Chapter 9, Problem 11SE

a.

To determine

Check whether there is any difference in the mean drainage times for the different channel designs or not.

Answer to Problem 11SE

There is sufficient evidence to conclude that there is a significant difference in the mean drainage times with different channel type at $\alpha =0.05$ level of significance.

Explanation of Solution

Given info:

The design variable is the channel type and the response is the drainage time. The table provides the drainage time corresponding to the channel type.

Calculation:

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5$

Alternative hypothesis:

$H_a:$ At least two of the ${\mu }_i{}^{\prime }s$ differ from each other.

The ANOVA table can be obtained as follows:

Software procedure:

Step by step procedure to obtain One-Way ANOVA using the MINITAB software:

• Choose Stat > ANOVA > One-Way.
• In Response, enter the column of Drainage time.
• In Factor, enter the column of Channel type.
• In Confidence level, enter 0.95.
• Click OK.

Output using the MINITAB software is given below:

From the ANOVA table, it is clear that P-value is 0.001 and the F-value is 8.71.

Since, the level of significance is not specified; the prior level of significance $\alpha =0.05$ can be used.

Decision:

If $P\text{-value}\le \alpha ,$ reject the null hypothesis $H_0$

If $P\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$

Conclusion:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(=0.001\right)<\alpha \left(=0.05\right)$.

By rejection rule, reject the null hypothesis.

There is sufficient evidence to conclude that there is a significant difference in the mean drainage times with different channel type at $\alpha =0.05$ level of significance.

b.

To determine

Identify the pairs of designs that can conclude to have differing mean drainage times.

Answer to Problem 11SE

There is sufficient evidence to conclude that the channels 3 and 4 differ from channels 1,2, and 5 at $\alpha =0.05$ level of significance.

Explanation of Solution

Calculation:

State the hypotheses:

Null hypothesis:

$H_0\text{:}$ There is no difference in mean drainage times.

Alternative hypothesis:

$H_1:$ There is difference in mean drainage times.

Decision:

By Tukey-Kramer method for multiple comparisons,

If $\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|>q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},$ reject the null hypothesis $H_0$

If $\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|\le q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},$ fail to reject the null hypothesis $H_0$

Here $I=5,J_1=J_2=J_3=J_4=J_5=4,\alpha =0.05,MSE=29$

From Appendix A table A.9, the upper 5% point of the $q_{5,15,0.05}$ distribution is 4.51.

For comparing channel 1 and 2:

The 5% critical value is,

$q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}=q_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_1}+\frac{1}{J_2}\right)},$

Substitute $q_{5,15,0.05}=4.51,J_1\text{=}{J}_2=\text{4and }MSE=29$ in the above equation.

$\begin{array}{c}{q}_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_1}+\frac{1}{J_2}\right)}=\left(4.51\right)\sqrt{\frac{29}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}\\ =\left(4.51\right)\sqrt{\left(14.5\right)\left(0.5\right)}\\ =\left(4.51\right)\sqrt{7.25}\\ =\left(4.51\right)\left(2.6926\right)\\ =12.1436\end{array}$

The sample means are,

$\begin{array}{c}{\overline{X}}_1=\frac{41.4+43.4+50+41.2}{4}\\ =\frac{176}{4}\\ =44\end{array}$

$\begin{array}{c}{\overline{X}}_2=\frac{37.7+49.3+52.1+37.3}{4}\\ =\frac{176.4}{4}\\ =44.1\end{array}$

Now,

$\begin{array}{c}\left|{\overline{X}}_1-{\overline{X}}_2\right|=\left|44-44.1\right|\\ =0.1\end{array}$

Which is less than 4.51.

Thus, fail to reject the null hypothesis $H_0$.

Hence, for channel 1 and 2 there is no difference in mean drainage times.

For comparing channel 1 and 3:

The 5% critical value is,

$q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},=q_{4,12,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_1}+\frac{1}{J_3}\right)},$

Substitute $q_{5,15,0.05}=4.51,J_1\text{=}{J}_3=\text{4and }MSE=29$

$\begin{array}{c}{q}_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_1}+\frac{1}{J_3}\right)}=\left(4.51\right)\sqrt{\frac{29}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}\\ =\left(4.51\right)\sqrt{\left(14.5\right)\left(0.5\right)}\\ =\left(4.51\right)\sqrt{7.25}\\ =\left(4.51\right)\left(2.6926\right)\\ =12.1436\end{array}$

The sample means are,

$\begin{array}{c}{\overline{X}}_3=\frac{32.6+33.7+34.8+22.5}{4}\\ =\frac{123.6}{4}\\ =30.9\end{array}$

Now,

$\begin{array}{c}\left|{\overline{X}}_1-{\overline{X}}_3\right|=\left|44-30.9\right|\\ =13.1\end{array}$

Which is greater than 4.51.

Thus, reject the null hypothesis $H_0$.

Hence, for channel 1 and 3 there is difference in mean drainage times.

For comparing channel 1 and 4:

The 5% critical value is,

$q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},=q_{4,12,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_1}+\frac{1}{J_4}\right)},$

Substitute $q_{5,15,0.05}=4.51,J_1\text{=}{J}_4=\text{4and }MSE=29$

$\begin{array}{c}{q}_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_1}+\frac{1}{J_4}\right)}=\left(4.51\right)\sqrt{\frac{29}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}\\ =\left(4.51\right)\sqrt{\left(14.5\right)\left(0.5\right)}\\ =\left(4.51\right)\sqrt{7.25}\\ =\left(4.51\right)\left(2.6926\right)\\ =12.1436\end{array}$

The sample means are,

$\begin{array}{c}{\overline{X}}_4=\frac{27.3+29.9+32.3+24.8}{4}\\ =\frac{114.3}{4}\\ =28.575\end{array}$

Now,

$\begin{array}{c}\left|{\overline{X}}_1-{\overline{X}}_4\right|=\left|44-28.575\right|\\ =15.425\end{array}$

Which is greater than 4.51.

Thus, reject the null hypothesis $H_0$.

Hence, for channel 1 and 4 there is difference in mean drainage times.

For comparing channel 1 and 5:

The 5% critical value is,

$q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},=q_{4,12,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_1}+\frac{1}{J_5}\right)},$

Substitute $q_{5,15,0.05}=4.51,J_1\text{=}{J}_5=\text{4and }MSE=29$

$\begin{array}{c}{q}_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_1}+\frac{1}{J_5}\right)}=\left(4.51\right)\sqrt{\frac{29}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}\\ =\left(4.51\right)\sqrt{\left(14.5\right)\left(0.5\right)}\\ =\left(4.51\right)\sqrt{7.25}\\ =\left(4.51\right)\left(2.6926\right)\\ =12.1436\end{array}$

The sample means are,

$\begin{array}{c}{\overline{X}}_5=\frac{44.9+47.2+48.5+37.1}{4}\\ =\frac{177.7}{4}\\ =44.425\end{array}$

Now,

$\begin{array}{c}\left|{\overline{X}}_1-{\overline{X}}_5\right|=\left|44-44.425\right|\\ =0.425\end{array}$

Which is less than 4.51.

Thus, fail to reject the null hypothesis $H_0$.

Hence, for channel 1 and 5 there is no difference in mean drainage times.

For comparing channel 2 and 3:

The 5% critical value is,

$q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},=q_{4,12,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_2}+\frac{1}{J_3}\right)},$

Substitute $q_{5,15,0.05}=4.51,J_2\text{=}{J}_3=\text{4and }MSE=29$

$\begin{array}{c}{q}_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_2}+\frac{1}{J_3}\right)}=\left(4.51\right)\sqrt{\frac{29}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}\\ =\left(4.51\right)\sqrt{\left(14.5\right)\left(0.5\right)}\\ =\left(4.51\right)\sqrt{7.25}\\ =\left(4.51\right)\left(2.6926\right)\\ =12.1436\end{array}$

Now,

$\begin{array}{c}\left|{\overline{X}}_2-{\overline{X}}_3\right|=\left|44.1-30.9\right|\\ =13.2\end{array}$

Which is greater than 4.51.

Thus, reject the null hypothesis $H_0$.

Hence, for channel 2 and 3 there is difference in mean drainage times.

For comparing channel 2 and 4:

The 5% critical value is,

$q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},=q_{4,12,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_2}+\frac{1}{J_4}\right)},$

Substitute $q_{5,15,0.05}=4.51,J_2\text{=}{J}_4=\text{4and }MSE=29$

$\begin{array}{c}{q}_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_2}+\frac{1}{J_4}\right)}=\left(4.51\right)\sqrt{\frac{29}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}\\ =\left(4.51\right)\sqrt{\left(14.5\right)\left(0.5\right)}\\ =\left(4.51\right)\sqrt{7.25}\\ =\left(4.51\right)\left(2.6926\right)\\ =12.1436\end{array}$

Now,

$\begin{array}{c}\left|{\overline{X}}_2-{\overline{X}}_4\right|=\left|44.1-28.575\right|\\ =15.525\end{array}$

Which is greater than 4.51.

Thus, reject the null hypothesis $H_0$.

Hence, for channel 2 and 4 there is difference in mean drainage times.

For comparing channel 2 and 5:

The 5% critical value is,

$q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},=q_{4,12,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_2}+\frac{1}{J_5}\right)},$

Substitute $q_{5,15,0.05}=4.51,J_2\text{=}{J}_5=\text{4and }MSE=29$

$\begin{array}{c}{q}_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_2}+\frac{1}{J_5}\right)}=\left(4.51\right)\sqrt{\frac{29}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}\\ =\left(4.51\right)\sqrt{\left(14.5\right)\left(0.5\right)}\\ =\left(4.51\right)\sqrt{7.25}\\ =\left(4.51\right)\left(2.6926\right)\\ =12.1436\end{array}$

Now,

$\begin{array}{c}\left|{\overline{X}}_2-{\overline{X}}_5\right|=\left|44.1-44.425\right|\\ =0.325\end{array}$

Which is less than 4.51.

Thus, fail to reject the null hypothesis $H_0$.

Hence, for channel 2 and 5 there is no difference in mean drainage times.

For comparing channel 3 and 4:

The 5% critical value is,

$q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},=q_{4,12,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_3}+\frac{1}{J_4}\right)},$

Substitute $q_{5,15,0.05}=4.51,J_3\text{=}{J}_4=\text{4and }MSE=29$

$\begin{array}{c}{q}_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_3}+\frac{1}{J_4}\right)}=\left(4.51\right)\sqrt{\frac{29}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}\\ =\left(4.51\right)\sqrt{\left(14.5\right)\left(0.5\right)}\\ =\left(4.51\right)\sqrt{7.25}\\ =\left(4.51\right)\left(2.6926\right)\\ =12.1436\end{array}$

Now,

$\begin{array}{c}\left|{\overline{X}}_3-{\overline{X}}_4\right|=\left|30.9-28.575\right|\\ =2.325\end{array}$

Which is less than 4.51.

Thus, fail to reject the null hypothesis $H_0$.

Hence, for channel 3 and 4 there is no difference in mean drainage times.

For comparing channel 3 and 5:

The 5% critical value is,

$q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},=q_{4,12,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_3}+\frac{1}{J_5}\right)},$

Substitute $q_{5,15,0.05}=4.51,J_3\text{=}{J}_5=\text{4and }MSE=29$

$\begin{array}{c}{q}_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_3}+\frac{1}{J_5}\right)}=\left(4.51\right)\sqrt{\frac{29}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}\\ =\left(4.51\right)\sqrt{\left(14.5\right)\left(0.5\right)}\\ =\left(4.51\right)\sqrt{7.25}\\ =\left(4.51\right)\left(2.6926\right)\\ =12.1436\end{array}$

Now,

$\begin{array}{c}\left|{\overline{X}}_3-{\overline{X}}_5\right|=\left|30.9-44.425\right|\\ =13.525\end{array}$

Which is greater than 4.51.

Thus, reject the null hypothesis $H_0$.

Hence, for channel 3 and 5 there is difference in mean drainage times.

For comparing channel 4 and 5:

The 5% critical value is,

$q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)},=q_{4,12,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_4}+\frac{1}{J_5}\right)},$

Substitute $q_{5,15,0.05}=4.51,J_4\text{=}{J}_5=\text{4and }MSE=29$

$\begin{array}{c}{q}_{5,15,0.05}\sqrt{\frac{MSE}{2}\left(\frac{1}{J_4}+\frac{1}{J_5}\right)}=\left(4.51\right)\sqrt{\frac{29}{2}\left(\frac{1}{4}+\frac{1}{4}\right)}\\ =\left(4.51\right)\sqrt{\left(14.5\right)\left(0.5\right)}\\ =\left(4.51\right)\sqrt{7.25}\\ =\left(4.51\right)\left(2.6926\right)\\ =12.1436\end{array}$

Now,

$\begin{array}{c}\left|{\overline{X}}_4-{\overline{X}}_5\right|=\left|28.575-44.425\right|\\ =15.85\end{array}$

Which is greater than 4.51.

Thus, reject the null hypothesis $H_0$.

Hence, for channel 4 and 5 there is difference in mean drainage times.

Conclusion:

There is sufficient evidence to conclude that the channels 3 and 4 differ from channels 1,2, and 5 at $\alpha =0.05$ level of significance.