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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

A confidence interval has two bounds, the lower and the upper

Lower bound:

$\pi - M$

Upper bound:

$\pi + M$

In this problem, we have that:

$\pi = 0.26, M = 0.05$

Lower bound:

$\pi - M = 0.26 - 0.05 = 0.21$

Upper bound:

$\pi + M = 0.26 + 0.05 = 0.31$

The 95% confidence interval for the percent of all coffee drinkers who would say they are addicted to coffee is between 21% and 31%.

4 0

3 years ago

8 = 4 - 2x<br> - 2x +5 = 17<br> 12= - 3 ( x+ 5)

max2010maxim [7]

$8 = 4 - 2x\\8 + 2x = 4\\2x = -4\\x = -2$

$- 2x +5 = 17\\-2x =12\\x = -6$

$12= - 3 ( x+ 5)\\12 = -3x -15\\-3x =27\\x= -9$

5 0

3 years ago