Question

Manufacturing integrated circuits is an enormously complicated task, as
there are many process variables that can be manipulated (thickness of this,
width of that, doping level of something else, etc). Generally speaking, many
copies of a circuit are put onto a single wafer, which is made all at once. We
have an experiment where we are varying two factors, each at two levels.
We have 20 wafers and assign each of the four factor/level combinations to
five wafers at random and make the wafer. We then choose three circuits at
random on each wafer and measure the performance of the circuit. Construct
a skeleton ANOVA table (that is, just the sources and degrees of freedom).

Answer 1

Soln

	Factor 2
Factor 1	Level 1	Level 2
Level 1	1	1
	2	2
	3	3
	4	4
	5	5
Level 2	1	1
	2	2
	3	3
	4	4
	5	5

Null and Alternative Hypothesis:

We will have three hypotheses:

Factor 1

H₀: µ_{Level 1} = µ _{Level 2}

H₁: Not all Means are equal

Factor 2

H₀: µ_{Level 1} = µ _{Level 2}

H₁: Not all Means are equal

H₀: An interaction is absent

H₁: Interaction is present

Alpha = 0.05

Degress of Freedom:

Df_{Factor 1}(A) = a-1 = 2-1 = 1

Df_{Factor 2} (B) = b-1 = 2-1 = 1

df _{Factor 1 * Factor 2} (A*B) = (a-1) * (b-1) = 1*1 = 1

df _error = N – ab = 20 – 2*2 = 16

df _total= N – 1 = 20 – 1 = 19

Critical Values:

Factor 1 (df_{Factor 1}(A), df _error): (1,16) = 4.49

Factor 2 (df_{Factor 2} (B),df _error): (1,16) = 4.49

Interaction (df _{Factor 1 * Factor 2} (A*B), df _error) : (1,16) = 4.49

[Factor 1] If F is greater than 4.49, reject the null hypothesis

[Factor 2] If F is greater than 4.49, reject the null hypothesis

[Interaction] If F is greater than 4.49, reject the null hypothesis

ANOVA Table Format

	SS	Df	MS	F
Factor 1		1
Factor 2		1
Interaction		1
Error		16
Total		19