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.

5 0

3 years ago

LOOK AT THEBLINE PLOT: 0, 1/4, 1/2, 3/4, 1

Sloan [31]

First add them all up.
0+1/4+1/2+3/4+1=2.5
Because there are 5 beakers you divide 2.5 or 2 1/2 by 5 and you get .5 or 1/2
there would be 1/2 water in each beaker.

7 0

3 years ago