Chapter 14, Problem 10RE

a.

To determine

To check: The assumption for equal variance is satisfied.

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 $n_1=4,n_2=5,n_3=4,n_4=6$ .

The total number in all the samples combined is,

  $\begin{array}{c}N=4+5+4+6\\ =19\end{array}$

From the given data the sample means are,

  $\begin{array}{c}\overline{x_1}=\frac{438+619+732+638}{4}\\ =606.75\\ \overline{x_2}=\frac{857+1014+1153+883+1053}{5}\\ =992\end{array}$

Further solve,

  $\begin{array}{c}\overline{x_3}=\frac{925+786+1179+786}{4}\\ =919\\ \overline{x_4}=\frac{893+891+917+695+675+595}{6}\\ =777.6667\end{array}$

The grand mean is,

  $\begin{array}{c}\overline{\overline{x}}=\frac{606.75+992+919+777.6667}{4}\\ =823.8542\end{array}$

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

  $\begin{array}{c}{\displaystyle \sum _{i=1}^4{x}_{1i}^2}={438}^2+{619}^2+{732}^2+{638}^2\\ =1517873\end{array}$

The standard of deviation of sample 1 is,

  $\begin{array}{c}{s}_1^2=\frac{1}{n_1-1}\left({\displaystyle \sum _{i=1}^4{x}_{1i}^2}-n_1\overline{x_1^2}\right)\\ =\frac{1}{4-1}\left(1517873-4{\left(606.75\right)}^2\right)\\ s_1=368145.5625\end{array}$

The value of ${\displaystyle \sum _{i=1}^5{x}_{2i}^2}$ is,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^5{x}_{2i}^2}={857}^2+{1014}^2+{1153}^2+{883}^2+{1053}^2\\ =4980552\end{array}$

The standard of deviation of sample 2 is,

  $\begin{array}{c}{s}_2^2=\frac{1}{n_2-1}\left({\displaystyle \sum _{i=1}^5{x}_{2i}^2}-n_2\overline{x_2^2}\right)\\ =\frac{1}{5-1}\left(4980552-5{\left(992\right)}^2\right)\\ s_2=122.711\end{array}$

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

  $\begin{array}{c}{\displaystyle \sum _{i=1}^4{x}_{3i}^2}={925}^2+{786}^2+{1179}^2+{786}^2\\ =3481258\end{array}$

The standard of deviation of sample 3 is,

  $\begin{array}{c}{s}_3^2=\frac{1}{n_3-1}\left({\displaystyle \sum _{i=1}^4{x}_{3i}^2}-n_3\overline{x_3^2}\right)\\ =\frac{1}{4-1}\left(3481258-4{\left(919\right)}^2\right)\\ s_3=185.3052\end{array}$

The value of ${\displaystyle \sum _{i=1}^6{x}_{4i}^2}$ is,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^6{x}_{4i}^2}={893}^2+{891}^2+{917}^2+{695}^2+{675}^2+{595}^2\\ =3724894\end{array}$

The standard of deviation of sample 4 is,

  $\begin{array}{c}{s}_4^2=\frac{1}{n_4-1}\left({\displaystyle \sum _{i=1}^6{x}_{4i}^2}-n_4\overline{x_4^2}\right)\\ =\frac{1}{6-1}\left(3724894-6{\left(777.6667\right)}^2\right)\\ s_4=138.7811\end{array}$

The comparison between forth and first sample is,

  $\frac{138.7811}{122.8695}=1.1295$

The comparison between forth and second sample is,

  $\frac{138.7811}{185.3052}=0.7489$

Since, the result is less than two.

Thus, the assumption of equal variance is satisfied.

b.

To determine

To find: The AVOVA table.

Answer to Problem 10RE

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 treatment sum of squares is,

  $\begin{array}{c}SSTr=n_1{\left(\overline{x_1}-\overline{\overline{x}}\right)}^2+n_2{\left(\overline{x_2}-\overline{\overline{x}}\right)}^2+n_3{\left(\overline{x_3}-\overline{\overline{x}}\right)}^2+n_4{\left(\overline{x_4}-\overline{\overline{x}}\right)}^2\\ =4{\left(606.75-823.8542\right)}^2+5{\left(992-823.8542\right)}^2+4{\left(919-823.8542\right)}^2+6{\left(777.6667-823.8542\right)}^2\\ =378912.5897\end{array}$

The error sum square is,

  $\begin{array}{c}SSE=\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2+\left(n_3-1\right)s_3^2+\left(n_4-1\right)s_4^2\\ =\left(4-1\right)\left(15096.9167\right)+\left(5-1\right)\left(15058\right)+\left(4-1\right)\left(34338\right)+\left(6-1\right)\left(19260.2044\right)\\ =304837.7721\end{array}$

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,

  $\begin{array}{c}N-I=19-4\\ =15\end{array}$

The treatment mean sum of square is,

  $\begin{array}{c}MSTr=\frac{SSTr}{I-1}\\ =\frac{378912.5897}{3}\\ =126304.1966\end{array}$

The error mean sum of square is,

  $\begin{array}{c}MSE=\frac{SSE}{N-I}\\ =\frac{304837.7721}{15}\\ =20322.5181\end{array}$

The value of test statistic is,

  $\begin{array}{c}F=\frac{MStr}{MSE}\\ =\frac{126304.1966}{20322.5181}\\ =6.21499\end{array}$

The ANOVA table is shown below.

Source of variation	Degree of freedom	Sum of square	Mean of square	F-value
Treatments	3	378912.5897	126304.1966	6.21499
Error	15	304837.7721	20322.5181
Total	18

Table-1

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