Question

Suppose that prior to conducting a coin-flipping experiment, we suspect that the coin is fair. 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? (note that the z-score was rounded to three decimal places in the calculation)

Answer 1

Answer:

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

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

$0.18 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.18\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.18}$

$(\sqrt{n})^{2} = (\frac{2.575*0.5}{0.18})^{2}$

$n = 51.16$

Rounding up

We have to flip the coin at least 52 times.