The fraction of defective integrated circuits produced in a photolithography process is being studied. a random sample of 300 circuits is tested, revealing 17 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. less-than-or-equal-to p less-than-or-equal-to (b) calculate a 95% upper confidence bound on the fraction of defective circuits. round the answer to 4 decimal places. p less-than-or-equal-to

(a) The confidence interval is: 0.0304 ≤ π ≤ 0.0830.

(b) Upper confidence bound = 0.0787

Step-by-step explanation:

(a) The confidence interval for p (proportion) can be calculated as

NOTE: π is the proportion ot the population, but it is unknown. It can be estimated as p.

For a 95% two-sided confidence interval, z=±1.96, so

The confidence interval is: 0.0304 ≤ π ≤ 0.0830.

(b) The confidence interval now has only an upper limit, so z is now 1.64.

The confidence interval is: -∞ ≤ π ≤ 0.0787.

i think it is c, not positive though

step-by-step explanation:

hreryryrranswer:

step-by-step explanation: