7.80+7.73+8}{3}\\ =7.8433\end{array}$

The grand mean is,

  $\begin{array}{c}\overline{\overline{x}}=\frac{7.76+7.7667+7.78+7.8433}{4}\\ =7.7875\end{array}$

The value of ${\displaystyle \sum _{i=1}^3{x}_{1i}^2}$ is,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^3{x}_{1i}^2}={7.88}^2+{7.72}^2+{7.68}^2\\ =180.6752\end{array}$

The standard of deviation of sample 1 is,

  $\begin{array}{c}{s}_1^2=\frac{1}{n_1-1}\left({\displaystyle \sum _{i=1}^3{x}_{1i}^2}-n_1\overline{x_1^2}\right)\\ =\frac{1}{3-1}\left(180.6752-3{\left(7.76\right)}^2\right)\\ s_1=0.1058\end{array}$

The value of ${\displaystyle \sum _{i=1}^3{x}_{2i}^2}$ is,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^3{x}_{2i}^2}={7.81}^2+{7.64}^2+{7.85}^2\\ =180.9882\end{array}$

The standard of deviation of sample 2 is,

  $\begin{array}{c}{s}_2^2=\frac{1}{n_2-1}\left({\displaystyle \sum _{i=1}^3{x}_{2i}^2}-n_2\overline{x_2^2}\right)\\ =\frac{1}{3-1}\left(180.9882-3{\left(7.7667\right)}^2\right)\\ s_2=0.1082\end{array}$

The value of ${\displaystyle \sum _{i=1}^3{x}_{3i}^2}$ is,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^3{x}_{3i}^2}={7.84}^2+{7.63}^2+{7.87}^2\\ =181.6194\end{array}$

The standard of deviation of sample 3 is,

  $\begin{array}{c}{s}_3^2=\frac{1}{n_3-1}\left({\displaystyle \sum _{i=1}^3{x}_{3i}^2}-n_3\overline{x_3^2}\right)\\ =\frac{1}{3-1}\left(181.6194-3{\left(7.78\right)}^2\right)\\ s_3=0.1308\end{array}$

The value of ${\displaystyle \sum _{i=1}^3{x}_{4i}^2}$ is,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^3{x}_{4i}^2}={7.8}^2+{7.73}^2+8^2\\ =184.5929\end{array}$

The standard of deviation of sample 4 is,

  $\begin{array}{c}{s}_4^2=\frac{1}{n_4-1}\left({\displaystyle \sum _{i=1}^3{x}_{4i}^2}-n_4\overline{x_4^2}\right)\\ =\frac{1}{3-1}\left(184.5929-3{\left(7.8433\right)}^2\right)\\ s_4=0.1427\end{array}$

The treatment sum of squares is,

  $\begin{array}{c}SSTr=n_1{\left(\overline{x_1}-\overline{\overline{x}}\right)}^2+n_2{\left(\overline{x_2}-\overline{\overline{x}}\right)}^2+n_3{\left(\overline{x_3}-\overline{\overline{x}}\right)}^2+n_4{\left(\overline{x_4}-\overline{\overline{x}}\right)}^2\\ =3{\left(7.76-7.7875\right)}^2+3{\left(7.7667-7.7875\right)}^2+3{\left(7.78-7.7875\right)}^2+3{\left(7.8433-7.7875\right)}^2\\ =0.01307\end{array}$

The error sum square is,

  $\begin{array}{c}SSE=\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2+\left(n_3-1\right)s_3^2+\left(n_4-1\right)s_4^2\\ =\left(3-1\right)\left(0.0112\right)+\left(3-1\right)\left(0.0117\right)+\left(3-1\right)\left(0.0171\right)+\left(3-1\right)\left(0.02035\right)\\ =0.1207\end{array}$

The degree of freedom for treatment sum of square is,

  $\begin{array}{c}I-1=4-1\\ =3\end{array}$

The degree of freedom for error sum of square is,

  $\begin{array}{c}N-I=12-4\\ =8\end{array}$

The treatment mean sum of square is,

  $\begin{array}{c}MSTr=\frac{SSTr}{I-1}\\ =\frac{0.01307}{3}\\ =0.004357\end{array}$

The error mean sum of square is,

  $\begin{array}{c}MSE=\frac{SSE}{N-I}\\ =\frac{0.1207}{8}\\ =0.01509\end{array}$

The value of test statistic is,

  $\begin{array}{c}F=\frac{MStr}{MSE}\\ =\frac{0.004357}{0.01509}\\ =0.2887\end{array}$

The ANOVA table is shown below.

Source of variation	Degree of freedom	Sum of square	Mean of square	F-value
Treatments	3	0.01307	0.004357	0.2887
Error	8	0.1207	0.01509
Total	11

Table-1

Therefore, the AVOVA table is shown in Table-1.