Question

. Suppose that we want to estimate the true proportion of defectives in a very large shipment of adobe bricks, and that we want to be at least 95% confident that the error is at most 0.04. How large a sample will we need if:

Answer 1

A sample of 601 bricks will be 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:

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

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 large a sample will we need?

A sample of n bricks will be needed.

n is found when M = 0.04.

We have no estimate for the proportion of defective bricks, so we work with the worst case scenario, which is
`M = 0.04`

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

`0.04 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.04√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.04)`

`(√(n))^(2) = ((1.96*0.5)/(0.04))^(2)`

`n = 600.25`

Rounding up

A sample of 601 bricks will be needed.