Suppose that prior to conducting a coin-flipping

Ответ:

zah16

24.01.2022

We have to flip the coin at least 52 times.

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

The margin of error is:

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

For this problem, we have that:

The coin is fair, so $\pi = 0.5$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so Z = 2.575 .

How many times would we have to flip the coin in order to obtain a 99% confidence interval of width of at most .18 for the probability of flipping a head?

At least n times.

n is found when M = 0.18 . So