Question

A machine that produces a major part for an airplane engine is monitored closely. In the past, 10% of the parts produced would be defective. With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is _________.

Answer 1

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

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:

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

For this problem, we have that:

`p = 0.1`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

We need a sample size of at least n, in which n is found M = 0.04.

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

`0.04 = 1.96\sqrt{(0.1*0.9)/(n)}`

`0.04√(n) = 0.588`

`√(n) = (0.588)/(0.04)`

`√(n) = 14.7`

`(√(n))^(2) = (14.7)^(2)`

`n = 216`

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.