Question

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 10 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.

Answer 1

Answer:

The correct answer is

(0.0128, 0.0532)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence interval $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$

For this problem, we have that:

In a random sample of 300 circuits, 10 are defective. This means that $n = 300$ and $\pi = \frac{10}{300} = 0.033$

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

So $\alpha$ = 0.05, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 - 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0128$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 + 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0532$

The correct answer is

(0.0128, 0.0532)