Question

One division of a large defense contractor manufactures telecommunication equipment for the military. Management wants to determine the percentage of rework for electrical components manufactured by the entire company. The Quality Control Department wants to estimate the true percentage of rework for electrical components to within 7.4%, with 99% confidence.

(A) How many components should they sample at least?

Answer 1

The should sample at least 303 components.

Explanation:

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

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

99% confidence level

So
`\alpha = 0.01`, z is the value of Z that has a pvalue of
`1 - (0.01)/(2) = 0.995`, so
`Z = 2.575`.

(A) How many components should they sample at least?

This is n when
`M = 0.074`

We do not know the true proportion, so we use
`\pi = 0.5`, which is the case in which we are going to need to sample the largest number of components.

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.074 = 2.575\sqrt{(0.5*0.5)/(n)}`

`0.074√(n) = 2.575*0.5`

`√(n) = (2.575*0.5)/(0.074)`

`(√(n))^(2) = ((2.575*0.5)/(0.074))^(2)`

`n = 302.7`

Rounding up

The should sample at least 303 components.