Question

The company ships the components in lots of 200. Lots containing more than 20 defective components may be returned. Find a 95% confidence interval for the proportion of lots that will be returned. Use the normal approximation to compute this proportion. Round the answers to four decimal places. The 95% confidence interval is

Answer 1

[ 0.0584, 1.0416 ]

Explanation:

Data provided in the question:

sample size, n = 200

Number of defective components = 20

thus,

p =
`(20)/(200)`= 0.1

Now,

Confidence interval for p is given as:

⇒ p ±
`z\sqrt(p(1-p))/(n)`

here, z = 1.96 for 95% confidence interval

therefore,

⇒ 0.1 ±
`1.96\sqrt(0.1(1-0.1))/(200)`

or

⇒ 0.1 ± 0.0416

or

confidence interval = [ 0.1 - 0.0416, 0.1 + 0.0416 ]

or

confidence interval = [ 0.0584, 1.0416 ]