Question

In an accelerated failure test, components are operated under extreme conditions so that a substantial number will fail in a rather short time. In such a test involving two types of microchips, 620 chips manufactured by an existing process were tested, and 125 of them failed. Then, 820 chips manufactured by a new process were tested, and 130 of them failed. Find a 90% confidence interval for the difference between the proportions of failures for chips manufactured by the two processes. (Round the final answers to four decimal places.) The 90% confidence interval is ( . , ).

Answer 1

Answer:

The 90% confidence interval is  $0.0097 < p_1 -p_2 < 0.0773$

Step-by-step explanation:

From the question we are told that

   The first  sample size is $n_1 = 620$

    The number of chips that failed is  k = 125

   The second sample size is $n_2 = 820$

    The number of chips that fail is  $u = 130$

Generally the confidence level is  90% , hence the level of significance is  

    $\alpha = (100 - 90)\%$

=>  $\alpha = 0.10$

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

   $Z_{\frac{\alpha }{2} } = 1.645$

Generally the first sample proportion is  

       $\r p _1 = \frac{125}{620}$

=>     $\r p _1 = 0.202$

Generally the second  sample proportion is  

       $\r p _1 = \frac{130}{820}$

=>     $\r p _1 = 0.1585$

Generally the standard error is mathematically represented as

$SE = \sqrt{\frac{\r p_1 (1 - \r p_1)}{n_1} +\frac{\r p_2 (1 - \r p_2)}{n_2} }$    

=> $SE = \sqrt{\frac{0.202(1 - 0.202)}{620} +\frac{0.1585 (1 - 0.1585)}{820} }$  

=>  $SE = 0.0206$  

Generally the margin of error is mathematically represented as  

      $E = Z_{\frac{\alpha }{2} } * SE$

=>     $E = 1.645 * 0.0206$

=>    $E =0.0338$

Generally 95% confidence interval is mathematically represented as  

      $(\r p_1 -\r p_2) -E < p_1 - p_2 < (\r p_1 -\r p_2) +E$

=>    $(0.202 -0.1585) -0.0338 < p_1 - p_2 < (0.202 -0.1585) +0.0338$

=>    $0.0097 < p_1 - p_2 < 0.0773$