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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 :

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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is of:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

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

This means that [tex]n = 30, \pi = \frac{3}{30} = 0.1[/tex]

Standard 95% confidence level

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

Margin of error:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]M = 1.96\sqrt{\frac{0.1*0.9}{30}}[/tex]

[tex]M = 0.1074[/tex]

The margin of error of his experiment is of 0.1074.