Chapter 14, Problem 11RE

To determine

To find: The pairs of means which are different by Tukey-Kramer test.

Answer to Problem 11RE

The pairs of means which are different by Tukey-Kramer test are $1,2$ and $1,3$ .

Explanation of Solution

Given information:

The value of $\alpha$ is $0.05$ and 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 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 mean of 1 and 2 sample is,

  $\begin{array}{c}{q}_{1,2}=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{\left|606.75-992\right|}{\sqrt{\frac{20322.5181}{2}\left(\frac{1}{4}+\frac{1}{5}\right)}}\\ =5.6972\end{array}$

The mean of 1 and 3 sample is,

  $\begin{array}{c}{q}_{1,3}=\frac{\left|\overline{x_1}-\overline{x_3}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_1}+\frac{1}{n_3}\right)}}\\ =\frac{\left|606.75-919\right|}{\sqrt{\frac{20322.5181}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =4.3807\end{array}$

The mean of 1 and 4 sample is,

  $\begin{array}{c}{q}_{1,4}=\frac{\left|\overline{x_1}-\overline{x_4}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_1}+\frac{1}{n_4}\right)}}\\ =\frac{\left|606.75-777.6667\right|}{\sqrt{\frac{20322.5181}{2}\left(\frac{1}{4}+\frac{1}{6}\right)}}\\ =2.6266\end{array}$

The mean of 2 and 3 sample is,

  $\begin{array}{c}{q}_{2,3}=\frac{\left|\overline{x_2}-\overline{x_3}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_2}+\frac{1}{n_3}\right)}}\\ =\frac{\left|992-919\right|}{\sqrt{\frac{20322.5181}{2}\left(\frac{1}{5}+\frac{1}{4}\right)}}\\ =1.07955\end{array}$

The mean of 2 and 4 sample is,

  $\begin{array}{c}{q}_{2,4}=\frac{\left|\overline{x_2}-\overline{x_4}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_2}+\frac{1}{n_4}\right)}}\\ =\frac{\left|992-777.6667\right|}{\sqrt{\frac{20322.5181}{2}\left(\frac{1}{5}+\frac{1}{6}\right)}}\\ =3.5112\end{array}$

The mean of 3 and 4 sample is,

  $\begin{array}{c}{q}_{3,4}=\frac{\left|\overline{x_3}-\overline{x_4}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_3}+\frac{1}{n_4}\right)}}\\ =\frac{\left|919-777.6667\right|}{\sqrt{\frac{20322.5181}{2}\left(\frac{1}{4}+\frac{1}{6}\right)}}\\ =2.1720\end{array}$

The critical value q for the student zed range distribution at $\alpha =0.05$ level of significance is $4.08$ .

The comparison of pairwise test statistic values with critical values is shown in table below.

Means	Test statistic	Critical value	Decision
1,2	5.6972	4.08	reject null hypothesis.
1,3	4.3807	4.08	reject null hypothesis.
1,4	2.6266	4.08	Do not reject null hypothesis.
2,3	1.07955	4.08	Do not reject null hypothesis.
2,4	3.5112	4.08	Do not reject null hypothesis.
3,4	2.1720	4.08	Do not reject null hypothesis.

Therefore, the pairs of means which are different by Tukey-Kramer test are $1,2$ and $1,3$ .