Answer to Question #235510 in Statistics and Probability for blossomqt

Question

1.    The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 13 defectives.

a.    Calculate a 95% two-sided CI on the fraction of non-defective circuits produced by this particular tool.

b.    Calculate a 90% upper confidence bound on the fraction of defective circuits.

Answer 1

Number of defective = 13

Total number of circuits = 300

a) Proportion of defective integrated circuits

$\hat{p} = \frac{x}{n}= \frac{13}{300}= 0.0433$

We will use Noramal distribution

$np= 300 \times 0.0433 = 13 \\ n(1-p) = 300 \times (1-0.0433) = 287$

95% two sided CI will be:

$\hat{p} -Z_{0.025} \sqrt{ \frac{\hat{p}(1- \hat{p})}{n}} ≤ p ≤ \hat{p} + Z_{0.025} \sqrt{ \frac{\hat{p}(1- \hat{p})}{n}} \\ Z_{0.025} = 1.96 \;(from \;table) \\ 0.0433 - 1.96 \times 0.011 ≤ p ≤ 0.0433 + 1.96 \times 0.011 \\ 0.02274 ≤ p ≤ 0.06486$

b) 95% upper confidence bound

$p ≤ \hat{p} + Z_{0.05} \sqrt{\frac{\hat{p}(1- \hat{p})}{n}} \\ Z_{0.05}= 1.645 \\ p ≤ 0.0433 + 1.945 \sqrt{\frac{0.0433 \times 0.9567}{300}} \\ p ≤ 0.0627$