3. A new process has been developed for applying pho­ toresist to 125-mm silicon wafers used in manufac­ turing integrated circuits. Ten wafers were tested, and the following photoresist thickness measurements (in angstroms x 1,000) were observed: 13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002. a. Test the hypothesis that mean thickness is 13.4 x 1,000 A. Use a = 0.05 and assume a two-sided alternative. b. Find a 99% two-sided confidence interval on mean photoresist thickness. Assume that thickness is normally distributed.

Answer:

a

The null hypothesis is  [tex]H_o : \mu  =  13.4000[/tex]

The alternative hypothesis is  [tex]H_a :  \mu \ne  13.4000[/tex]

The null hypothesis is rejected

b

The  99% confidence level is   [tex]13.3930  < \mu  < 13.3994 [/tex]

Step-by-step explanation:

From the question we are told that

  The sample size is  n =  10

   The  population mean is  [tex]\mu =  13.4 000 \  angstroms[/tex]

   The level of significance is  [tex]\alpha =  0.05[/tex]

   The  sample data is  

13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002

Generally the sample mean is mathematically represented as

        [tex]\= x =  \frac{13.3987+ 13.3957\cdots +13.4002 }{10}[/tex]

=>     [tex]\= x = 13.3962 [/tex]

Generally the sample standard deviation  is mathematically represented as

    [tex]\sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }[/tex]

=> [tex]\sigma = \sqrt{\frac{ (13.3987 - 13.3962)^2 +  (13.3987 - 13.3962)^2 + \cdots + (13.3987 -13.4002)^2  }{10} }[/tex]

=>  [tex]\sigma =0.0039 [/tex]

The null hypothesis is  [tex]H_o : \mu  =  13.4000[/tex]

The alternative hypothesis is  [tex]H_a :  \mu \ne  13.4000[/tex]

Generally the test statistics is mathematically represented as

      [tex]z  =  \frac{\= x - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

=>    [tex]z  =  \frac{13.3962 -  13.4000}{\frac{0.0039}{\sqrt{10} } }[/tex]

=>   [tex]z =  3.08[/tex]

Generally the p-value is mathematically represented as

       [tex]p-value  = 2P( z   >  3.08)[/tex]

From the z-table  [tex]P(z >  3.08)= 0.001035[/tex]

So

       [tex]p-value  = 2* 0.001035[/tex]

      [tex]p-value  = 0.00207[/tex]

So from the obtained value we see that

     [tex]p-value  < \alpha[/tex]

Hence the null hypothesis is rejected

Consider the b question

Given that the confidence level is  99%  then the level of significance is

    [tex]\alpha =  (100 -99)\%[/tex]

=> [tex]\alpha =  0.01[/tex]

Generally from the normal distribution table critical value  of  [tex]\frac{\alpha }{2}[/tex] is  

    [tex]Z_{\frac{\alpha }{2} } =  2.58[/tex]

Generally the margin of error is mathematically represented as  

     [tex]E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }[/tex]

=>   [tex]E = 2.58  *  \frac{0.0039}{\sqrt{10} }[/tex]

=>    [tex]E = 0.00318[/tex]

Generally the 99% confidence interval is mathematically represented as

     [tex]13.3962   - 0.00318  < \mu  < 13.3962   +  0.00318 [/tex]

=>   [tex]13.3930  < \mu  < 13.3994 [/tex]

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