Chapter 10, Problem 37SE

Numerous factors contribute to the smooth running of an electric motor (“Increasing Market Share Through Improved Product and Process Design: An Experimental Approach,” Quality Engineering, 1991: 361–369). In particular, it is desirable to keep motor noise and vibration to a minimum. To study the effect that the brand of bearing has on motor vibration, five different motor bearing brands were examined by installing each type of bearing on different random samples of six motors. The amount of motor vibration (measured in microns) was recorded when each of the 30 motors was running. The data for this study follows. State and test the relevant hypotheses at significance level .05, and then carry out a multiple comparisons analysis if appropriate.

To determine

State the relevant hypotheses and test the same at level of significance 0.05.

Perform a multiple comparisons analysis, if appropriate.

Answer to Problem 37SE

The relevant hypotheses are:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5=0$

Alternative hypothesis:

$H_a:{\mu }_i\ne 0$, for at least one $i=1,2,3,4,5$.

The F test at level $\alpha =0.05$ suggests that the true average motor vibrations differ between at least two of the five brands.

A multiple comparison Tukey’s test reveals that the brands 5, 3 and 1 are similar, brands 5, 3 and 1 are similar, the brands 1 and 4 are similar and the brands 4 and 2 are similar, in terms of motor vibrations. There is significant difference of brand 5 with brands 4 and 2, and also significant difference of brand 3 with brand 2.

Explanation of Solution

Given info:

The data relates to 6 observations on running motor vibration (microns), corresponding to 5 different motor bearing brands. The sample means are also provided for each brand, where ${\overline{X}}_{1.}=13.68$, ${\overline{X}}_{2.}=15.95$, ${\overline{X}}_{3.}=13.67$, ${\overline{X}}_{4.}=14.73$ and ${\overline{X}}_{5.}=13.08$.

Calculation:

Suppose there are I treatments and J observations corresponding to each treatment in a designed experiment, resulting in a total of IJ observations. Then, the following properties hold true:

• Degrees of freedom (df): The treatment df is $I-1$, the total df is $IJ-1$. The error df is the difference between these, that is, $I\left(J-1\right)$.
• Sum of squares: The total sum of squares (SST), treatment sum of squares (SSTr) and error sum of squares (SSE) are related as: $\text{SST}=\text{SSTr}+\text{SSE}$.
• Mean of squares: The mean of squares is the ratio of the sum of squares to the corresponding df. The mean square error (MSE) is, $\text{MSE}=\frac{\text{SSE}}{I\left(J-1\right)}$. The mean square treatment (MSTr) is, $\text{MSTr}=\frac{\text{SSTr}}{I-1}$.
• The F-statistic: The F- statistic is the ratio of MSTr and MSE, that is, $F=\frac{\text{MSTr}}{\text{MSE}}$.

Here, each motor bearing brand is a treatment.

Denote $X_{ij}$ as the j^th observation corresponding to the i^th treatment, for $j=1,2,\mathrm{...},J\left(=6\right)$, corresponding to each $i=1,2,\mathrm{...},I\left(=5\right)$.

State the test hypotheses.

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5=0$

That is, the effects of all the treatments are similar.

Alternative hypothesis:

$H_a:{\mu }_i\ne 0$, for at least one $i=1,2,3,4,5$.

That is, the effects of all the treatments are not similar.

Test statistic:

The suitable test statistic is the F- statistic, which is the ratio of the mean square treatment (MSTr) and the mean square error (MSE), that is,

$F=\frac{\text{MSTr}}{\text{MSE}}$.

Degrees of freedom:

In this case, number of treatments (types of boxes) is $I=5$. Thus, the treatment df is:

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

The number of observations corresponding to each treatment is $J=6$.

The total df is:

$\begin{array}{c}IJ-1=\left(5\times 6\right)-1\\ =30-1\\ =29.\end{array}$

The error df is:

$\begin{array}{c}I\left(J-1\right)=5\times \left(6-1\right)\\ =5\times 5\\ =25.\end{array}$

The numerator degrees of freedom is:

$\begin{array}{c}{\nu }_1=I-1\\ =4.\end{array}$

The denominator degrees of freedom is:

$\begin{array}{c}{\nu }_2=I\left(J-1\right)\\ =25.\end{array}$

Calculation for test statistic:

The sample total corresponding to the i^th treatment is:

$X_{i.}={\displaystyle \sum _{j=1}^{J_i}{X}_{ij}},i=1,2,\mathrm{...},I$.

The sample means are ${\overline{X}}_{1.}=13.68$, ${\overline{X}}_{2.}=15.95$, ${\overline{X}}_{3.}=13.67$, ${\overline{X}}_{4.}=14.73$, ${\overline{X}}_{5.}=13.08$.

Thus, the corresponding sample totals are:

$\begin{array}{c}{X}_{1.}=J_i{\overline{X}}_{1.}\\ =6\times 13.68\\ =\mathrm{82.1.}\end{array}$

$\begin{array}{c}{X}_{2.}=J_i{\overline{X}}_{1.}\\ =6\times 15.95\\ =\mathrm{95.7.}\end{array}$

$\begin{array}{c}{X}_{3.}=J_i{\overline{X}}_{1.}\\ =6\times 13.67\\ =\mathrm{82.0.}\end{array}$

$\begin{array}{c}{X}_{4.}=J_i{\overline{X}}_{1.}\\ =6\times 14.73\\ =\mathrm{88.4.}\end{array}$

$\begin{array}{c}{X}_{5.}=J_i{\overline{X}}_{1.}\\ =6\times 13.08\\ =\mathrm{78.5.}\end{array}$

The grand total is:

$\begin{array}{c}{X}_{\cdot \cdot }={\displaystyle \sum _{i=1}^I{X}_{i.}}\\ =82.1+95.7+82.0+88.4+78.5\\ =\mathrm{426.7.}\end{array}$.

Thus, the grand mean is:

$\begin{array}{c}\overline{X}\mathrm{..}=\frac{1}{IJ}{\displaystyle \sum _{i=1}^I{\displaystyle \sum _{j=1}^J{X}_{ij}}}\\ =\frac{1}{IJ}{\displaystyle \sum _{i=1}^I{X}_{i.}}\\ =\frac{1}{IJ}{X}_{\mathrm{..}}\\ =\frac{426.7}{30}\end{array}$

$=\mathrm{14.22.}$

The mean square of treatments, MSTr is:

$\begin{array}{c}\text{MSTr}=\frac{\text{SSTr}}{I-1}\\ =\frac{J}{I-1}{\displaystyle \sum _{i=1}^I{\left({\overline{X}}_{i.}-{\overline{X}}_{\mathrm{..}}\right)}^2}\\ =\frac{6}{4}\left[\begin{array}{l}{\left(13.68-14.22\right)}^2+{\left(15.95-14.22\right)}^2+{\left(13.67-14.22\right)}^2\\ +{\left(14.73-14.22\right)}^2+{\left(13.08-14.22\right)}^2\end{array}\right]\\ =\frac{3}{2}\left[0.2916+2.9929+0.3025+0.2601+1.2996\right]\end{array}$

$=7.72$.

Thus, the sum of squares of treatment is:

$\begin{array}{c}\text{SSTr}=\text{MSTr}\times \left(I-1\right)\\ =7.72\times 4\\ =\mathrm{30.88.}\end{array}$

The total sum of squares, SST is:

$\text{SST}={\displaystyle \sum _{i=1}^I{\displaystyle \sum _{j=1}^J{X}_{ij}{}^2}}-\frac{1}{IJ}{X}^2{}_{\cdot \cdot }$

The calculation for ${\displaystyle \sum _{j=1}^J{X}_{ij}{}^2}$ is shown in the following table:

Brand 1	$X_{1j}$	13.1	15	14	14.4	14	11.6
	$X^2{}_{1j}$	171.61	225	196	207.36	196	134.56	$\begin{array}{l}{\displaystyle \sum _{j=1}^J{X}_{1j}{}^2}\\ =1,130.53\end{array}$
Brand 2	$X_{2j}$	16.3	15.7	17.2	14.9	14.4	17.2
	$X^2{}_{2j}$	265.69	246.49	295.84	222.01	207.36	295.84	$\begin{array}{l}{\displaystyle \sum _{j=1}^J{X}_{2j}{}^2}\\ =1,533.23\end{array}$
Brand 3	$X_{3j}$	13.7	13.9	12.4	13.8	14.9	13.3
	$X^2{}_{3j}$	187.69	193.21	153.76	190.44	222.01	176.89	$\begin{array}{l}{\displaystyle \sum _{j=1}^J{X}_{3j}{}^2}\\ =1,124.00\end{array}$
Brand 4	$X_{4j}$	15.7	13.7	14.4	16	13.9	14.7
	$X^2{}_{4j}$	246.49	187.69	207.36	256	193.21	216.09	$\begin{array}{l}{\displaystyle \sum _{j=1}^J{X}_{4j}{}^2}\\ =1,306.84\end{array}$
Brand 5	$X_{5j}$	13.5	13.4	13.2	12.7	13.4	12.3
	$X^2{}_{5j}$	182.25	179.56	174.24	161.29	179.56	151.29	$\begin{array}{l}{\displaystyle \sum _{j=1}^J{X}_{5j}{}^2}\\ =1,028.19\end{array}$

Thus,

$\begin{array}{c}\text{SST}={\displaystyle \sum _{i=1}^I{\displaystyle \sum _{j=1}^J{X}_{ij}{}^2}}-\frac{1}{IJ}{X}^2{}_{\cdot \cdot }\\ =\left(1,130.53+1,533.23+1,124.00+1,306.84+1,028.19\right)-\frac{{\left(426.7\right)}^2}{30}\\ =6,122.79-\frac{182,072.9}{30}\\ =6,122.79-6,069.096\end{array}$

$=53.6937$.

Now,

$\text{SST}=\text{SSTr}+\text{SSE}$.

Thus,

$\begin{array}{c}\text{SSE}=\text{SST}-\text{SSTr}\\ =53.6937-30.88\\ =\mathrm{22.8137.}\end{array}$

As a result,

$\begin{array}{c}\text{MSE}=\frac{\text{SSE}}{I\left(J-1\right)}\\ =\frac{22.8137}{25}\\ =\mathrm{0.91.}\end{array}$

Thus, the F-statistic is:

$\begin{array}{c}F=\frac{\text{MSTr}}{\text{MSE}}\\ =\frac{7.72}{0.91}\\ =\mathrm{8.48.}\end{array}$

Level of significance:

The level of significance here is $\alpha =0.05$.

Critical value:

The critical value for the $F_{{\nu }_1,{\nu }_2}$ distribution at level of significance $\alpha$ is $F_{\alpha ;{\nu }_1,{\nu }_2}$, which is the value of the $F_{{\nu }_1,{\nu }_2}$-distribution, the probability above which is $\alpha$.

Here, the critical value would be $F_{0.05;4,25}$. From the Table A.9, “Critical Values for F Distributions”, $F_{0.05;4,25}=2.76$.

P-value:

The P-value is the area to the right of the F-statistic value f, under the $F_{{\nu }_1,{\nu }_2}$ distribution curve, that is, $P\text{-value}=P\left(F_{{\nu }_1,{\nu }_2}>f\right)$

Hence, the P-value is $P\text{-value}=P\left(F_{4,25}>8.48\right)$.

Here, the test statistic value, $f=8.48$, which is greater than the value 6.49, corresponding to $F_{0.001;4,25}$, thus, evidently, greater than $F_{0.05;4,25}=2.76$.

That is, $f\left(=8.48\right)>F_{0.001;4,25}\left(=6.49\right)>F_{0.05;4,25}\left(=2.76\right)$.

Rejection rule:

If the P-value is less than the level of significance $\alpha$, such that $P\text{-value}<\alpha$, then reject the null hypothesis at level $\alpha$.

Conclusion:

Here, the P-value is less than 0.001, which is less than the significance level 0.05.

That is, $P\text{-value}<0.001<0.05$.

Thus, the decision is “reject the null hypothesis”.

Therefore, the data provide sufficient evidence to conclude that the effects of all the treatments are not similar.

Thus, the F test reveals at $\alpha =0.05$ that there is significant difference between the effects of at least two of the hormones on plant growth .

Hence, it is appropriate to use a multiple comparisons method, in order to investigate the differences amongst the means.

Multiple comparisons method- Tukey’s procedure:

For Tukey’s test, first find the value of $w=Q_{\alpha ,I,I\left(J-1\right)}\cdot \sqrt{\frac{\text{MSE}}{J}}$.

The level of significance is 0.05.

From Table A.10 “Critical Values for Studentized Range Distribution”, the value of $Q_{0.05;5,25}$ is unavailable. Consider the value $Q_{0.05;5,24}=4.17$ and $Q_{0.05;5,30}=4.10$.

It can be observed that at level of significance 0.05, for a particular value of first df, I, the critical value reduces by $0.07\left(=4.17-4.10\right)$ for increase in the second df, $I\left(J-1\right)$.

Assuming that the change is approximately uniform over the interval of the two consecutive second df values, 24 and 30, the unitary method gives that for unit increase in the second df from 24 to 25 reduces the critical value by approximately 0.011.

Hence, $Q_{0.05;5,25}\approx 4.158$.

The value of w is,

$\begin{array}{c}w=4.158\sqrt{\frac{0.91}{6}}\\ =4.158\times 0.389\\ \approx \mathrm{1.62.}\end{array}$

The averages for the brands are arranged in ascending order and their corresponding differences are calculated as:

Brand	5		3		1		4		2
Mean $\overline{x_i}$	13.08		13.67		13.68		14.73		15.95
Difference		0.59		0.01		1.05		1.22

The difference between means for each pair of consecutive brands is less than $w=1.62$. Thus, each consecutive pair forms a group.

It is evident the difference between the means for brands 5 and 1 is $0.60\left(=0.59+0.01\right)$ , which is less than w. Thus, brands 5, 3 and 1 form a group.

Now, the difference between the means for brands 5 and 4 is $1.65\left(=0.59+0.01+1.05\right)$ which is greater than w. Evidently, brands 5 and 1 are significantly different. The brand 5 must also be significantly different from the brands with higher means, such as brands 4 and 2.

The difference between brands 3 and 4 is $1.06\left(=0.01+1.05\right)$, which is less than w. Thus, brands 3, 1 and 4 form a group.

The difference between the means for brands 3 and 2 is $2.28\left(=0.01+1.05+1.22\right)$ which is greater than w. Evidently, brands 3 and 2 are significantly different.

Now, the between brands 1 and 2 is $2.27\left(=1.05+1.22\right)$, which is greater than w. Evidently, brands 1 and 2 are significantly different.

Join the consecutive means, whose differences are less than w by a line segment. Discontinue the line segment wherever the difference exceeds w and repeat.

Thus, the following arrangement is found:

$\begin{array}{l}\text{ 5 3 1 4 2}\\ \underset{\_}{\text{13}\text{.08 13}\text{.67 13}\text{.68}}\text{ 14}\text{.73 15}\text{.95}\\ \overline{\underset{\_}{}}\\ \overline{}\end{array}$

The brands 5, 3 and 1 have similar motor vibrations, brands 5, 3 and 1 have similar motor vibrations, the brands 1 and 4 have similar motor vibrations and the brands 4 and 2 have similar motor vibrations. There is significant difference of brand 5 with brands 4 and 2, and also significant difference between brands 3 and 2.