Design And Analysis Of Experiments
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Design and Analysis of Experiments Student: Nguyen Thi Hang ID: R9605005
Mark:
2.21. In semiconductor manufacturing wet chemical etching is often used to remove silicon from the backs of wafers prior to metallization. The etch rate is an important characteristic of this process. Two different etching solutions are being evaluated. Eight randomly selected wafers have been etched in each solution, and the observed etch rates (in mils/min) are as follow. Solution 1 9.9 9.4 10.0 10.3 10.6 10.3 9.3 9.8
Solution 2 10.2 10.0 10.7 10.5 10.6 10.2 10.4 10.3
(a) Do the data indicate that the claim that both solutions have the same meant etch rate is valid? Use α = 0.05 and assume equal variances. The experiment was carried out following: “two different etching solutions are being evaluated, Eight randomly selected wafers have been etched in each solution”, thus we can use the Two-Sample t-Test for comparing tow treatment means. We have two hypothesizes: H0: µ1 = µ2 H1: µ1 ≠ µ2
With α = 0.05 is used, the Minitab two-sample t-test software is applied for this comparison, the results are shown on below table: Two-sample T for Solution 1 vs Solution 2 Solution 1 Solution 2
N 8 8
Mean 9.950 10.363
StDev 0.450 0.233
SE Mean 0.16 0.082
Difference = mu (Solution 1) - mu (Solution 2) Estimate for difference: -0.412500 95% CI for difference: (-0.811832, -0.013168) T-Test of difference = 0 (vs not =): T-Value = -2.30 P-Value = 0.044 DF = 10
From this table, we find that P - value = 0.044 < α = 0.05, it means hypothesis H0 is rejected or solution’s mean differs. (b) Find a 95% percent confidence interval on the difference in mean etch rates 95% CI for difference: (-0.811832, -0.013168) (c) Use normal probability plots to investigate the adequacy of the
assumptions of normality and equal variance.
Percent (cumulative normal probability x 100)
Normal probability plots 99
Variable Solution1 Solution2
95
Mean StDev N AD P 9.95 0.4504 8 0.216 0.764 10.36 0.2326 8 0.158 0.919
90 80 70 60 50 40 30 20 10 5
1
8
9
10 rate (mils/ min)
11
12
Mark:
2.21. In semiconductor manufacturing wet chemical etching is often used to remove silicon from the backs of wafers prior to metallization. The etch rate is an important characteristic of this process. Two different etching solutions are being evaluated. Eight randomly selected wafers have been etched in each solution, and the observed etch rates (in mils/min) are as follow. Solution 1 9.9 9.4 10.0 10.3 10.6 10.3 9.3 9.8
Solution 2 10.2 10.0 10.7 10.5 10.6 10.2 10.4 10.3
(a) Do the data indicate that the claim that both solutions have the same meant etch rate is valid? Use α = 0.05 and assume equal variances. The experiment was carried out following: “two different etching solutions are being evaluated, Eight randomly selected wafers have been etched in each solution”, thus we can use the Two-Sample t-Test for comparing tow treatment means. We have two hypothesizes: H0: µ1 = µ2 H1: µ1 ≠ µ2
With α = 0.05 is used, the Minitab two-sample t-test software is applied for this comparison, the results are shown on below table: Two-sample T for Solution 1 vs Solution 2 Solution 1 Solution 2
N 8 8
Mean 9.950 10.363
StDev 0.450 0.233
SE Mean 0.16 0.082
Difference = mu (Solution 1) - mu (Solution 2) Estimate for difference: -0.412500 95% CI for difference: (-0.811832, -0.013168) T-Test of difference = 0 (vs not =): T-Value = -2.30 P-Value = 0.044 DF = 10
From this table, we find that P - value = 0.044 < α = 0.05, it means hypothesis H0 is rejected or solution’s mean differs. (b) Find a 95% percent confidence interval on the difference in mean etch rates 95% CI for difference: (-0.811832, -0.013168) (c) Use normal probability plots to investigate the adequacy of the
assumptions of normality and equal variance.
Percent (cumulative normal probability x 100)
Normal probability plots 99
Variable Solution1 Solution2
95
Mean StDev N AD P 9.95 0.4504 8 0.216 0.764 10.36 0.2326 8 0.158 0.919
90 80 70 60 50 40 30 20 10 5
1
8
9
10 rate (mils/ min)
11
12
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