Question

Jason wants to know the percentage of M & M's that are green. He chooses a sample of 30 M & M's from a bag, and 3 of them are green. What is the margin of error of his experiment (round to 4 decimal places)? .1074 .1074 .9845 .9845 .1235 .1235 .8824

Answer 1

The margin of error of his experiment is of 0.1074.

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is of:

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

30 M & M's from a bag, and 3 of them are green.

This means that
`n = 30, \pi = (3)/(30) = 0.1`

Standard 95% confidence level

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

Margin of error:

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

`M = 1.96\sqrt{(0.1*0.9)/(30)}`

`M = 0.1074`

The margin of error of his experiment is of 0.1074.