Question

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 1

Answer:

a

The null hypothesis is  $H_o : \mu = 13.4000$

The alternative hypothesis is  $H_a : \mu \ne 13.4000$

The null hypothesis is rejected

b

The  99% confidence level is   $13.3930 < \mu < 13.3994$

Step-by-step explanation:

From the question we are told that

  The sample size is  n =  10

   The  population mean is  $\mu = 13.4 000 \ angstroms$

   The level of significance is  $\alpha = 0.05$

   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

        $\= x = \frac{13.3987+ 13.3957\cdots +13.4002 }{10}$

=>     $\= x = 13.3962$

Generally the sample standard deviation  is mathematically represented as

    $\sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }$

=> $\sigma = \sqrt{\frac{ (13.3987 - 13.3962)^2 + (13.3987 - 13.3962)^2 + \cdots + (13.3987 -13.4002)^2 }{10} }$

=>  $\sigma =0.0039$

The null hypothesis is  $H_o : \mu = 13.4000$

The alternative hypothesis is  $H_a : \mu \ne 13.4000$

Generally the test statistics is mathematically represented as

      $z = \frac{\= x - \mu}{\frac{\sigma}{\sqrt{n} } }$

=>    $z = \frac{13.3962 - 13.4000}{\frac{0.0039}{\sqrt{10} } }$

=>   $z = 3.08$

Generally the p-value is mathematically represented as

       $p-value = 2P( z > 3.08)$

From the z-table  $P(z > 3.08)= 0.001035$

So

       $p-value = 2* 0.001035$

      $p-value = 0.00207$

So from the obtained value we see that

     $p-value < \alpha$

Hence the null hypothesis is rejected

Consider the b question

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

    $\alpha = (100 -99)\%$

=> $\alpha = 0.01$

Generally from the normal distribution table critical value  of  $\frac{\alpha }{2}$ is  

    $Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as  

     $E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=>   $E = 2.58 * \frac{0.0039}{\sqrt{10} }$

=>    $E = 0.00318$

Generally the 99% confidence interval is mathematically represented as

     $13.3962 - 0.00318 < \mu < 13.3962 + 0.00318$

=>   $13.3930 < \mu < 13.3994$