In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , 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:

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

95% confidence interval

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

In this problem, we have that:

$p = 0.5, M = 0.1$

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

$0.1 = 1.96*\sqrt{\frac{0.5*0.5}{n}}$

$0.1\sqrt{n} = 0.98$

$\sqrt{n} = 9.8$

$(\sqrt{n})^{2} = (9.8)^{2}$

$n = 96$

The minimum number of cities we need to contact is 96.

8 0

2 years ago

RIGHT ANSWER WILL BE MARK BRAINLIEST

nirvana33 [79]

...............the answer is D

6 0

2 years ago

Read 2 more answers

Deshawn paid $9.20 for coffee and left a $2.30 tip.If he the same rate later today while out at dinner,how much will his tip be

xenn [34]

Around 4.50 would be the tip

8 0

3 years ago