The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 12 defectives.(a) Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool. Round the answers to 4 decimal places.< p\l>(b) Calculate a 95% upper confidence bound on the fraction of defective circuits. Round the answer to 4 decimal places

(a) 0.0178 <= p <= 0.0622

(b) p <= 0.0586

Step-by-step explanation:

We have that the sample proportion is:

p = 12/300 = 0.04

(to)

For 95% confidence interval alpha = 0.05, so critical value of z will be 1.96

Therefore, we have that the interval would be:

p + - z * (p * (1-p) / n) ^ (1/2)

replacing we have:

0.04 + - 1.96 * (0.04 * (1-0.04) / 300) ^ (1/2)

0.04 + - 0.022

Therefore the interval would be:

0.04 - 0.022 <= p <= 0.04 + 0.022

0.0178 <= p <= 0.0622

(b)

For upper bounf z-critical value for 95% confidence interval is 1.645, so upper bound is:

p + z * (p * (1-p) / n) ^ (1/2)

replacing:

0.04 + 1.645 * (0.04 * (1-0.04) / 300) ^ (1/2)

0.04 + 0.0186 = 0.0586

p <= 0.0586

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11