Question

Suppose a quality control engineer wishes to estimate the proportion of defective items in a production line. i. (4 points) To estimate the proportion within 0.02 and with 98% confidence, what is the minimum required sample size if the company currently has no idea about the

Answer 1

The minimum sample size required is 3385.

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)}`

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.327`.

To estimate the proportion within 0.02 and with 98% confidence, what is the minimum required sample size if the company currently has no idea about the

The minimum sample size required is n.

We find n when
`M = 0.02`

We don't know the true proportion, so we use
`\pi = 0.5`, which is when the largest minimum sample size will be needed.

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

`0.02 = 2.327\sqrt{(0.5*0.5)/(n)}`

`0.02√(n) = 2.327*0.5`

`√(n) = (2.327*0.5)/(0.02)`

`(√(n))^(2) = ((2.327*0.5)/(0.02))^(2)`

`n = 3384.33`

Rounding up

The minimum sample size required is 3385.