Question

The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of sizes n = 160 units. Suppose that today%u2019s sample contains 14 defectives.

How many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within 2% (using the information from today%u2019s sample--that is using the result that ?

Answer 1

A sample of 767 is needed.

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 given by:

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

For this problem, we have that:

14 defectives out of 160, so
`\pi = (14)/(160) = 0.0875`

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`.

How many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within 2% of the percentage?

We would need a sample of n.

n is fround when
`M = 0.02`. So

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

`0.02 = 1.96\sqrt{(0.0875*0.9125)/(n)}`

`0.02√(n) = 1.96√(0.0875*0.9125)`

`√(n) = (1.96√(0.0875*0.9125))/(0.02)`

`(√(n))^2 = ((1.96√(0.0875*0.9125))/(0.02))^2`

`n = 766.8`

Rounding up

A sample of 767 is needed.