Hi Please , l want to you to answer the All questions ( 1 , 2 , 3 , 4 , 5 ) in t

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Please , l want to you to answer the All questions ( 1 , 2 , 3 , 4 , 5 ) in the Quiz page

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Quiz 4 Number 3.32 (page 136) Write model and define all terms in the model. 2. Test the hypothesis about the variance component wafer position. Make sure you define the null and alternate hypothesis, report the p-value and draw conclusions. 3.. Estimate the random error term (o2) 4. Estimate the variability due to wafer position (or) 5. The total variation due to the model is V(y)- o2 of What percent of the total variation is due to the wafer position?

Explanation / Answer

Solution

Part 1 Model

Let xij = film thickness uniformity of jth replicate with Wafer Position i, j = 1, 2, 3; i = 1, 2, 3, 4

Then the model is: xij = µ + i + ij where µ = common effect, i = effect of ith wafer position and ij = random error which is assumed to be Normally distributed with mean 0 and variance 2.

Part 2

Test – ANOVA with one-way classification

Null hypothesis H0: 1 = 2 = 3 = 4 = 0, i.e., wafer effect is nil.

Vs Alternative HA: H0 is false.

The ANOVA calculations are shown in the table below:

Wafer position

xij

Row

Total (xi.)

Rep1

Rep2

Rep3

1

2.76

5.67

4.49

12.92

2

1.43

1.70

2.19

5.32

3

2.34

1.97

1.47

5.78

4

0.94

1.36

1.65

3.95

Row total = xi. = sum(xij)(j = 1,2,3 )

Grand Total = GT = sum(xi.)(i = 1,2,3,4) = 27.97

Correction Factor =(GT)2/12(total number of observations)

                              = 65.1934

Total Sum of Squares, SST = sum(xij2) – CF = 21.4373

Sum of Squares due to Wafer Position, SSW = sum(xi.2/3) – CF = 16.2198

[3 being the numberof replicates per wafer position]

Error Sum of Squares = SST – SSW = 5.2175

ANOVA TABLE

Source of

variation

Degrees of

Freedom

Sum of

squares

Mean sum

of squares

F

Wafer position

3

SSW = 16.2198

SSW/3 = 5.4066

MSSW/MSSE =

20.73

Error

20

SSE = 5.2175

SSE/20 = 0.2609

Total

23

SST = 21.4373

SST/23 = 0.9321

degrees of freedom: Wafer position: No of wafer position - 1; Total: Total number of observations and fro error it is DF for Total - DF for Wafer position

Upper 5% point of F-distribution with degrees of freedom 3, 20 is 8.64[from Standard Statistical Tables]

Since calculated value of F (20.73) > 8.64, H0 is rejected.

=> there is evidence to suggest that wafer position has an effect on thickness uniformity.

Part (3)

Estimate of random error = 0.2609 (MSSE)

Part (4)

Estimate of variability due to wafer position = 5.4066 (MSSW)

Part (5)

Estimate of total variability in the model = 5.6675 (MSSE + MSSW)

Wafer position

xij

Row

Total (xi.)

Rep1

Rep2

Rep3

1

2.76

5.67

4.49

12.92

2

1.43

1.70

2.19

5.32

3

2.34

1.97

1.47

5.78

4

0.94

1.36

1.65

3.95

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