Semiconductor wafer and device process yields are some of the most closely guard

Question

Semiconductor wafer and device process yields are some of the most closely guarded trade secrets in the semiconductor industry, and they measure how well a given semiconductor process can fabricate and package non-defective integrated circuits. Suppose a process used to manufacture CMOS chips produces the following sampling of process yields:

X = {68.5%, 77.2%, 78.6%, 74.3%, 75.4%, 77.4%, 48.6%, 69.7%, 72.7%, 76.2%}

Does the sample provided give any evidence that the mean yield is not 74%? Assume the standard deviation is 2.5%.

a) Findthemeanvalueofthesample.

b) Writeoutthedescriptionofthehypothesistestusingthestandardnotationand state whether it is one-sided or two-sided.

c) TestthehypothesisusingtheP-Valueapproach(hint:youneedtocomputethe z-standard value for the sample mean). What does the computed P-value suggest about the hypothesis?

d) Test the hypothesis using the fixed interval approach. Assume = 0.04. What is / are the z-values of the boundaries of the fixed interval? What does this test suggest about the hypothesis?

e) Suppose the true yield is actually 76%. What is the probability of a type II error under these circumstances? What is the power of this test? How many samples would you need to increase the power of the test to 95%?

f) Compute the 95% confidence intervals for the mean value of the sample. What do the confidence intervals suggest about the hypothesis?

Explanation / Answer

A)

Using technology to find the descriptive statistics,

X = sample mean =    71.86   [ANSWER]

*****************

b)      

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   74  
Ha:    u   =/   74  
              
As we can see, this is a    two   tailed test. [ANSWER]

*******************

c)

Getting the test statistic, as              
              
X = sample mean =    71.86          
uo = hypothesized mean =    74          
n = sample size =    10          
s = standard deviation =    2.5          
              
Thus, z = (X - uo) * sqrt(n) / s =    -2.706909677          
              
Also, the p value is              
              
p =    0.006791273   [P VALUE]

Hence, as P is very small, there is much evidence that the null hypothesis could be incorrect.

********************

d)      

Thus, getting the critical z, as alpha =    0.04   ,      
alpha/2 =    0.02          
zcrit =    +/-   2.053748911      

As |z| > 2.0537, then there is significant evidence at 0.04 level that the the mean yield is not 74% at 0.04 level. [CONCLUSION]

*******************************************

Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit, and please still include the background information. Thanks!

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.