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

Answer:

The margin of error of his experiment is of 0.1074.

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

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

This means that $n = 30, \pi = \frac{3}{30} = 0.1$

Standard 95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

Margin of error:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 1.96\sqrt{\frac{0.1*0.9}{30}}$

$M = 0.1074$

The margin of error of his experiment is of 0.1074.