Assuming that no interaction takes place between the factors, we can calculate the appropriate sums of squares, representing the grand total by $T$ .

The total sum of squares is given by

$S{S}_T=202^2+\dots +182^2+\dots +251^2-\frac{3524^2}{4\times 4}=789420-776161=13259$

The sum of squares for the treatments is given by

$S{S}_{Tr}=\frac{913^2+\dots +853^2}{5}-\frac{3524^2}{4\times 4}=3102808-776161=846$

The sum of the squares for the blocks is given by

$S{S}_{Bl}=\frac{750^2+\dots +1006^2}{4}-\frac{3524^2}{4\times 4}=786540-776161=10379$

The sum of squares for the errors is given by

$S{S}_E=S{S}_T-S{S}_{Tr}-S{S}_{Bl}=13259-846-10379=2034$

These calculations give rise to the following ANOVA table

Conclusions

1. Blocks

   From $F$ -tables, $F_{3,9}=3.86$ and since $1.25<3.86$ we have no evidence to reject the null hypothesis that there are no systematic differences between measurements from the different laboratories.

2. Treatments

   From the $F$ -tables, $F_{3,9}=3.86$ and since $15.31>3.86$ we have sufficient evidence to reject the hull hypothesis that the extrusion process has no effect on the tensile strength of the tubing. Hence our conclusion (at the 5% level of significance) is that the extrusion process does affect the tensile strength of the aluminium tubing.

The treatment totals are as follows:

1. Subtracting 80 from every observation will have no effect on the analysis of variance. The data now become as follows.

   Control	Cell (block)										Total
   Scheme	1	2	3	4	5	6	7	8	9	10
   A	0.27	$-0.56$	1.59	$-0.22$	0.39	1.92	2.87	2.04	3.41	4.52	16.23
   B	4.31	3.33	6.57	4.49	4.15	5.45	5.05	3.62	5.96	5.62	48.55
   C	3.59	0.36	4.55	0.03	1.59	0.75	2.61	5.20	4.29	1.60	24.57
   Total	8.17	3.13	12.71	4.30	6.13	8.12	10.53	10.86	13.66	11.74	89.35

   The total sum of squares for the 30 observations is given by

   $S{S}_T=0.27^2+\dots +1.60^2-\frac{89.35^2}{30}=388.5413-\frac{89.35^2}{30}=122.400$

   The sum of squares for the three control schemes is given by

   $S{S}_S=\frac{16.23^2+48.55^2+24.57^2}{10}-\frac{89.35^2}{30}=\frac{3224.2003}{10}-\frac{89.35^2}{30}=56.306$

   The sum of squares for the ten cells (blocks) is given by

   $S{S}_B=\frac{8.17^2+\dots +11.74^2}{3}-\frac{89.35^2}{30}=\frac{913.3349}{3}-\frac{89.35^2}{30}=38.331$ The residual sum of squares is given by

   $S{S}_E=S{S}_T-S{S}_S-S{S}_B=122.400-56.306-38.331=27.763$

   These calculations lead to the following ANOVA table.

   Source of Variation	Deg. of Freedom	Sum of Squares	Mean square	Variance Ratio
   Control Schemes	2	56.306	28.153	18.25
   Cells (blocks)	9	38.311	4.259	2.76
   Residuals (error)	18	27.763	1.542
   Total	29	122.400

   The following conclusions may be drawn.

   Control Schemes

   From $F$ -tables $F_{2,18}=3.55$ and since 18.25 $>$ 3.55 we have sufficient evidence to reject the null hypothesis that there is no difference in effect between the control schemes. It appears that Control Scheme $B$ gives the greatest current efficiency.

   Cells (blocks)

   From $F$ -tables $F_{9,18}=2.46$ and since 2.76 $>$ 2.46 we do not have sufficient evidence to reject the null hypothesis that there is no difference in mean current efficiency between the cells. Allowing for the cell effect may, nevertheless, have improved the sensitivity of the test for control scheme effects.

2. Since there is only one observation on each combination of batch and extrusion method, we must assume that there is no interaction between the factors. We can calculate the sums of squares as follows.

   Batch	Tensile Strength Method
   (Block)	1	2	3	4	Total
   1	202	186	182	180	750
   2	200	195	224	199	818
   3	259	243	225	223	950
   4	252	227	276	251	1006
   Total	913	851	907	853	3524

   The total sum of squares is given by

   $S{S}_T=202^2+\dots +251^2-\frac{3524^2}{16}=789420-776161=13259$

   The sum of squares for the extrusion methods is given by

   $S{S}_M=\frac{913^2+\dots +853^2}{4}-\frac{3524^2}{16}=777007-776161=846$

   The sum of squares for the blocks (batches) is given by

   $S{S}_B=\frac{750^2+\dots +1006^2}{4}-\frac{3524^2}{16}=786540-776161=10379$

   The residual sum of squares is given by

   $S{S}_E=S{S}_T-S{S}_M-S{S}_B=13259-846-10379=2034$

   These calculations lead to the following ANOVA table.

   Source of Variation	Deg. of Freedom	Sum of Squares	Mean Square	Variance Ratio
   Extrusion Methods	3	846	282.00	1.25
   Batches (blocks)	3	10379	3459.67	15.31
   Residual (error)	9	2034	226.00
   Total	15	13259

   The following conclusions may be drawn.

   Extrusion Methods

   From $F$ -tables $F_{3,9}=3.86$ and since 1.25 $<$ 3.86 we do not have sufficient evidence to reject the null hypothesis that the choice of extrusion method has no effect on the tensile strength of the tubes.

   Batches (blocks)

   From $F$ -tables $F_{3,9}=3.86$ and since 15.31 $>$ 3.86 we have sufficient evidence to reject the null hypothesis that mean tensile strength is the same in all batches. We conclude that mean tensile strength does differ between batches.

3. The treatment totals are as follows:

   Mechanism	A	B	C	D
   Total	14.11	14.88	15.16	14.09

   The row totals are as follows:

   Angle	1	2	3	4
   Total	15.82	14.91	13.97	13.54

   The column totals are as follows:

   Dummy	1	2	3	4
   Total	14.53	13.65	15.98	14.08

   The grand total is 58.24.

   The total sum of squares is

   $3.84^2+\cdots +3.41^2-\frac{58.24^2}{16}=2.1568.$

   The treatments sum of squares is

   $\frac{1}{4}\left(14.11^2+\cdots +14.09^2\right)-\frac{58.24^2}{16}=0.22145.$

   The angles sum of squares is

   $\frac{1}{4}\left(15.82^2+\cdots +13.54^2\right)-\frac{58.24^2}{16}=0.77465.$

   The dummies sum of squares is

   $\frac{1}{4}\left(14.53^2+\cdots +14.08^2\right)-\frac{58.24^2}{16}=0.76895.$

   The analysis of variance table is as follows.

   Source of variation	Degrees of freedom	Sum of squares	Mean square	Variance ratio
   Treatments (mechanisms)	3	0.22145	0.07382	1.131
   Angles	3	0.77465	0.25822	3.955
   Dummies	3	0.76895	0.25632	3.926
   Residual	6	0.39175	0.06529
   Total	15	2.15680

   The upper 5% point of the $F_{3,6}$ distribution is 4.76. Thus we can draw the following conclusions.

   1. Treatments (mechanisms)

      The variance ratio is not significant at the 5% level so we conclude that there is not sufficient evidence to reject the null hypothesis that the mechanisms give equal mean maximum accelerations.

   2. Angles

      The variance ratio is not significant at the 5% level so we conclude that there is not sufficient evidence to reject the null hypothesis that the angles have no effect on mean maximum accelerations.

   3. Dummies

      The variance ratio is not significant at the 5% level so we conclude that there is not sufficient evidence to reject the null hypothesis that the dummies have no effect on mean maximum accelerations.

   The assumptions made are as follows.

   • The observation in each cell is taken from a normal distributions.
   • Each of these normal distributions has the same variance.
   • The observations are independent.
   • The effects of the factors are additive. That is, there are no interactions.

   To allow investigation of the possibility of interactions we would need to make more than one observation in each cell.