Chapter 11.4, Problem 39E

The accompanying data resulted from a 2³ experiment with three replications per combination of treatments designed to study the effects of concentration of detergent (A), concentration of sodium carbonate (B), and concentration of sodium carboxymethyl cellulose (C) on the cleaning ability of a solution in washing tests (a larger number indicates better cleaning ability than a smaller number).

Factor Levels

1. a. After obtaining cell totals x_ijk., compute estimates of β₁, ${\gamma }_{11}^{AC}$, and ${\gamma }_{12}^{AC}$.
2. b. Use the cell totals along with Yates’s method to compute the effect contrasts and sums of squares. Then construct an ANOVA table and test all appropriate hypotheses using α = .05.
3. c. Suppose a low water temperature has been used to obtain the data. The entire experiment is then repeated with a higher water temperature to obtain the following data. Use Yates’s algorithm on the entire set of 48 observations to obtain the sums of squares and ANOVA table, and then test appropriate hypotheses at level .05.

Condition	Observations
d	144, 154, 158
ad	239, 227, 244
bd	232, 242, 246
abd	364, 362, 346
cd	194, 162, 203
acd	284, 295,291
bed	291, 287, 297
abed	411,406, 395

To determine

Compute the estimates of ${\beta }_1$, ${\gamma }_{11}^{AC}$ and ${\gamma }_{21}^{AC}$.

Answer to Problem 39E

The estimates of ${\beta }_1$, ${\gamma }_{11}^{AC}$ and ${\gamma }_{21}^{AC}$ are 54.38, –2.21 and 2.21 respectively.

Explanation of Solution

Given info:

Inan $2^3$ experiment to study the effects of concentration of detergents A, concentration of sodium carbonate B and sodium carboxymethyl cellulose C on the cleaning ability. The treatment combinations were replicated three times for each combination of treatments.

Calculation:

Yates method:

• Write the combination of treatments in the standard order in the beginning of the table called treatment column.
• Find the total effect obtained from each treatment combination and write it in the second column $x_{ijk\cdot }$.
• Find columns $\left(1\right),\left(2\right),\mathrm{...},\left(n\right)$

1. 1. Calculate column $\left(1\right)$ by using the column $x_{ijk\cdot }$ and the procedure is given below:
2. 2. Upper half is obtained by adding the pairs of values in the column $x_{ijk\cdot }$
3. 3. Lower half is obtained by taking the differences in pairs. That is, by subtracting the first value of pairs with the second value.
4. 4. Similarly the columns $\left(2\right),\left(3\right),\mathrm{...},\left(n\right)$ are obtained by proceeding in the same manner as mentioned above.

• Finally, this process is repeated for n times in $2^n$ factorial experiment.
• Sum of squares due to contrast is obtained by using the formula given below:
• $SS=\frac{{\left(\text{contrast}\right)}^2}{24}$

The table below shows the Yates method of computation for the given data:

Treatment (1)	$x_{ijk\cdot }$ (2)	1 (3)	2 (4)	Effect contrast (5)	$SS=\frac{{\left(\text{contrast}\right)}^2}{24}$ (6)
$\left(1\right)$	315	927	2,478	5,485
$\left(a\right)$	612	1,551	3,007	1,307	71,177.04
$\left(b\right)$	584	1,163	680	1,305	70,959.38
$\left(ab\right)$	967	1,844	627	199	1,650.042
$\left(c\right)$	453	297	624	529	11,660.04
$\left(ac\right)$	710	383	681	–53	117.0417
$\left(bc\right)$	737	257	86	57	135.375
$\left(abc\right)$	1,107	370	113	27	30.375

Where,

The column $x_{ijk\cdot }$ is obtained by adding the given values in pairs. That is,

For $\left(1\right)$ it is,

$106+93+116=315$

For $\left(a\right)$ it is,

$198+200+214=612$

Similarly the remaining values are obtained.

The column (3) named as 1 is calculated by using the following method:

For $\left(1\right)$ it is the addition of first two values from the column (2) named as $x_{ijk\cdot }$ and it is shown below:

$315+612=927$

The same method is followed for the remaining 7 rows.

The second half of column (3) named as 1is calculated by find the differences in pairs of values from the column (2) named as $x_{ijk\cdot }$

$612-315=297$

Then,

$967-584=383$

The fourth column named as 2, fifth column named as 3 and sixth column named as effect contrast were also calculated by using the method as given above.

Finally, the $SS=\frac{{\left(\text{contrast}\right)}^2}{24}$ is calculated by taking the square of Effect contrast and then dividing it by 24 since n is three.

The general formula for SS is $SS=\frac{{\left(\text{contrast}\right)}^2}{8\left(n\right)}$, the denominator n is multiplied by a factor 8 because there are eight treatment combinations in $2^3$ experiment and n is three since there are threeobservations for each treatment combinations.

$\begin{array}{c}{\widehat{\beta }}_1={\overline{x}}_{\cdot 2\cdot }-{\overline{x}}_{\cdot \cdot \cdot }\\ =\frac{x_{121\cdot }+x_{121\cdot }+x_{122\cdot }+x_{222\cdot }-x_{111\cdot }-x_{211\cdot }-x_{112\cdot }-x_{212\cdot }}{8n}\\ =\frac{584+967+737+1,107-315-612-453-710}{8\left(3\right)}\\ =\frac{1,305}{24}\end{array}$

$\approx 54.38$

$\begin{array}{c}{\widehat{\gamma }}_{11}^{AC}=\frac{x_{111\cdot }-x_{211\cdot }+x_{121\cdot }-x_{221\cdot }-x_{112\cdot }+x_{212\cdot }-x_{122\cdot }+x_{222\cdot }}{8n}\\ =\frac{315-612+584-967-453+710-737+1,107}{\left(8\right)\left(3\right)}\\ =\frac{-53}{24}\\ =-2.21\end{array}$

Thus, the value of ${\widehat{\gamma }}_{11}^{AC}$ is –2.21.

As,

$\begin{array}{c}{\widehat{\gamma }}_{21}^{AC}=-{\widehat{\gamma }}_{11}^{AC}\\ =-\left(-2.21\right)\\ =2.21\end{array}$

Thus, the value of ${\widehat{\gamma }}_{21}^{AC}$ is 2.21.

To determine

Compute the effect contrasts and sum of squares.

Construct the ANOVA table and test all the hypotheses using 5% level of significance.

Answer to Problem 39E

The effect contrasts and sum of squares are given below:

Treatment (1)	Effect contrast (5)	$SS=\frac{{\left(\text{contrast}\right)}^2}{24}$ (6)
$\left(1\right)$	5,485
$\left(a\right)$	1,307	71,177.04
$\left(b\right)$	1,305	70,959.38
$\left(ab\right)$	199	1,650.042
$\left(c\right)$	529	11,660.04
$\left(ac\right)$	–53	117.0417
$\left(bc\right)$	57	135.375
$\left(abc\right)$	27	30.375
Total		155,729.29

The ANOVA table is given below:

Effect	Degrees of freedom	Sum of squares	Mean sum of squares	F-ratio
$\left(a\right)$	1	71,177.04	71,177.04	436.56
$\left(b\right)$	1	70,959.38	70,959.38	435.23
$\left(ab\right)$	1	1,650.042	1,650.042	10.12
$\left(c\right)$	1	11,660.04	11,660.04	71.52
$\left(ac\right)$	1	117.0417	117.0417	0.72
$\left(bc\right)$	1	135.375	135.375	0.83
$\left(abc\right)$	1	30.375	30.375	0.19
Error	16	2,608.7	163.04
Total	48	158,337.6

There is sufficient evidence to conclude that at least one of the main or the interaction effects is significant at 5% level of significance and those significant effects are A, B, C, D and AB.

Explanation of Solution

Given info:

Use the Yates table.

Calculation:

The effect contrasts and sum of squares are given below:

Treatment (1)	$x_{ijk\cdot }$ (2)	1 (3)	2 (4)	Effect contrast (5)	$SS=\frac{{\left(\text{contrast}\right)}^2}{24}$ (6)
$\left(1\right)$	315	927	2,478	5,485
$\left(a\right)$	612	1,551	3,007	1,307	71,177.04
$\left(b\right)$	584	1,163	680	1,305	70,959.38
$\left(ab\right)$	967	1,844	627	199	1,650.042
$\left(c\right)$	453	297	624	529	11,660.04
$\left(ac\right)$	710	383	681	–53	117.0417
$\left(bc\right)$	737	257	86	57	135.375
$\left(abc\right)$	1,107	370	113	27	30.375
Total					155,729.29

Effect contrast is calculated by adding the numbers in pairs for the first half values and finding the difference in pairs for the remaining half values from the column (4).

For first half,

$2,478+3,007=5,485$.

Similar method is following for the remaining pairs.

For second half,

$3,007-2,478=529$.

Similar method is following for the remaining pairs.

The $SS=\frac{{\left(\text{contrast}\right)}^2}{24}$ is calculated by taking the square of Effect contrast and then dividing it by 24 since n is three.

The general formula for SS is $SS=\frac{{\left(\text{contrast}\right)}^2}{8\left(n\right)}$, the denominator n is multiplied by a factor 8 because there are eight treatment combinations in $2^3$ experiment and n is three since there are three observations for each treatment combinations.

Total sum of squares is calculated as follows:

$\text{SST}={\displaystyle \sum {\displaystyle \sum {\displaystyle \sum {\displaystyle \sum x_{ijkl}^2}}}}-\frac{{\left(x_{\cdot \cdot \cdot \cdot }\right)}^2}{8n}$

${\displaystyle \sum {\displaystyle \sum {\displaystyle \sum {\displaystyle \sum x_{ijkl}^2}}}}$ is total obtained by squaring each response for a treatment and adding together and finally taking the grand total for all the treatments.

$x_{\cdot \cdot \cdot \cdot \cdot }$ is the addition of all $x_{ijk\cdot }$ values.

Substitute ${\displaystyle \sum {\displaystyle \sum {\displaystyle \sum {\displaystyle \sum x_{ijkl}^2}}}}$ as 1,411,889, n as 3 and

$x_{\cdot \cdot \cdot \cdot \cdot }$ is $5,485\left(=315+612+\mathrm{...}+737+1,107\right)$

$\begin{array}{c}\text{SST}=1,411,889-\frac{{\left(5,485\right)}^2}{8\left(3\right)}\\ =1,411,889-\frac{30,085,225}{24}\\ =1,411,889-1,253,551.042\\ \approx 158,337.96\end{array}$

Thus, the SST is 158,337.96.

Sum of squares due to error is calculated as follows:

$\begin{array}{c}\text{SSE}=\text{SST}-\text{SS}\left(\text{Remaining factors}\right)\\ =158,337.96-155,729.29\\ =2,608.67\\ \approx 2,608.7\end{array}$

Thus, the sum of squares due to error is 2,608.7.

Mean sum of squares is calculated as follows:

$\begin{array}{c}\text{MSE}=\frac{\text{SSE}}{16}\\ =\frac{2,608.7}{16}\\ \approx 163.04\end{array}$

Thus, the mean sum of squares is 163.04.

The ANOVA table is given below:

Effect	Degrees of freedom	Sum of squares	Mean sum of squares	F-ratio
$\left(1\right)$	1
$\left(a\right)$	1	71,177.04	71,177.04	436.56
$\left(b\right)$	1	70,959.38	70,959.38	435.23
$\left(ab\right)$	1	1,650.042	1,650.042	10.12
$\left(c\right)$	1	11,660.04	11,660.04	71.52
$\left(ac\right)$	1	117.0417	117.0417	0.72
$\left(bc\right)$	1	135.375	135.375	0.83
$\left(abc\right)$	1	30.375	30.375	0.19
Error	16	2,608.7	163.04
Total	48	158,337.6

Testing the hypotheses:

Null hypothesis:

$H_0:$ All the main and the interaction effects are not significant.

Alternative hypothesis:

$H_0:$ At least one of the main or the interaction effects is significant.

Critical value:

Software procedure:

Step-by-step procedure to obtain the critical value is given below:

• Click on Graph, select View Probability and click OK.
• Select F, enter 1 in numerator df and 16 in denominator df.
• Under Shaded Area Tab select Probability under Define Shaded Area By and select right tail.
• Choose Probability as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

Thus, the critical value is 4.494.

Conclusion:

The test statisticfor all the main effects and interaction AB is greater than the critical value is 4.494.

The test statisticis greaterthan the critical value.

Thus, the null hypothesis is rejected.

Hence, there is sufficient evidence to conclude that at least one of the main or the interaction effects is significant at 5% level of significance and those significant effects are A, B, C, D and AB.

Also, the remaining factors AC, BC and ABC are not significant at 5% level of significance.

To determine

Construct the ANOVA table and test all the hypotheses using 1% level of significance.

Answer to Problem 39E

The ANOVA table is given below:

Effect	Degrees of freedom	Sum of squares	Mean sum of squares	F-ratio
$\left(1\right)$	1
$\left(a\right)$	1	136,640	136,640	749.48
$\left(b\right)$	1	139,644.2	139,644.2	765.96
$\left(ab\right)$	1	2,173.521	2,173.521	11.92
$\left(c\right)$	1	24,616.02	24,616.02	135.02
$\left(ac\right)$	1	2.520833	2.520833	0.01
$\left(bc\right)$	1	165.0208	165.0208	0.91
$\left(abc\right)$	1	42.1875	42.1875	0.23
$\left(d\right)$	1	20,377.52	20,377.52	111.77
$\left(ad\right)$	1	58.52083	58.52083	0.32
$\left(bd\right)$	1	9.1875	9.1875	0.05
$\left(abd\right)$	1	117.1875	117.1875	0.64
$\left(cd\right)$	1	17.52083	17.52083	0.1
$\left(acd\right)$	1	188.0208	188.0208	1.03
$\left(bcd\right)$	1	13.02083	13.02083	0.07
$\left(abcd\right)$	1	204.1875	204.1875	1.12
Error	32	5,834.024	182.313
Total	48

There is sufficient evidence to conclude that at least one of the main or the interaction effects is significant at 5% level of significance and those significant effects are A, B, C, D and AB.

Explanation of Solution

Given info:

A fourth factor called low water temperature is added to the given data and its effects are observed.

Calculation:

The table below shows the Yates method of computation for the given data:

Condition (1)	$x_{ijkl\cdot }$ (2)	1 (3)	2 (4)	3 (5)	Effect contrast (6)	$SS=\frac{{\left(\text{contrast}\right)}^2}{48}$ (7)
$\left(1\right)$	315	927	2,478	5,485	11,959
$\left(a\right)$	612	1,551	3,007	6,474	2561	136,640
$\left(b\right)$	584	1,163	2,958	1,307	2589	139,644.2
$\left(ab\right)$	967	1,844	3,516	1,254	323	2,173.521
$\left(c\right)$	453	1,166	680	1,305	1087	24,616.02
$\left(ac\right)$	710	1,792	627	1,284	–11	2.520833
$\left(bc\right)$	737	1,429	606	199	89	165.0208
$\left(abc\right)$	1,107	2,087	648	124	–45	42.1875
$\left(d\right)$	456	297	624	529	989	20,377.52
$\left(ad\right)$	710	383	681	558	–53	58.52083
$\left(bd\right)$	720	257.00	626.00	–53.00	–21.00	9.1875
$\left(abd\right)$	1,072	370.00	658.00	42.00	–75.00	117.1875
$\left(cd\right)$	559	254.00	86.00	57.00	29.00	17.52083
$\left(acd\right)$	870	352.00	113.00	32.00	95.00	188.0208
$\left(bcd\right)$	875	311	98	27	–25	13.02083
$\left(abcd\right)$	1,212	337	26	–72	–99	204.1875
Total						324,268.646

All the calculations are same as given in part (a) but there is addition of extra rows because a factor D is added to the experiment.

Finally, the $SS=\frac{{\left(\text{contrast}\right)}^2}{48}$ is calculated by taking the square of Effect contrast and then dividing it by 16 since n is three.

The general formula for SS is $SS=\frac{{\left(\text{contrast}\right)}^2}{16\left(n\right)}$, the denominator n is multiplied by a factor 16 because there are sixteen treatment combinations in $2^4$ experiment and n is three since there are three observations for each treatment combinations.

Total sum of squares is calculated as follows:

$\text{SST}={\displaystyle \sum {\displaystyle \sum {\displaystyle \sum {\displaystyle \sum {\displaystyle \sum x_{ijklm}^2}}}}}-\frac{{\left(x_{\cdot \cdot \cdot \cdot \cdot }\right)}^2}{16n}$

${\displaystyle \sum {\displaystyle \sum {\displaystyle \sum {\displaystyle \sum {\displaystyle \sum x_{ijklm}^2}}}}}$ is total obtained by squaring each response for a treatment and adding together and finally taking the grand total for all the treatments.

$x_{\cdot \cdot \cdot \cdot \cdot \cdot }$ is the addition of all $x_{ijkl\cdot }$ values.

Substitute ${\displaystyle \sum {\displaystyle \sum {\displaystyle \sum {\displaystyle \sum {\displaystyle \sum x_{ijklm}^2}}}}}$ as 3,308,143, n as 3 and

$x_{\cdot \cdot \cdot \cdot \cdot \cdot }$ is $11,959\left(=315+612+\mathrm{...}+875+1,212\right)$

$\begin{array}{c}\text{SST}=3,308,143-\frac{{\left(11,956\right)}^2}{16\left(3\right)}\\ =3,308,143-\frac{142,945,936}{48}\\ =3,308,143-2,978,040.333\\ =330,102.67\end{array}$

Sum of squares due to error is calculated as follows:

$\begin{array}{c}\text{SSE}=\text{SST}-\text{SS}\left(\text{Remaining factors}\right)\\ =330,102.67-324,268.646\\ =5,834.024\end{array}$

Thus, the sum of squares due to error is 5,834.024.

Mean sum of squares is calculated as follows:

$\begin{array}{c}\text{MSE}=\frac{\text{SSE}}{32}\\ =\frac{5,834.024}{32}\\ \approx 182.313\end{array}$

Thus, the mean sum of squares is 182.313.

The ANOVA table is given below:

Effect	Degrees of freedom	Mean sum of squares	F-ratio
$\left(1\right)$	1
$\left(a\right)$	1	136,640	749.48
$\left(b\right)$	1	139,644.2	765.96
$\left(ab\right)$	1	2,173.521	11.92
$\left(c\right)$	1	24,616.02	135.02
$\left(ac\right)$	1	2.520833	0.01
$\left(bc\right)$	1	165.0208	0.91
$\left(abc\right)$	1	42.1875	0.23
$\left(d\right)$	1	20,377.52	111.77
$\left(ad\right)$	1	58.52083	0.32
$\left(bd\right)$	1	9.1875	0.05
$\left(abd\right)$	1	117.1875	0.64
$\left(cd\right)$	1	17.52083	0.1
$\left(acd\right)$	1	188.0208	1.03
$\left(bcd\right)$	1	13.02083	0.07
$\left(abcd\right)$	1	204.1875	1.12
Error	32	182.313
Total	48

Testing the hypothesis:

Null hypothesis:

$H_0:$ All the main and the interaction effects are not significant.

Alternative hypothesis:

$H_a:$ At least one of the main or the interaction effects is significant.

Critical value:

Software procedure:

Step-by-step procedure to obtain the critical value is given below:

• Click on Graph, select View Probability and click OK.
• Select F, enter 1 in numerator df and 32 in denominator df.
• Under Shaded Area Tab select Probability under Define Shaded Area By and select right tail.
• Choose Probability as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

Thus, the critical value is 4.149.

Conclusion:

The test statistic for all the main effects and interaction AB is greater than the critical value is 4.494.

The test statisticis greater than the critical value.

Thus, the null hypothesis is rejected.

Hence, there is sufficient evidence to conclude that at least one of the main or the interaction effects is significant at 5% level of significance and those significant effects are A, B, C, D and AB.

Also, the remaining factors AC, BC and ABC are not significant at 5% level of significance.