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

Answer:

[ 0.0584, 1.0416 ]

Step-by-step explanation:

Data provided in the question:

sample size, n = 200

Number of defective components = 20

thus,

p = $\frac{20}{200}$= 0.1

Now,

Confidence interval for p is given as:

⇒ p ± $z\sqrt\frac{p(1-p)}{n}$

here, z = 1.96 for 95% confidence interval

therefore,

⇒ 0.1 ± $1.96\sqrt\frac{0.1(1-0.1)}{200}$

or

⇒ 0.1 ± 0.0416

or

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

or

confidence interval = [ 0.0584, 1.0416 ]