Question 3

Fixed effects in ANOVA refer to the factors having various levels which are introduced on purpose by the researcher. In sharp contrast, random effects in ANOVA refer to those factors which are obtaining during sampling and are not intended by the researcher.

Question 4

OFAT stands for One Factor At a Time and essentially implies method where only one factor can be changed at a time while the others remain constant. This is in sharp contrast with DOE where two or more factors are changed simultaneously.

Advantages of OFAT over DOE

1) Easier to understand and implement.

2) Preferred by non-experts having limited knowledge of statistical inference

3) In certain situations, OFAT is preferable to DOE.

Disadvantages of OFAT over DOE

1) Interaction estimation cannot be done through OFAT.

2) The amount of runs required to achieve same precision as DOE is quite high.

3) The optimal conditions may be difficult to achieve.

Question 5

The response variable is essentially the dependent variable which tends to alter on account of the treatment or changes in the independent variable and is measures by the experimenter for drawing conclusion.

Question 6

The result from the single factor experiment regarding adhesive and bonding strength of paper is highlighted below. 

Level of significance = 0.05

Computation

Correction factor (CF) = (135.1)² / 12 = 1521.001

Total sum of square TSS = 1591.39 -1521.001 = 70.389

Sum of squares between rows SSR = (6312.45/4)- 1521.001 = 57.11167

Sum of squares between column SSC =(4597.29/3) -1521.001 =11.429

Square due to error SSE = TSS –SSR –SSC =70.389-57.11167-11.429 = 1.848

Two Way ANOVA Test Result

Critical values of F from F distribution table

• For treatment, the critical values of F for (2, 6) from F distribution table at 0.05 significance level comes out to be 5.1433.
• For varieties, the critical value F for (3,6) from F distribution table at 0.05 significance level comes out to be 4.7571.

It can be seen that calculated F values is higher than critical values (92.697>5.1433) and (12.367>4.7571) and hence sufficient evidence is present to reject the null hypothesis and to accept the alternative hypothesis. Thus, at least one group mean is not same.

Question 7

ANOVA Table

• Degree of freedom for single factor in experiment = 70.88 /14.18 = 5
• Number of degree of freedom for error = 23-5 = 18
• Sum of square for the factor = 108.63-37.75 = 70.88
• The mean square for error = 37.75/18 =2.10
• The value of test statistic = 14.18/2.10 =6.67
• 6 level of factor has been used in this experiment.
• 4 times the student has replicated the experiment.
• The sum of square for the block is 12 and hence, the ANOVA display would be changed.

ANOVA display

Reduction in the estimation of the experiment error = (2.10-1.72)/2.10 = 18%

Question 8

1. A) In the given case, the response variable is the product grain size while the independent variable is the stirring rate and the underlying process. The experiment also contains a nuisance or unwanted variable in the form of furnace type since none of the four furnaces are same and comparable. Thereby the experiment seeks to understand the relation between the variables of choice while accounting for the nuisance variable.
2. B) The analysis of variance model would be used as the preferred model for this experiment. This would tend to outline two type of effects namely fixed effects and random effects. The fixed effect is caused by the changes in process (stirring speed) with regards to the grain size. Further, the random effect would highlight the magnitude of effect which is being produced on account of the nuisance variable or the differences in furnace. This model can successfully predict whether the underlying relationship between the variables of interest is statistically significant or not.
3. C) The respective degree of freedoms are summarised as indicated below.

Question 9

There are three main assumptions for ANOVA model as highlighted below.

1) Normality – Highlighted by residual normal distribution

2) Randomness

3) Homoscedasticity implying equality of variances

Based on the given diagrams, following conclusions can be made.

1) Normality – This assumption is satisfied as the normal probability plot for residuals is almost linear which is indicative of the distribution being normal. Additional evidence is provided by the histogram which seems to be approximately symmetric and fitting in a bell curve indicating fulfilment of this assumption.

2) Randomness – This assumption is also fulfilled which is apparent from the random positions occupied by the points in the scatter plot.

3) Homoscedasticity – Considering the versus fit scatter plot, it is apparent that over the entire plot, the width of the points tend to remain constant and there is no major variation in the width. This highlights that this assumption is also fulfilled.