Chapter 14.1, Problem 19E

Pesticide danger: Using the data in Exercise 17, perform the Tukey—Kramer test to determine which pairs of means, if any, differ. Use the $\alpha =0.05$ level of significance.

To determine

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

Answer to Problem 19E

The pairs of means which are different by Tukey-Kramer testare $1,3$ , $1,5$ and $2,5$ .

Explanation of Solution

Given information:

The value of $\alpha$ is $0.05$ and the given data is,

Duration	Amount Absorbed
1	1.7	1.5	1.2	1.5
2	1.7	1.6	1.9	1.9
4	1.9	1.7	2.1	2.0
10	2.2	1.9	1.7	1.6
24	2.1	2.2	2.5	2.3

Calculation:

The sample size are $4$ .

The total number in all the samples combined is,

  $\begin{array}{c}N=4+4+4+4+4\\ =20\end{array}$

From the given data the sample means are,

  $\begin{array}{c}\overline{x_1}=\frac{1.7+1.5+1.2+1.5}{4}\\ =1.475\\ \overline{x_2}=\frac{1.7+1.6+1.9+1.9}{4}\\ =1.775\end{array}$

Further solve,

  $\begin{array}{c}\overline{x_3}=\frac{1.9+1.7+2.1+2.0}{4}\\ =1.925\\ \overline{x_4}=\frac{2.2+1.9+1.7+1.6}{4}\\ =1.85\end{array}$

Further solve,

  $\begin{array}{c}\overline{x_5}=\frac{2.1+2.2+2.5+2.3}{4}\\ =2.275\end{array}$

The grand mean is,

  $\begin{array}{c}\overline{\overline{x}}=\frac{1.475+1.775+1.925+1.85+2.275}{5}\\ =1.86\end{array}$

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

  $\begin{array}{c}{\displaystyle \sum _{i=1}^4{x}_{1i}^2}={1.7}^2+{1.5}^2+{1.2}^2+{1.5}^2\\ =8.83\end{array}$

The standard of deviation of sample 1 is,

  $\begin{array}{c}{s}_1^2=\frac{1}{n_1-1}\left({\displaystyle \sum _{i=1}^4{x}_{1i}^2}-n_1\overline{x_1^2}\right)\\ =\frac{1}{4-1}\left(8.83-4{\left(1.475\right)}^2\right)\\ =\frac{1}{4-1}\left(8.83-4\times 2.17563\right)\\ s_1=0.2062\end{array}$

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

  $\begin{array}{c}{\displaystyle \sum _{i=1}^4{x}_{2i}^2}={1.7}^2+{1.6}^2+{1.9}^2+{1.9}^2\\ =12.67\end{array}$

The standard of deviation of sample 2 is,

  $\begin{array}{c}{s}_2^2=\frac{1}{n_2-1}\left({\displaystyle \sum _{i=1}^4{x}_{2i}^2}-n_2\overline{x_2^2}\right)\\ =\frac{1}{4-1}\left(12.67-4{\left(1.775\right)}^2\right)\\ =\frac{1}{4-1}\left(12.67-4\times 3.1506\right)\\ s_2=0.15\end{array}$

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

  $\begin{array}{c}{\displaystyle \sum _{i=1}^4{x}_{3i}^2}={1.9}^2+{1.7}^2+{2.1}^2+2^2\\ =14.91\end{array}$

The standard of deviation of sample 3 is,

  $\begin{array}{c}{s}_3^2=\frac{1}{n_3-1}\left({\displaystyle \sum _{i=1}^4{x}_{3i}^2}-n_3\overline{x_3^2}\right)\\ =\frac{1}{4-1}\left(14.91-4{\left(1.925\right)}^2\right)\\ =\frac{1}{4-1}\left(14.91-4\times 3.7056\right)\\ s_3=0.1709\end{array}$

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

  $\begin{array}{c}{\displaystyle \sum _{i=1}^4{x}_{4i}^2}={2.2}^2+{1.9}^2+{1.7}^2+{1.6}^2\\ =13.9\end{array}$

The standard of deviation of sample 4 is,

  $\begin{array}{c}{s}_4^2=\frac{1}{n_4-1}\left({\displaystyle \sum _{i=1}^4{x}_{4i}^2}-n_4\overline{x_4^2}\right)\\ =\frac{1}{4-1}\left(13.9-4{\left(1.85\right)}^2\right)\\ =\frac{1}{4-1}\left(13.9-4\times 3.4225\right)\\ s_4=0.2646\end{array}$

The value of ${\displaystyle \sum _{i=1}^4{x}_{5i}^2}$ is,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^4{x}_{5i}^2}={2.1}^2+{2.2}^2+{2.5}^2+{2.3}^2\\ =20.79\end{array}$

The standard of deviation of sample 5 is,

  $\begin{array}{c}{s}_5^2=\frac{1}{n_5-1}\left({\displaystyle \sum _{i=1}^4{x}_{5i}^2}-n_5\overline{x_5^2}\right)\\ =\frac{1}{4-1}\left(20.79-4{\left(2.275\right)}^2\right)\\ =\frac{1}{4-1}\left(20.79-4\times 5.1756\right)\\ s_5=0.1709\end{array}$

The treatment sum of squares is,

  $\begin{array}{c}SSTr=n_1{\left(\overline{x_1}-\overline{\overline{x}}\right)}^2+n_2{\left(\overline{x_2}-\overline{\overline{x}}\right)}^2+n_3{\left(\overline{x_3}-\overline{\overline{x}}\right)}^2+n_4{\left(\overline{x_4}-\overline{\overline{x}}\right)}^2+n_5{\left(\overline{x_5}-\overline{\overline{x}}\right)}^2\\ =4{\left(1.475-1.86\right)}^2+4{\left(1.775-1.86\right)}^2+4{\left(1.925-1.86\right)}^2+4{\left(1.85-1.86\right)}^2\\ +4{\left(\mathrm{1.2.275}-1.86\right)}^2\\ =4\left(0.1482+0.00723+0.00423+0.0001+0.1722\right)\\ =1.3278\end{array}$

The error sum square is,

  $\begin{array}{c}SSE=\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2+\left(n_3-1\right)s_3^2+\left(n_4-1\right)s_4^2+\left(n_5-1\right)s_5^2\\ =\left(4-1\right)\left(0.0425\right)+\left(4-1\right)\left(0.0225\right)+\left(4-1\right)\left(0.0292\right)+\left(4-1\right)\left(0.07\right)\\ +\left(4-1\right)\left(0.0292\right)\\ =3\left(0.1934\right)\\ =0.5802\end{array}$

The degree of freedom for treatment sum of square is,

  $\begin{array}{c}I-1=5-1\\ =4\end{array}$

The degree of freedom for error sum of square is,

  $\begin{array}{c}N-I=20-5\\ =15\end{array}$

The treatment mean sum of square is,

  $\begin{array}{c}MSTr=\frac{SSTr}{I-1}\\ =\frac{1.3278}{4}\\ =0.33195\end{array}$

The error mean sum of square is,

  $\begin{array}{c}MSE=\frac{SSE}{N-I}\\ =\frac{0.5802}{15}\\ =0.03868\end{array}$

The mean of 1 and 2 sample is,

  $\begin{array}{c}{q}_{1,2}=\frac{\left|\overline{x_1}-\overline{x_2}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{\left|1.475-1.775\right|}{\sqrt{\frac{0.03868}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =\frac{0.3}{\sqrt{0.01934\left(0.5\right)}}\\ =3.0506\end{array}$

The mean of 1 and 3 sample is,

  $\begin{array}{c}{q}_{1,3}=\frac{\left|\overline{x_1}-\overline{x_3}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_1}+\frac{1}{n_3}\right)}}\\ =\frac{\left|1.475-1.925\right|}{\sqrt{\frac{0.03868}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =\frac{0.45}{\sqrt{0.01934\left(0.5\right)}}\\ =4.576\end{array}$

The mean of 1 and 4 sample is,

  $\begin{array}{c}{q}_{1,4}=\frac{\left|\overline{x_1}-\overline{x_4}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_1}+\frac{1}{n_4}\right)}}\\ =\frac{\left|1.475-1.85\right|}{\sqrt{\frac{0.03868}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =\frac{0.375}{\sqrt{0.01934\left(0.5\right)}}\\ =3.8133\end{array}$

The mean of 1 and 5 sample is,

  $\begin{array}{c}{q}_{1,5}=\frac{\left|\overline{x_1}-\overline{x_5}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_1}+\frac{1}{n_5}\right)}}\\ =\frac{\left|1.475-2.275\right|}{\sqrt{\frac{0.03868}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =\frac{0.8}{\sqrt{0.01934\left(0.5\right)}}\\ =8.1351\end{array}$

The mean of 2 and 3 sample is,

  $\begin{array}{c}{q}_{2,3}=\frac{\left|\overline{x_2}-\overline{x_3}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_2}+\frac{1}{n_3}\right)}}\\ =\frac{\left|1.775-1.925\right|}{\sqrt{\frac{0.03868}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =\frac{0.15}{\sqrt{0.01934\left(0.5\right)}}\\ =1.5253\end{array}$

The mean of 2 and 4 sample is,

  $\begin{array}{c}{q}_{2,4}=\frac{\left|\overline{x_2}-\overline{x_4}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_2}+\frac{1}{n_4}\right)}}\\ =\frac{\left|1.775-1.85\right|}{\sqrt{\frac{0.03868}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =\frac{0.075}{\sqrt{0.01934\left(0.5\right)}}\\ =0.7627\end{array}$

The mean of 2 and 5 sample is,

  $\begin{array}{c}{q}_{2,5}=\frac{\left|\overline{x_2}-\overline{x_5}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_2}+\frac{1}{n_5}\right)}}\\ =\frac{\left|1.775-2.275\right|}{\sqrt{\frac{0.03868}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =\frac{0.5}{\sqrt{0.01934\left(0.5\right)}}\\ =5.0844\end{array}$

The mean of 3 and 4 sample is,

  $\begin{array}{c}{q}_{3,4}=\frac{\left|\overline{x_3}-\overline{x_4}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_3}+\frac{1}{n_4}\right)}}\\ =\frac{\left|1.925-1.85\right|}{\sqrt{\frac{0.03868}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =\frac{0.075}{\sqrt{0.01934\left(0.5\right)}}\\ =0.7627\end{array}$

The mean of 3 and 5 sample is,

  $\begin{array}{c}{q}_{3,5}=\frac{\left|\overline{x_3}-\overline{x_5}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_3}+\frac{1}{n_5}\right)}}\\ =\frac{\left|1.925-2.275\right|}{\sqrt{\frac{0.03868}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =\frac{0.35}{\sqrt{0.01934\left(0.5\right)}}\\ =3.5591\end{array}$

The mean of 4 and 5 sample is,

  $\begin{array}{c}{q}_{4,5}=\frac{\left|\overline{x_4}-\overline{x_5}\right|}{\sqrt{\frac{MSE}{2}\left(\frac{1}{n_4}+\frac{1}{n_5}\right)}}\\ =\frac{\left|1.85-2.275\right|}{\sqrt{\frac{0.03868}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}}\\ =\frac{0.425}{\sqrt{0.01934\left(0.5\right)}}\\ =4.3217\end{array}$

The critical value q for the student zed range distribution at $\alpha =0.05$ level of significance is $4.37$ .

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

Means	Test statistic	Critical value	Decision
1,2	3.0506	4.37	Do not reject null hypothesis.
1,3	4.576	4.37	reject null hypothesis.
1,4	3.8133	4.37	Do not reject null hypothesis.
1,5	8.1351	4.37	reject null hypothesis
2,3	1.5253	4.37	Do not reject null hypothesis.
2,4	0.7627	4.37	Do not reject null hypothesis.
2,5	5.0844	4.37	reject null hypothesis
3,4	0.7627	4.37	Do not reject null hypothesis.
3,5	3.5591	4.37	Do not reject null hypothesis.
4,5	4.3217	4.37	Do not reject null hypothesis.

Therefore, the pairs of means which are different by Tukey-Kramer test are $1,3$ , $1,5$ and $2,5$ .