Question

A new process has been developed for applying photoresist film to 125 mm silicon wafers used in manufacturing integrated circuits. Ten wafers were tested, and the following photoresist film thickness measurements were observed: 13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946 and 13.4002 angstroms (1 angstrom = 1x 10- metre). The photoresist film process has shown from past results that the distribution is normal. Use a = 0.05 and s = .00391 a. Test the hypothesis that the mean thickness is 13.4 angstroms (Therefore a 2 tail test). Solve by working through the 5 step process and by using the "P value" approach b. Find the 99% confidence interval on the mean photoresist film thickness. Use a = 0.01. a. Test the Hypothesis Step 1: The Set-Up i. Describe the population parameter of interest. li. State the null and alternative hypothesis. Step 2: The Hypothesis Test Criteria i. Check the assumptions. ii. identify the probability distribution and test statistic to be used. ill. Determine the level of significance, a Step 3: The Sample Evidence i. Collect the sample information. ii. Calculate the value of the Test Statistic.

Answer 1

a)

Ho :   µ =   13.4                  
Ha :   µ ╪   13.4       (Two tail test)          
                          
Level of Significance ,    α =    0.05                  
population std dev ,    σ =    0.00391                 
Sample Size ,   n =    10                  
Sample Mean,    x̅ = ΣX/n =    13.396                       
                          
Standard Error , SE = σ/√n =   0.0039   / √    10   =   0.0012      
Z-test statistic= (x̅ - µ )/SE = (   13.396   -   13.4   ) /    0.0012   =   -3.1
                             
                          
p-Value   =   0.0020   [ Excel formula =NORMSDIST(z) ]              
Decision:   p-value<α, Reject null hypothesis

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b)
Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.0039   / √   10   =   0.001236
margin of error, E=Z*SE =   2.5758   *   0.00124   =   0.00318
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    13.40   -   0.003185   =   13.392995
Interval Upper Limit = x̅ + E =    13.40   -   0.003185   =   13.399365
99%   confidence interval is (   13.39   < µ <   13.40   )

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