An experiment was conducted in a low-pressure vapor deposition of polysilicon to

ID: 3375422 • Letter: A

Question

An experiment was conducted in a low-pressure vapor deposition of polysilicon to investigate the effect of wafer position on film thickness uniformity. Wafer positions in the reactor were randomly selected. Three replicates of the experiment were run. The data are presented in the table below. (a) State the hypothesis and conduct an analysis of variance at a-0.05 to test the hypothesis that there is a difference in film thickness due to wafer position. (b) Plot and analyze the residuals from the experiment and comment on model adequacy Uniformit Replicate Wafer Position 2.76 4.17 3.49 2.38 2.70 3.24 2.44 1.97 1.37 4 0.94 1.46 1.85

Explanation / Answer

Solution

Final ANOVA table is given below. Concept Base and Detailed Calculations follow:

ANOVA

? =

0.05

Source

Fcrit

p-value

Row

7.447158

2.482386

8.398972

4.06618

0.007455

Error

2.364467

0.295558

Total

9.811625

Decision: Since F > Fcrit, (also p-value < ?), null hypothesis of no wafer position effect is rejected.

Conclusion: There is enough evidence to suggest that the claim that ‘There is a difference in film thickness due to wafer position.’ is valid ANSWER

Detailed Calculations

xij

xi.

sumxij^2

2.76

4.17

3.49

10.42

37.1866

28.77

2.38

2.7

3.24

8.32

23.452

68.976

2.44

1.97

1.37

5.78

11.7114

SST

9.8116

0.94

1.46

1.85

4.25

6.4377

SSR

7.4472

SSE

2.3645

Concept Base

Let xij represent the jth observation (replicate) in the ith row (wafer position), j = 1,2, 3; i = 1,2, 3, 4

Then the ANOVA model is: xij = µ + ?i + ?ij, where µ = common effect, ?i = effect of ith row, and ?ij is the error component which is assumed to be Normally Distributed with mean 0 and variance ?2.

Claim: There is a difference in film thickness due to wafer position.

Null hypothesis: H0: ?1 = ?2 = ?3 = ?4 = 0 Vs Alternative: HA: H0 is false [at least one ?i is different from others, i.e., claim]

Now, to work out the solution,

Terminology:

Row total = xi.= sum over j of xij

Grand total = G = sum over i of xi.

Correction Factor = C = G2/N, where N = total number of observations = r x n =

Total Sum of Squares: SST = (sum over i,j of xij2) – C

Row Sum of Squares: SSR = {(sum over i of xi.2)/(n)} – C

Error Sum of Squares: SSE = SST – SSR

Mean Sum of Squares = Sum of squares/Degrees of Freedom

Fcrit: upper ?% point of F-Distribution with degrees of freedom n1 and n2, where n1 is the DF for Row and n2 is the DF for Error

Significance: Fobs is significant if Fobs > Fcrit

DONE